Class 10

Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Take π = \(\frac{22}{7}\) unless stated otherwise

Question 1.
A metallic sphere of radius 42 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
We have,
Radius of sphere (r₁) = 42 cm
Volume of sphere = \(\frac{4}{3}\)πr₁³
= \(\frac{4}{3}\)π × (42)³ cm³
Radius of the cylinder (r₂) = 6 cm
Let height of cylinder be h cm. Then
Volume of cylinder = πr₂²h
= π × 6² × h
Volume of cylinder = Volume of sphere
⇒ π × 6² × h = \(\frac{4}{3}\)π × (4.2)³
⇒ h = \(\frac{\frac{4}{3} \pi \times 4.2 \times 4.2 \times 4.2}{\pi \times 6 \times 6}\)
⇒ h = 4 × 0.7 × 0.7 × 1.4
⇒ h = 2.74 cm
Hence, height of the cylinder = 2.74 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
We have,
Radii of metallic sphere are 6 cm, 8 cm, and 10 cm respectively,.
Volume of 1st sphere (V₁) = \(\frac{4}{3}\)π × (6)³
Volume of IInd sphere (V₂) = \(\frac{4}{3}\)π × (8)³
Volume of IIIrd sphere (V₃) = \(\frac{4}{3}\)π × (10)³
Let the radius of resulting sphere be R cm.
Then
Volume of resulting sphere = V₁ + V₂ + V₃
⇒ \(\frac{4}{3}\)πR³ = \(\frac{4}{3}\)π × (6)³ + \(\frac{4}{3}\)π × (8)³ + \(\frac{4}{3}\)π × (10)³
⇒ \(\frac{4}{3}\)πR³ = \(\frac{4}{3}\)π [6³ + 8³ + 10³]
⇒ R³ = 216 + 512 + 1000
⇒ R³ = 1728.
⇒ R³ = 12 × 12 × 12 = 12³
⇒ R = 12
Hence, radius of resulting sphere = 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
We have,
Radius of well (r) = \(\frac{7}{2}\) = 3.5 m
Depth of the well (h) = 20 m
Volume of earth taken out of the well = πr²h

= \(\frac{22}{7}\) × 3.5 × 3.5 × 20 = 770 m³
Let the height of the platform be h₁ m
Base dimensions of the platform = 22m by 14 m
Volume of earth to form platform = Volume of earth taken put of the well
⇒ 22 × 14 × h₁ = 770
⇒ h₁ = \(\frac{770}{22 \times 14}\) = 2.5 m.
Hence, height of the platform = 2.5 m.

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. [CBSE 2011]
Solution:
We have,
Radius of the well (r) = \(\frac{3}{2}\) = 1.5 m
Depth of the well (h) = 14 m

Volume of the earth dug out = πr²h
= π × (1.5)² × 14
Radius of the embankment (R) = 1.5 + 4 = 5.5 m
Let height of the embankment be h’ m
Volume of the earth used for making embankment = Volume of the earth dug out
⇒ πR²h’ – πr²h’ = π × (1.5)² × 14
⇒ πh’ (R² – r²) = π × 1.5 × 1.5 × 14
⇒ πh’ (5.5² – 1.5²) = π × 1.5 × 1.5 × 14
⇒ πh’ [(5.5 + 1.5) (5.5 – 1.5)] = π × 1.5 × 1.5 × 14
⇒ πh’ × 7 × 4 = π × 1.5 × 1.5 × 14
h’ = \(\frac{\pi \times 1.5 \times 1.5 \times 14}{\pi \times 7 \times 4}\)
⇒ h’ = \(\frac{2.25}{2}\)
⇒ h’ = 1.125 m
Hence, height of the embankment = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
We have
Radius of cylinder (R) = \(\frac{12}{2}\) = 6 cm
Height of cylinder (H) = 15 cm
Volume of ice cream which filled in cylinder = πR²H
= π × 6² × 15 cm³
Radius of conical part (r) = \(\frac{6}{2}\) = 3 cm
Height of conical part (A) = 12 cm
Radius of hemispherical part = 3 cm

Volume of one ice cream cone = Volume of conical part + Volume of hemispherical part
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³
= \(\frac{1}{3}\) πr² (h + 2r)
= \(\frac{1}{3}\) π × 3² (12 + 2 × 3) d
= π × 3 × 18 cm³
Let the number of cones made be n
Volume of ice cream of n cones = Volume of ice cream which filled from cylinder
⇒ π × 3 × 18 × n = π × 6² × 15
n = \(\frac{\pi \times 6 \times 6 \times 15}{3 \times 18 \times \pi}\)
⇒ n = 10.
Hence, number of cones = 10.

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm ?
Solution:
We have,
Radius of the coin (r) = \(\frac{1.75}{2}\) cm
Thickness of coin (h) = 2 mm = 0.2 cm
Volume of 1 coin = πr²h
= \(\frac{22}{7} \times\left(\frac{1.75}{2}\right)^2\) × 0.2
Let number of coins made be n.
Volume of n coins = n × \(\frac{22}{7} \times\left(\frac{1.75}{2}\right)^2\) × 0.2 cm³
Volume of the cuboid = 5.5 × 10 × 3.5 cm³
Volume of n coins = Volume of the cuboid

⇒ n = 400.
Hence, required number of coins = 400.

Question 7.
A cylindrical bucket 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
We have,
Radius of cylindrical bucket (r₁) = 18 cm
Height of cylindrical bucket (h₁) = 32 cm
Volume of sand in the bucket = πr₁²h₁
= π × 18² × 32 cm³
Height of the conical heap (h₂) = 24
Let radius of conical heap be r₂ cm
Volume of conical heap of sand = \(\frac{1}{3}\) πr₂²h₂
= \(\frac{1}{3}\) × π × r₂² × 24
Volume of conical heap of sand = Volume of sand in the bucket
⇒ \(\frac{1}{3}\) π × r₂² × 24 = π × 18² × 32
r₂² = \(\frac{\pi \times 18 \times 18 \times 32 \times 3}{\pi \times 24}\)
⇒ r₂² = 18 × 18 × 4
⇒ r₂² = 18² × 2²
⇒ r₂ = \(\sqrt{18^2 \times 2^2}\)
⇒ r₂ = 18 × 2 = 36 cm
Slant height of conical heap = \(\sqrt{h_2^2+r_2^2}\)
= \(\sqrt{24^2+36^2}\)
= \(\sqrt{576+1296}\)
= \(\sqrt{1872}\)
= \(\sqrt{12 \times 12 \times 13}\)
= 12√13 cm.

Hence, radius and slant height of conical heap are 36 cm and l2Ji cm respectively.

Question 8.
Water in a canal, 6 m wide and 15 m deep, is flowing with a speed of 10 km/hr.
How much area will it irrigate in 30 minutes, if 8
cm of standing water is needed?
Solution:
We have,
Depth of water of canal = 1.5 m
Width of canal = 6m
Speed of canal = 10 km/h = 10 × 1000 m/h = 10000 m/h
Volume of water flowing through canal in 60 minutes = 10000 × 6 × 1.5 m³
Volume of water flowing through canal in 30 minutes = \(\frac{10000 \times 6 \times 1.5 \times 30}{60}\)
= 5000 × 6 × 15 m³
Height of standing water = 8 cm = \(\frac{8}{100}\) m
Let required area be x m²
Volume of standing water = Volume of water flowing through canal in 30 minutes
\(\frac{x \times 8}{100}\) = 5000 × 6 × 1.5
[∵ Volume of standing water = area × height]
⇒ x = \(\frac{5000 \times 6 \times 1.5 \times 100}{8}\)
⇒ x = 562500 m².
⇒ Area = 562500 m² = \(\frac{562500}{10000}\)
= 56.25 hectares.
Hence, area will irrigate = 562500 m² or 56.25 hectares.

Question 9.
A farmer cormects a pipe of internal diameter 20 cm from i» nal i,t’ r cylindrical tank in her field, which i’ I m in dia rnet and 2 m deep. If water flows through hc pipe at tiw rate of 3 km/h, in how much time will the Lank be filled?
Solution:
We have,
Internal diameter of pipe = 20 cm
Radius of the pipe = \(\frac{20}{2}\) = 10 cm
= \(\frac{10}{100}\) = \(\frac{1}{10}\) m.
Speed of water = 3 km/h = 3000 m/h
Volume of water that flows 1h rough the pipe in 1h = π × (\(\frac{1}{10}\))² × 3000
= 30π m³
Radius of cylindrical tank (r) = \(\frac{10}{2}\) = 5 m
Depth of the tank (h) = 2 m
Volume of the tank = πr²h
= π × 5² × 2
= 50π m³
Time taken to filled the tank = \(\frac{\text { Volume of the tank }}{\text { Volume of water flows in } 1 \mathrm{hr}}\)
= \(\frac{50 \pi}{30 \pi}=\frac{5}{3}\) h
= 1 h 40 minutes.
Hence, required time = 1 h 40 minutes OR 100 minutes.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Unless stated otherwise, take π = \(\frac{22}{7}\).

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and height of the cone is euqal to its radius. Find the volume of the solid in terms of it.
Solution.
We have,
Radius of the hemisphere (r) = 1 cm
Radius of the cone (r) = 1 cm
Height of the cone (h)=1 cm
Volume of the solid = Volume of the hemisphere + Volume of the cone
= \(\frac{2}{3}\) πr³ + \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) π² (2r + h)

= \(\frac{1}{3}\) π × 1² (2 × 1 + 1)
= \(\frac{1}{3}\) π × 1 × 3
Hence, volume of the solid = π cm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same)
Solution:
We have,
Radius of two conical parts (r) = \(\frac{3}{2}\) = 15 cm
Radius of the cylindrical part (r) = 15 cm
Height of the conical part (h) = 2 cm
Height of the cylindrical part (H) = 12 – (2 + 2) = 8 cm.
Volume of the air contained in the model = Volume of the cylindrical part + Volume of two conical parts
= πr²H + \(\frac{2 \times \pi r^2 h}{3}\)
= πr² [H + \(\frac{2 \times h}{3}\)]

= π × 1.5² [8 + \(\frac{2 \times 2}{3}\)]
= π × 2.25 × \(\frac{28}{3}\)
= \(\frac{22}{7}\) × 2.25 × \(\frac{28}{3}\)
Hence, volume of the air contained in the model = 66 cm³

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see in figure.)

Solution.
We have,
Radius of a cylindrical part (r) = 1.4 cm
Radius of hemispherical part (r) = 1.4 cm
Length of the cylindrical part of a gulab jamun (h) = [5 – (1.4 + 1.4)]
= 5 – 2.8 = 2.2 cm
Volume of gulab jamun = Volume of the cylindrical part + Volume oftwo hemispherical part
= πr²h + 2 × \(\frac{2}{3}\)πr³
= πr² (h + \(\frac{4}{3}\)r)
= \(\frac{22}{7}\) × (1.4)² (2.2 + \(\frac{4}{3}\) × 1.4)
= 6.16 × 4.0666 = 25.05 cm³
Volume of 45 gulab jamuns = 25.05 × 45 = 1127.25 cm³

Volume of the syrup = 30% of 1127.25
= \(\frac{30}{100}\) × 1127.25 = 338.18 cm³
Hence, volume of the syrup = 338 cm³ (approx).

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand, (see in figure.)

Solution.
We have,
Dimensions of the cuboidal part are 15 cm × 10 cm × 3.5 cm
Volume of the cuboid = 15 × 10 × 3.5 = 525 cm³
Radius of each conical depression (r) = 0.5 cm
Depth (height) of each conical depression (h) = 1.4 cm
Volume of wood taken out to make four cavities = 4 × volume of conical depression
= 4 × \(\frac{1}{3}\) πr²h
= 4 × \(\frac{1}{3} \times \frac{22}{7}\) × 0.5² × 1.4
= \(\frac{88 \times 0.25 \times 0.2}{3}\)
= 1.47 cm³ (approx)
Volume of the wood in the entire stand = 525 – 1.47 = 523.53 cm³
Hence, volume of the wood in the entire stand = 523.53 cm³

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to brim. When lead shots, each of which is a sphere of radius 0-5 cm are dropped into the vessel, one fourth of the water flows out. Find the member of lead shots dropped in the vessel.
Solution.
We have,
Radius of the base of the cone (r) = 5 cm
Height of the cone (h) = 8 cm
Volume of water of cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 5² × 8
⇒ \(\frac{200 \pi}{3}\) cm³
Volume of the water that flows out of the cone = \(\frac{1}{4} \times \frac{200 \pi}{3}\)
= \(\frac{50 \pi}{3}\) cm³

Radius of spherical lead (r₁) = 0.5 cm
Volume of spherical lead = \(\frac{4}{3} \pi r_1^3=\frac{4}{3} \pi \times(0.5)^3\)
= \(\frac{\pi \times 0.5}{3}=\frac{\pi}{6}\) cm³
Let n spherical lead shots are dropped into the vessel.
So that \(\frac{1}{4}\) th of the water contianed in the vessel flows out.
∴ Volume of n spherical lead shots = Volume of water that flows out
⇒ n \(\frac{\pi}{6}=\frac{50}{3} \pi\)
⇒ n = \(\frac{50 \pi}{3} \times \frac{6}{\pi}\)
⇒ n = 100
Hence, number of lead shots dropped in the vessel = 100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Solution.
We have,
First cylindrical part has radius 12 cm and height 220 cm
Its volume = π × 12² × 220 = 31680π cm³
Ilnd cylindrical part has radius 8 cm and height 60 cm
Its volume = π × 8² × 60 = 3840π cm³
Total volume of iron pole = 31680π + 3840π = 35520π
= 35520 × 3.14 = 111532.8 cm³

Total mass of the iron pole = 111532.8 × 8 gm
= \(\frac{111532 \cdot 8 \times 8}{1000}\) kg
= 892.2624 kg = 892.26 kg
Hence, weight of iron pole = 892.26 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:
We have,
Radius of the cylinder (r) = 60 cm
Height of the cylinder (H) = 180 cm
Volume of water that the cylinder contain = πr²h
= π × 60² × 180
= 648000π cm³
Radius of the conical part (r) = 60 cm
Radius of the hemispherical part (r) = 60 cm
Height of the conical part (h) = 120 cm
Volume of the solid = Volume of conical part + Volume of hemispherical part
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³
= \(\frac{1}{3}\) πr² (h + 2r)
= \(\frac{1}{3}\) π × 60² (120 + 2 × 60)
= \(\frac{1}{3}\) π × 3600 × 240
= 288000π cm³
Volume of water left in the cylinder = Volume of water that cylinder contains – Volume of solid
= 648000 π – 288000 π
= 360000π
= 360000 × \(\frac{22}{7}\) cm³
= \(\frac{360000 \times 22}{7 \times 100 \times 100 \times 100}\) m³
[∵ 1 cm³ = \(\frac{1}{100} \times \frac{1}{100} \times \frac{1}{100}\) m³]
= \(\frac{22 \times 36}{700}\) m³
= 1.1314 m³
= 1.131 m³ (approx)
Hence, Volume of water out in the cylinder = 1.131 m³ (approx).

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter ; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements and π = 3.14.

Solution.
We have,
Radius of cylindrical neck (r) = \(\frac{2}{2}\) = 1 cm
Height of cylindrical neck (A) = 8 cm
Radius of the spherical part (R) = \(\frac{8.5}{2}\) cm
Amount of water in the vessel = Volume of the cylindrical neck + Volume of the spherical part
= πr²h + \(\frac{4}{3}\) πR³
= π(r²h + \(\frac{4}{3}\) R³)
= 3.14 [1² × 8 + \(\frac{4}{3} \times\left(\frac{8 \cdot 5}{2}\right)^3\)]
= 3.14 (8 + 102.354)
= 3.14 × 110.354
= 346.51 cm³
Hence, volume find by the child is not correct.
Correct answer is 346.51 cm³.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
Find the area of shaded region in the given figure, If PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Solution:
We have,
PQ = 24 cm.
PR = 7 cm.
∠QPR = 90° [∵ Angle in a semicircle is 90°]
In right angled ∆RPQ, we have
RQ² = RP² + PQ
⇒ RQ² = 7² + 24²
⇒ RQ² = 49 + 576
⇒ RQ² = 625
⇒ RQ = √625 = 25 cm
Radius (RO) = \(\frac{25}{2}\) cm
Area of shaded region= Area of semicircle – Area of ∆RPQ
= \(\frac{1}{2}\) πr² – \(\frac{1}{2}\) × PQ × RP
= \(\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 24 \times 7\)
= 245.54 – 84
= 161.54 cm².
Hence, area of shaded region = 161.54 cm²

Question 2.
Find the area of shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
We have,
Radii of two concentric circles = 7 cm, 14 cm and ∠AOC = 40°
Area of shaded region = Area of sector AOC – Area of sector BOD
= \(\frac{\pi \times 14^2 \times 40^{\circ}}{360^{\circ}}-\frac{\pi \times 7^2 \times 40^{\circ}}{360^{\circ}}\)
= \(\frac{\pi}{9}\) [14² – 7² ]
= \(\frac{\pi}{9}\) × (14 + 7) (14 – 7)
= \(\frac{22}{7 \times 9}\) × 21 × 7
= \(\frac{154}{3}\) cm²
Hence, area of shaded region = \(\frac{154}{3}\) cm².

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semi-circles.

Solution:
We have,
Side of square = 14 cm
radius of semicircle = \(\frac{14}{2}\) = 7 cm
Area of shaded region = Area of square – 2 × Area of semicircle
= 14 × 14 – 2 × \(\frac{1}{2}\) × π × r²
= 14 × 14 – 2 × \(\frac{1}{2}\) × 7 × 7
= 196 – 154 = 42 cm².
Hence, area of shaded region = 42 cm².

Question 4.
Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Solution.
We have,
Radius of circle = 6 cm
Side of equilateral triangle = 12 cm
Area of shaded region = Area of circle + Area of equilateral ∆OAB – Area of sector OCD
= π × 6² + \(\frac{\sqrt{3}}{4}\) × (12)² – \(\frac{60^{\circ}}{360^{\circ}}\) × π × 6²
= \(\frac{22}{7} \times 36+\frac{\sqrt{3} \times 144}{4}-\frac{1}{6} \times \frac{22}{7} \times 36\)
= \(\frac{22 \times 36}{7}-\frac{22 \times 6}{7}\) + 36√3
= \(\frac{660}{7}\) + 36√3
Hence, area of shaded region = (\(\frac{660}{7}\) + 36√3) cm².

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of remaining portion of the square.

Solution:
We have,
side of the square = 4 cm
Radius of quadrant = 1 cm
and radius of circle = \(\frac{2}{2}\) = 1 cm.
Area of the remaining portioñ of the square = Area of square – (4 × Area of quadrant of circle + Area of circle)
= 4 × 4 – \(\left(4 \times \frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(1)^2+\frac{22}{7} \times 1^2\right)\)
= 16 – \(\left(4 \times \frac{1}{4} \times \frac{22}{7}+\frac{22}{7} \times 1\right)\)
= 16 – (\(\frac{22}{7}+\frac{22}{7}\))
= 16 – \(\frac{44}{7}\)
= \(\frac{112-44}{7}=\frac{68}{7}\) cm².
Hence, area of remaining portion = \(\frac{68}{7}\) cm².

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design.

Solution:
We have,
Radius of the circle = 32 cm
O is the centre of the circle
∆ABC is an equilateral triangle.
Join OA, OB and OC.
Now ∠AOB = ∠BOC = ∠COA = 120°
In ∆OBC, we have
OB = OC [Equal radii of circle]
Draw OD ⊥ BC
∆ODB ≅ ∆ODC [By R.H.S. congruence]
⇒ ∠COD = \(\frac{120^{\circ}}{2}\) = 60°. [CPCT]
⇒ \(\frac{\mathrm{BD}}{\mathrm{OB}}\) = sin 60°
⇒ \(\frac{\mathrm{BD}}{32}=\frac{\sqrt{3}}{2}\)
⇒ BD = \(\frac{32 \sqrt{3}}{2}\) = 16√3
⇒ BC = 2 × BD
⇒ BC = 2 × 16√3 = 32√3 cm

Area of the shaded region (designed) = Area of the circle – Area of the ∆ABC
= πr² – \(\frac{\sqrt{3}}{4}\) (side)²
= π × (32)² – \(\frac{\sqrt{3}}{4}\) × (32√3)²
= \(\frac{22}{7}\) × 1024 – √3 × 768
= (\(\frac{22528}{7}\) – 768√3) cm²
Hence, area of the shaded region (designed) = (\(\frac{22528}{7}\) – 768√3) cm²

Question 7.
In the given figure, ABCD is a square of side 14 cm, with centres A, B, C and D, four circles are drawn such that each circles touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
We have,
Side of square = 14 cm
Radius of each circle = \(\frac{14}{2}\) = 7 cm
Area of the Shaded region = Area of square – 4 × quadrant area
= (side)² – 4 × \(\frac{1}{4}\) πr²
= 14 × 14 – 4 × \(\frac{1}{4}\) × \(\frac{22}{7}\) × 7²
= 196 – 22 × 7
= 196 – 154 = 42 cm².
Hence, area of shaded region = 42 cm².

Question 8.
In the given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(i) The distance around the track along its inner edge.
(ii) The area of the track.
Solution:
(i)

The distance around the track along its inner edge = Perimeter of PMS + PQ + Perimeter of QNR + SR
= (π × 30 + 106 + π × 30 + 106) m
= 212 + 60π
= 212 + 60 × \(\frac{22}{7}\)
= 212 + \(\frac{1320}{7}\)
= \(\frac{1484+1320}{7}\) = \(\frac{2804}{7}\)

(ii) Outer radius = 30 + 10 = 40 m
Area of the track = Area of rectangle PABQ + Area of rectangle DCRS + 2 × Area of semi circular ring
= 106 × 10 + 106 × 10 + 2 × \(\frac{1}{2}\) π(40² – 30²)
= 1060 + 1060 + \(\frac{22}{7}\) × (40 + 30) (40 – 30) 22 1
= 2120 + \(\frac{22}{7}\) × 700
= 2120 + 2200 = 4320 m².
Hence, distance around the track along its inner edge = \(\frac{2804}{7}\) m
area 0f track = 4320 m².

Question 9.
In the given figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
we have,
OA = 7 cm.
∴ AB = CD = 2 × 7 = 14 cm
OC = OB = OA = OD = 7 cm
diameter AB ⊥ diameter CD
Radius of smaller circle having centre = \(\frac{7}{2}\) cm.
Area of shaded region= Area of smaller circle having centre 0′ + (Area of semicircle ACB – area of ∆ABC)
= \(\pi \times\left(\frac{7}{2}\right)^2+\left[\frac{1}{2} \pi \times(7)^2-\frac{1}{2} \times 14 \times 7\right]\)
= \(\frac{22}{7} \times \frac{49}{4}+\left[\frac{1}{2} \times \frac{22}{7} \times 49-49\right]\)
= 38.5 + [77 – 49]
= 38.5 + 28 = 66.5 cm².
Hence, area of the shaded region= 66.5 cm².

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see in figure). Find the area of the shaded region. (use π = 3.14 and √3= 1.73205)

Solution.
Let the side of equilateral ∆ be x cm.
Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) × x²
⇒ 17320.5 = \(\frac{\sqrt{3}}{4}\) × x²
[given, area of triangle = 17320.5 cm²]
⇒ 17320.5 × 4 = √3x²
⇒ x² = \(\frac{17320 \cdot 5 \times 4}{\sqrt{3}}\)
⇒ x² = \(\frac{17320 \cdot 5}{1 \cdot 73205}\) × 4
⇒ x² = 10000 × 4
⇒ x² = 40000
⇒ x = \(\sqrt{40000}\) = 200 cm.
Then, radius of each circle = \(\frac{200}{2}\) = 100 cm
Area of shaded region = Area of ∆ABC – (Area of sector AMO + Area of sector BMN + Area of sector CNO)
= 17320.5 – \(\left(\frac{60^{\circ}}{360^{\circ}} \times \pi \times(100)^2+\frac{60^{\circ}}{360^{\circ}} \times \pi \times(100)^2+\frac{60^{\circ}}{360^{\circ}} \times \pi \times(100)^2\right)\)
= 17320.5 – [3 × \(\frac{1}{6}\) × π × (100)²]
= 17320.5 – \(\frac{\pi}{2}\) × 10000
= 17320.5 – \(\frac{3.14}{2}\) × 10000
= 17320.5 – 15700 = 1620.5 cm².
Hence, area of shaded region = 1620.5 cm².

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see in figure). Find the area of the remaining portion of the handker-chief.

Solution.
We have,
radius of circle = 7 cm
Diameter of the circle = 2 × 7 = 14 cm
Side of the square ABCD = 3 × 14 = 42 cm
Area of the remaining portion of handkerchief = Area of square ABCD – 9 × area of circle
= 42 × 42 – 9 × π × 7²
= 1764 – 9 × \(\frac{22}{7}\) × 7 × 7
= 1764 – 1386
= 378 cm².
Hence, area of the remaining portion of handkerchief = 378 cm².

Question 12.
In the given figure, OACB is a quadrant of a circle with centre 0 and radius 3’5 cm. If OD = 2 cm, find the area of the :
(i) Quadrant OACB
(ii) Shaded region

Solution.
We have,
Radius of a quadrant = 3.5 cm and OD = 2 cm
(i) Area of quadrant OACB = \(\frac{1}{4}\) × π × (3.5)2
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 35 × 35
= \(\frac{77}{8}\) cm².

(ii) In right ∆BOD, we have OB = 35cm, and OD = 2 cm
Area of ∆OBD = \(\frac{1}{2}\) × 35 × 2
= 3.5 cm² = \(\frac{7}{2}\) cm²
Area of shaded region= quadrant area – area of ∆BOD
= \(\frac{77}{8}-\frac{7}{2}\)
= \(\frac{77-28}{8}\)
= \(\frac{49}{8}\) cm².
Hence, arei of quadrant = \(\frac{77}{8}\) cm²
and area of shaded region = \(\frac{49}{8}\) cm².

Question 13.
In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (use π = 3.14)

Solution:
We have,
Side of square (OA) = 20 cm
In right angle triangle OAB.
OB² = OA² + AB² (By Pythagoras theorem)
OB² = 20² + 20²
OB² = 400 + 400
OB² = 800
OB = \(\sqrt{800}\) = 20√2 cm
∴ Radius of the quadrant 20√2 cm
Area of shaded region= Area of Quadrant – area of Square.
= \(\frac{1}{4}\) π × (20√2)² – 20 × 20
= \(\frac{1}{4}\) × 3.14 × 800 – 400
= 628 – 400 = 228 cm².
Hence, area of shaded region = 228 cm².

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see in figure). If ∠AOB = 30°, Find the area of the shaded region.

Solution.
We have,
Radius of sector AOB (r₁) = 21 cm
Radius of sector COD (r₂) = 7 cm
Area of shaded region= Area of sector AOB – Area of sector COD

Hence, area of shaded region = \(\frac{308}{3}\) cm².

Question 15.
In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
We have,
Radius of a quadrant = 14 cm
In right ∆BAC,
BC² = AB² + AC²
BC² = 14² + 14²
BC² = 2 × 14²
BC = 14√2 cm
Area of region (I) = Area of quadrant ABC – Area of ∆ABC
= \(\frac{1}{4}\) π × 14² – \(\frac{1}{2}\) × 14 × 14
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 14 × 14 – 7 × 14
= 154 – 98 = 56 cm².

Area of shaded region (II) = Area of the semicircle drawn on BC as diameter – Area of region I
= \(\frac{1}{2} \times \pi \times\left(\frac{14 \sqrt{2}}{2}\right)^2-56\)
= \(\frac{1}{2} \times \frac{22}{7} \times \frac{392}{4}\) – 56
= \(\frac{616}{4}\) – 56
= 154 – 56 = 98 cm².
Hence, area of shaded region = 98 cm².

Question 16.
Calculate the area of the designed region in given figure common between the two quadrants of circles of radius 8 cm each.

Solution:
We have,
Radius of the circle = 8 cm
Area of the designed region = 2 (Area of quadrant ∆BQD – Area of ∆ABD)

Hence, area of designed region = \(\frac{256}{7}\) cm².

Haryana State Board HBSE 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
Find the area of sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution.
We have,
Radius = 6 cm
θ = 60°
Area of the sector = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{6^2 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{6 \times 6}{6}=\frac{132}{7}\)
Hence, area of sector = \(\frac{132}{7}\) cm².

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let r be the radius of the circle whose circumference is 22 cm.
2πr = 22
r = \(\frac{22}{2 \pi}\)
r = \(\frac{22}{2 \times \frac{22}{-7}}\)
r = \(\frac{22 \times 7}{2 \times 22}\)
r = \(\frac{7}{2}\) cm
θ = 90°
[∵ sector is quadrant ∴ θ = \(\frac{360}{4}\) = 90°]
Therefore,area of the quadrant = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{90^{\circ}}{360^{\circ}}\)
= \(\frac{22 \times 7}{4 \times 4}\)
= \(\frac{77}{8}\) cm²
Hence, area of the quadrant = \(\frac{77}{8}\) cm²

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution.
Angle describe by the minute hand in 60 minutes = 360°
Angle, describe by the minute hand in 5 minutes = \(\frac{360^{\circ}}{60^{\circ}}\) × 5 = 30°
Now, θ = 30° and radius = 14 cm
Area swept by the minute hand in 5 minutes = Area of sector AOB

= \(\frac{11 \times 14}{3}\)
= \(\frac{154}{3}\) cm².
Hence, area swept by the minute hand = \(\frac{154}{3}\) cm².

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment
(ii) major sector. (use π = 3.14)
Solution.
We have,
Radius = 10 cm
and θ = 90°
Area of the minor segment ACB = \(\frac{\pi r^2 \theta}{360}\) – Area of ∆AOB
= \(\frac{3.14 \times 10^2 \times 90^{\circ}}{360^{\circ}}\) – \(\frac{1}{2}\) × 10 × 10
= \(\frac{3.14 \times 100}{4}\) – 50
= 785 – 50 = 28.5 cm²
Area of major sector ADB

= \(\frac{\pi r^2\left(360^{\circ}-90^{\circ}\right)}{360^{\circ}}\)
= \(\frac{3.14 \times 10^2 \times 270^{\circ}}{360^{\circ}}\)
= 314 × \(\frac{3}{4}\)
= 235.5 cm²
Hence, area of minor segment 285 cm² and area of major sector = 235.5 cm².

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 600 at the centre. Find:
(i) The length of the arc
(ii) Area of the sector formed by the arc
(iii) Area of the segment formed by the corresponding chord.
Solution.
We have,
Radius (r) = 21 cm
Sector angle (θ) = 60°
(i) Length of the arc = \(\frac{\theta}{360^{\circ}}\) × 2πr

= 22 cm
(ii) Area of the sector formed by the arc = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{21^2 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{21 \times 21 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{22 \times 3 \times 21 \times 1}{6}\)
= 11 × 21 = 231 cm²

(iii) In ∆OAB. OA = OB (equal radii)
⇒ ∠OAB = ∠OBA
Let ∠OAB = ∠OBA = x
∴ x + x + ∠AOB = 180°
⇒ 2x + 60° = 180°
⇒ 2x = 180° – 60° = 120°
⇒ x = \(\frac{120^{\circ}}{2}\) = 60°
Therefore, ∆OAB is an equilateral triangle of side 21 cm.
∴ Area of ∆OAB = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{\sqrt{3}}{4}\) × 21²
= \(\frac{441 \sqrt{3}}{4}\) cm²

Area of the segment formed by the corresponding chord = Area of segment ACB.
= Area of the sector OACBO – Area of ∆OAB
= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{441 \sqrt{3}}{4}\)
= \(=\frac{22}{7} \times \frac{21^2 \times 60^{\circ}}{360^{\circ}}-\frac{441 \sqrt{3}}{4}\)
= 11 × 21 – \(\frac{441 \sqrt{3}}{4}\)
= (231 – \(\frac{441 \sqrt{3}}{4}\)) cm²
Hence, length of the arc = 22 cm,
area of the sector = 231 cm²
and area of the segment (ACB) = (231 – \(\frac{441 \sqrt{3}}{4}\)) cm².

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor and major segments of the circle. [use π = 3.14 and √3 = 1.73]
Solution:
We heve,
Radius (r) = 15 cm
Sector angle (θ) = 60°
In ∆AOB
⇒ OA = OB (equal radii of the circle)
⇒ ∠OAB = ∠OBA
⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 2∠OAB + 60°= 180°
⇒ 2∠OAB = 180° – 60° = 120°
⇒ ∠OAB = \(\frac{120^{\circ}}{2}\) = 60°
Therefore, ∆AOB is an equilateral triangle.
Its side = 15 cm
Area of ∆AOB = \(\frac{\sqrt{3}}{4}\) × 15²
= \(\frac{225 \sqrt{3}}{4}\) cm².

Area of minor segment (ACB) = Area of sector – area of ∆AOB

= 117.75 – 97.31 = 2O.44 cm².
Area of major segment = nr² – Area of minor segment
= 3.14 × 15² – 20.44.
= 706.5 – 20.44 = 686.06 cm²
Hence, area of minor segment and major segment are 20.44 cm² and 686.06 cm² respectively.

Question 7.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. [use π = 3.14 and √3 = 1.73]
Solution.
We have,
Radius (r) = 12 cm
∠AOB = 120°
Area of segment (ACB) = \(\frac{\pi r^2 \theta}{360^{\circ}}-r^2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)
= \(\frac{3.14 \times 12^2 \times 120}{360}-12^2 \sin \frac{120^{\circ}}{2} \cos \frac{120^{\circ}}{2}\)
= 3.14 × 12 × 4 – 144 sin 60° cos 60°

= 3.14 × 48 – 144 × \(\frac{\sqrt{3}}{2} \times \frac{1}{2}\)
= 3.14 × 48 – 36√3
= 150.72 – 36 × 173
= 150.72 – 62.28
= 88.44 cm².
Hence, area of segment (ACB)= 88.44 cm².

Question 8.
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find
(i) The area of that part of the field in which the horse can graze.
(ii) The increase in the grazing area if the rope were 10m long instead of 5 m. (use π = 3.14)

Solution:
(i) We have,
Side of the square = 15 m
The area of the field in which the horse can graze represented in the figure by APR, a quadrant of the circle with radius 5 m.
The area which horse can graze = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 5^2 \times 90^{\circ}}{360^{\circ}}\)
= \(\frac{3 \cdot 14 \times 25}{4}\)
= 19.63 cm².

(ii) Increase in area when the rope is 10 m long = Area of quadrant ASQ – Area of quadrant APR
= \(\frac{\pi r^2 \theta}{360^{\circ}}\) – 19.63
= \(\frac{3.14 \times 10^2 \times 90^{\circ}}{360^{\circ}}\) – 19.63
= \(\frac{314}{4}\) – 19.63
= 78.5 – 19.63 = 58.87 cm².

Hence, area which horse can graze = 19.63 cm²
and increase in area = 58.87 cm².

Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also usal in making 5 centimeters which divide the circle into 10 equal sectors as shown in figure. Find:
(i) The total length of the silver wire required.
(ii) The area of each sector of the brooch.

Solution:
(i) We have,
Radius of the circle = \(\frac{35}{2}\) mm.
Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\)
= 110 mm
The total length of wire required = circumference of the circle + length of 5 diameters
= 110 + 5 × 35
= 110 + 175 = 285 mm.

(ii) Area of the circle = πr²
= \(\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}\)
= \(\frac{11 \times 5 \times 35}{2}\)
= 962.5 mm²

Since circle is divided into 10 equal sectors.
Therefore, area of each sector = \(\frac{\text { Area of the circle }}{10}\)
= \(\frac{962.5}{10}\)
= 96.25 mm².
Hence, (i) total length of wire required = 285 mm.
(ii) Area of each sector = 96.25 mm²

Question 10.
An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
We have,
Radius of circle = 45 cm
Area of the circle = πr²
= π × 45²
= 2025π
= \(\frac{2025 \times 22}{7}\)
= 6364.29 cm²
Since the ribs divide the circle into 8 sectors of equal area
Area between the two consecutive ribs = Area of each sector of the circle
= \(\frac{\text { Area of the circle }}{8}\)
= \(\frac{6364 \cdot 29}{8}\)
= 795.54 cm²
Hence, area between the two consecutive ribs of the umbrella = 795.54 cm²

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution.
We have,
One blade of a wiper sweeps a sector area of circle of a radius 25 cm.
The sector angle (θ) = 115°

Hence, total area cleaned at each sweep of the blades = \(\frac{158125}{126}\) cm²

Question 12.
To warn ships for underwater rocks, a light house spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. find the area of the sea over which the ships are warned, (use π = 3.14)
Solution.
We have,
Radius = 16.5 km

Sector angle (θ) = 80°
Area of the sea over which the ships are warned = \(\frac{\pi r^2 \theta}{360}\)
= \(\frac{3.14 \times 16.5 \times 16.5 \times 80}{360}\)
= \(\frac{3 \cdot 14 \times 272 \cdot 25 \times 2}{9}\)
= 189.97 km²
Hence, area of the sea over which the ships are warned = 189.97 km².

Question 13.
A round table cover has six equal designs as shown in figure. if the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use √3 = 1.7)

Solution:
We have,
r = 28 cm
and θ = \(\frac{360^{\circ}}{6}\) = 60°

In the figure OA = OB
∠OAB = ∠OBA
∴ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 2∠OAB + 60° = 180°
⇒ 2 ∠OAB = 180° – 60° = 120°
⇒ ∠OAB = \(\frac{120}{2}\) = 60°
∴ ∠AOB is an equilateral triangle having side 28 cm.
Area of shaded designed portion = Area of the sector OAB – Area of ∆OAB
= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{\sqrt{3}}{4} \times 28^2\)
= \(\frac{22}{7} \times \frac{28^2 \times 60^{\circ}}{360^{\circ}}-\frac{\sqrt{3}}{4} \times 28 \times 28\)
= \(\frac{22 \times 4 \times 28 \times 1}{6}\) – √3 × 7 × 28
= \(\frac{44 \times 28}{3}\) – 1.7 × 196
= 410.67 – 333.2 = 77.47 cm².
The total area of six shaded designed portion = 6 × 77.47 = 464.82 cm².
Cost of making designs at the rate of 0.35 ₹/cm² = 464.82 × 0.35 = ₹ 162.68
Hence, cost of the making designs = ₹ 162.68.

Question 14.
Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is
(A) \(\frac{p}{180^{\circ}}\) × 2πR
(B) \(\frac{p}{180^{\circ}}\) × πR
(C) \(\frac{p}{360^{\circ}}\) × 2πR
(D) \(\frac{p}{720^{\circ}}\) × 2πR²
Solution:
(D) \(\frac{p}{720^{\circ}}\) × 2πR²

Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\).

Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let a be length of an edge of the cube,
Volume of cube = 64 cm³ (given)
⇒ a³ = 64
⇒ a = \(\sqrt[3]{64}\)
⇒ a = 4 cm

Now, dimensions of the resulting cuboid made by joining two cubes are 8 cm × 4 cm × 4 cm.
∴ Length of cuboid = 8 cm
Breadth of cuboid = 4 cm
Height of cuboid = 4 cm
Surface area of the cuboid= 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2 (32 + 16 + 32)
= 2 × 80 = 160 cm²
Hence, surface area of the cuboid = 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
We have,
Diameter of the hemispherical portion = 14 cm
Radius of the hemispherical portion = \(\frac{14}{2}\) = 7 cm

Total height of the vessel = 13 cm
Height of the cylindrical portion = 13 – 7 = 6 cm
Total inner surface area of the vessel
= C.S.A. of hemisphrical portion + C.S.A. of cylindrical portion
= 2πr² + 2πrh
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 7 (7 + 6)
= 44 × 13 = 572 cm²
Hence, total inner surface area of the vessel = 572 cm²

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
We have,
Radius (r) = 3.5 cm
Total height of the toy = 15.5 cm
height of the conical part (h) = 15.5 – 3.5 = 12 cm

Slant height of the conical part (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{12^2+(3 \cdot 5)^2}\)
= \(\sqrt{(12)^2+(3 \cdot 5)^2}\)
= \(\sqrt{144+12 \cdot 25}\)
= \(\sqrt { 1 5 6 \cdot 2 5 }\) = 12.5 cm
Total surface area of the toy = 2πr² + πrl = πr (2r + l)
= \(\frac{22}{7}\) × 3.5 (2 × 3.5 + 12.5)
= 11 × 19.5 = 214.5 cm².
Hence, total surface area of the toy = 214.5 cm².

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Solution:
We have,
Edge of cubical block = 7 cm
Greatest diameter of the hemisphere = edge of cube = 7 cm

Surface area of the solid = The surface area of the cube + C.S.A. of the hemisphere – base area of the hemisphere
= 6 × (edge)² + 2πr² – πr²
= 6 × (7)² + πr²
= 6 × 49 + \(\frac{22}{7}\) × (3.5)²
= 294 + 38.5 = 332.5 cm²
Hence, surface area of the solid = 332.5 cm².

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
We have,
Diameter of the hemisphere = l
Edge of the cube = diameter of the hemsphere = l
radius of the hemisphere = \(\frac{l}{2}\)

Surface Area of the remaining solid = surface area of the cube + Inner C.S.A. of the hemisphere – Area of the top of the hemisphere
= 6l² + 2πr x (\(\frac{l}{2}\))² – π (\(\frac{l}{2}\))²
= 6l² + π × (\(\frac{l}{2}\))²
= 6l² + \(\frac{\pi l^2}{4}\)
= \(\frac{1}{4}\) (24l² + πl²)
= \( \frac{l^2}{4}\) (24 + π)
Hence, surface area of the remaining solid = \(\frac{1}{4}\) l² (24 + π).

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see in figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
We have,
Radius of the cylindrical part and hemispherical part (r) = \(\frac{5}{2}\) = 2.5 mm
Length of the cylindrical part (h) = 14 – (2.5 + 2.5)
= 14 – 5 = 9 mm
Surface area of the capsule = C.S.A. of the cylindrical part + C.S.A. of two hemispherical parts
= 2πrh + 2 × 2πr²
= 2πr (h + 2r)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 2 × 2.5)
= \(\frac{22}{7}\) × 5 × 14
= 220 mm².
Hence, surface area of the capsule = 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution.
We have,
Radius of conical portion radius (r) = \(\frac{4}{2}\) = 2 m
Slant height (l) = 2.8 m
For cylindrical portion Radius (r) = 2 m
Height (h) = 2.1 m
Area of canvas used for making the tent = Surface area of tent = C.S.A. of conical part + C.S.A. of cylindrical part
= πrl + 2πrh
= πr(l + 2h)

= \(\frac{22}{7}\) × 2(2.8 + 2 × 2.1)
= \(\frac{44}{7}\) × 7 = 44 m².
Cost of canvas at the rate of ₹ 500/m² = 44 × 500 = ₹ 22000
Hence, area of the canvas = 44 m²
and cost of canvas = Rs 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution.
We have,
Height of the cylinder = 2.4 cm
Diameter of the cylinder = 1.4 cm
∴ Radius of the cylinder = \(\frac{1.4}{2}\) = 0.7 cm

Slant height of the cone (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{(2 \cdot 4)^2+(0 \cdot 7)^2}\) = 2.5 cm
Total surface area of the remaining solid = C.S.A. of the cylinder + Area of base of the cylinder + C.S.A. of the cone
= 2πrh + πr² + πrl
= πr (2h + r + l)
= \(\frac{22}{7}\) × 0.7 (2 × 2.4 + 0.7 + 2.5)
= 2.2 (4.8 + 3.2)
= 2.2 × 8 = 17.6 cm²
= 18 cm², (apporx)
Hence, the total surface area of the remaining solid to the nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution.
We have,
Height of the cylinder (A) = 10 cm
Radius of cylinder and hemisphere (r) = 3.5 cm

Total surface area of the article = C.S.A of the cylinder + 2 (surface area of a hemisphere)
= 2πrh + 2 × 2πr²
= 2πr (h + 2r)
= 2 × \(\frac{22}{7}\) × 3.5 (10 + 2 × 3.5)
= 22 × 17
= 374 cm²
Hence, total surface area of the article = 374 cm².

Haryana State Board HBSE 10th Class Maths Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 10 Circles Ex 10.2

In question 1 to 3, choose the correct option and give justification :

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is :
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution :
OP ⊥ PQ [By theorem 10.1]
In right ∆OPQ, we have
OQ² = QP² + OP²
[By Pythagoras theorem]

25² = 24² + OP²
OP² = 25² – 24²
⇒ OP² = (25 + 24) (25 – 24)
⇒ OP² = 49
⇒ OP = √49 = 7
∴ Radius = 7 cm
So, correct option is (a).

Question 2.
In the given figure, if TP and TQ are two tangents to a circle with centre O, so that ∠POQ = 110°, then ∠PTQ is equal to :

Solution:
∠OPT = ∠OQT = 90° (By theorem 10.1)
In a quadrilateral POQT, we have
∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360°
⇒ 90° + 90° + 110° + ∠PTQ = 360°
⇒ 290° + ∠PTQ = 360°
⇒ ∠PTQ = 360° – 290°
⇒ ∠PTQ = 70°
The correct option is (b).

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to :
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution :
OA ⊥ AP and OB ⊥ PB (By theorem 10.1)
In AOAP and AOBP, we have
AO = OB (radii of the same circle)
∠OAP = ∠OBP (each is 90°)
PO = OP (common)
∆OAP = ∆OBP (By R.H.S. congruence)
∠APO = ∠BPO (By CPCT)

⇒ 40° + 90° + ∠POA = 180°
⇒ 130° + ∠POA = 180°
⇒ ∠POA = 180° – 130° = 50°
Hence, correct option is (a).

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Given : AB, CD are two tangents at the ends of a diameter PQ of a circle with centre O.

To Prove: AB || CD
Proof : OP ⊥ AB [By theorem 10.1]
OQ ⊥ CD[By theorem 10.1]
∠APQ = 90° and ∠CQP = 90°
∠APQ + ∠CQP = 90° + 90° = 180°
But these are co-interior angles.
∴ AB || CD Hence Proved.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Given :
A circle with centre O and a tangent AB at point P.
To Prove : OP passes through centre.

Proof : AB is a tangent at point P and OP is the radius of circle.
If we have some points m, n on tangent AB. Then we find that OP is the shortest line from O in comparison of Om and On.
Therefore, OP ⊥ AB.
Hence, perpendicular OP drawn to tangent line at P passess through the centre (O) of the circle.
Hence Proved.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution :
OB ⊥ AB (By theorem 10.1)
In right ∆ABO, we have
AO² = AB² + OB² [By Pythagoras theorem]

⇒ 5² = 4² + 6²
⇒ OB² = 5² – 4²
⇒ OB² = 25 – 16
⇒ OB² = 9
⇒ OB = √9 = 3

Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Let O be the centre of two concentric circles of radii 5 cm and 3 cm.
Let AB be a chord of the larger circle touching the smaller circle at P. Then
OP = 3 cm and AO = 5 cm OP ⊥ AB and AP = PB
In right ∆OPA, we have

AO² = AP² + OP²
5² = AP² + 3²
AP² = 5² – 3²
AP² = (5 + 3) (5 – 3)
AP² = 8 × 2 = 16
⇒ AP = √16 = 4
AB = 2 × AP = 2 × 4 = 8
Hence, length of chord is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure below). Prove that: AB + CD = AD + BC
OR
A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.
Solution:

Given : A quadrilateral ABCD is drawn to circumscribe a circle.
To Prove : AB + CD = AD + BC
Proof : By theorem 10.2
We know that tangents drawn from an exterior point to a circle are equal in length.
AP = AS …..(1) [Tangents from point A]
PB = BQ ….(2) [Tangents from point B]
CR = QC ….(3) [Tangents from Point C]
DR = DS ….(4) [Tangents from point D]
Adding (1), (2), (3) and (4), we get
AP + PB + CR + DR = AS + BQ + QC + DS
⇒ AB + CD = AS + DS + BQ + QC
⇒ AB + CD = AD + BC
Hence Proved.

Question 9.
In the given figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Given : Tangent XY || Tangents X’Y’ and another Tangent AB with point of contact C intersect XY at A and X’Y’ at B.
To Prove : ∠AOB = 90°
Construction : Join OC
Proof : In ∆APO and ∆ACO
AP = AC (By Theorem 10.2)
∠APO = ∠ACO = 90° (By theorem 10.1)
AO = AO (Common)
∆APO s ∆ACO (By RHS congruence)
∠PAO = ∠CAO (By CPCT)
⇒ 2 ∠CAO = ∠PAC ……………..(1)
Similarly, 2 ∠CBO = ∠QBC …………..(2)
Now, ∠PAC + ∠QBC = 180° [co-interior ∠s of same side]
2 ∠CAO + 2 ∠CBO = 180° [Using (1) and (2)]
⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90“
⇒ ∠BAO + ∠ABO = 90°
In A AOB, we have
∠BAO + ∠AOB + ∠ABO = 180°
⇒ ∠AOB + 90° = 180° [Using (3)]
⇒ ∠AOB = 180° – 90° = 90°.
Hence Proved.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
OR
In the given figure, O is the centre of a circle and two tangents CA, CB are drawn on the circle from a point C lying outside the circle. Prove that ∠AOB and ∠ACB are supplimentary.

Solution:
Given : PA and PB be two tangents drawn from an external point P to a circle with centre O. Also, the line segments OA and OB are drawn.
To Prove : ∠APB + ∠AOB = 180°
Proof: PA ⊥ OA and
PB ⊥ OB [By theorem 10.1]

i.e. ∠PAO = 90° and ∠PBO = 90°
∠PAO + ∠PBO = 180° ………………..(1)
We know that sum of ∠s of a quadrilateral 360°.
∠APB + ∠AOB + ∠PAO + ∠PBO = 360°
⇒ ∠APB + ∠AOB + 180° = 360°
⇒ ∠APB + ∠AOB = 360° – 180°
⇒ ∠APB + ∠AOB = 180°.
Hence Proved.

Question 11.
Prove that the parallelogram circumscri-bing a circle is a rhombus.
Solution:
Given: A parallelogram ABCD circumscribes a circle with centre O.
To Prove : AB = BC = CD = AD
Proof: By theorem 10.2 we know that lengths of tangents drawn from an exterior point to a circle are equal.
BQ = BP …(1) (Tangents from Point B]
QC = RC…(2) [Tangents from Point C]
SD = RD …(3) [Tangents from point D]
AS = AP …(4) [Tangents from point A]

Adding (1), (2), (3) and (4), we get
BQ + QC + SD + AS = BP + RC + RD + AP
⇒ BC + AD = BP + AP + RC + RD
⇒ BC + AD = AB + CD
⇒ 2BC = 2AB
[∵ Opposite sides of ||gm are equal]
⇒ BC = AB
Since adjacent sides of a ||gm are equal]
∴ AB = BC = CD = AD.
Therefore, ABCD is a rhombus.
Hence Proved.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
Let tangents AB, BC and AC touch the circle at points E, D and F respectively. Join OE, OF, OA, OC and OB.
By theorem 10.2, we know that lengths of tangents drawn from an external point to the circle are equal.
CF = CD = 6 cm
BE = BD = 8 cm
AF = AE
Let AF = AE = x cm
AC = (6 + x) cm and AB = (8 + x) cm
∴ s = \(\frac{a+b+c}{2}\)
[Where a, b and c are lengths of sides of ∆ABC]
s = \(\frac{14+6+x+8+x}{2}\)
s = \(\frac{28+2 x}{2}\)
s = 14 + x
Area of ∆ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{\begin{array}{r}
(14+x)[14+x-14\} \\
\{14+x-6-x\} \\
{[14+x-8-x]}
\end{array}}\)
= \(\sqrt{(14+x)(x)(8)(6)}\)
= \(\sqrt{48 x(14+x)}\) cm² ……………(1)
Now, OD ⊥ BC, OE ⊥ AB, OF ⊥ AC [By theorem 10.1]
Area of ∆ABC = Area of ∆BOC + Area of ∆AOB + Area of ∆AOC
= \(\frac{1}{2}\) (14) × 4 + \(\frac{1}{2}\) (8 + x) × 4 + \(\frac{1}{2}\) (6 + x) × 4
= 28 + 16 + 2x + 12 + 2x
= (56 + 4x) cm² ……………….(2)
From (1) and (2), we get
\(\sqrt{48 x(14+x)}\) = 56 + 4x
Squaring both sides, we get
48x (14 + x) = (56 + 4x)²
⇒ 48x (14 + x) = 16 (14 + x)²
⇒ 3x (14 + x) = (14 + x)²
⇒ 3x = \(\frac{(14+x)^2}{(14+x)}\)
⇒ 3x = 14 + x
⇒ 3x – x = 14
⇒ 2x = 14
⇒ x = \(\frac{14}{2}\) = 7
Length of AB = 8 + x = 8 + 7 = 15 cm
Length of AC = 6 + x = 6 + 7 = 13 cm
Hence, lengths of AB and AC are 15 cm and 13 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution :
Given : A quadrilateral ABCD circumscribes a circles with centre O.

To Prove: ∠AOB + ∠COD = 180
and ∠AOD + ∠BOC = 180°
Construction : Join OP, OQ, OR and OS.
Proof: In ∆OPB and ∆OQB,
∠OPB = ∠OQB (each is 90°) ,
PB = QB(By theorem 10.2)
OB = OB (Common)
∆OPB = ∆OQB (By RHS congruence)
Similarly,
∠1 = ∠2 (By CPCT)
∠3 = ∠4
∠5 = ∠6
∠7 = ∠8
∵ Sum of ∠s of a quadrilateral is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
⇒ 2 (∠2 + ∠3) + 2 (∠6 + ∠7) = 360
[∵ ∠1 = ∠2, ∠4 = ∠3, ∠5 = ∠6 and ∠8 = ∠7]
and similarly, 2 (∠1 + ∠8) + 2 (∠4 + ∠5) = 360°
⇒(∠2 + ∠3) + (∠6 + Z7) = 180° and (∠1 + ∠8) + (∠4 + ∠5) = 180°
⇒ ∠BOC + ∠AOD = 180° and ∠AOB + ∠COD = 180°.
Hence Proved.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of construction :

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction :
1. Draw a circle with centre O of radius 6 cm.
2. Mark a point A, 10 cm away from the centre.
3. Join AO and bisect it at M.

4. Draw a circle with M as a centre and radius equal to AM intersects the given circle at points P and Q.
5. Join AP and AQ.
Then AP and AQ are the required tangents. Lengths of AP and AQ are 8-2 cm. (approx.)
Justification : Join OP
∠APO = 90° [angle in a semi-circle]
AP ⊥ OP
Therefore, AP is a tangent to the given circle.
Similarly AQ is a also tangent to the given circle.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:
Steps of Construction :
1. O as the centre and draw two circles having radii 4 cm and 6 cm.
2. Taking point P on the larger circle.
3. Join OP and bisect it at M.
4. Taking M as the centre and radius equal to PM draw the circle which intersects the smaller circle at A and B.
5. Join PA and PB.
Then PA and PB are the required tangents.
On measuring the lengths of tangents
PA = PB = 4.47 (Approx.)
By calculation we have the length of tangent.
Joining AO
AO ⊥ PA [By theorem 10.1]
In right ∆PAO, we have
PO² = PA² + AO²
⇒ (6)² = PA² + (4)²

PA² = (6)² – (4)²
PA² = 36 – 16 = 20
⇒ PA = √20 = 4.472 cm
Justification : PA ⊥ OA
AO is the radius of the circle.
PA is tangent to a given circle.
Similarly PB is the tangent to the given circle.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction :
1. Draw a circle with centre O and radius equal to 3 cm.
2. Take points P and Q on extended diameter of the circle such that OP = OQ = 7 cm.
3. Mark mid point M₁ of OP and mid point of OQ as M₂.

4. Draw the circles as M₁ and M₂ centres and radii equal to PM₁ and OM₂ respectively, which intersects A and B (M₁ as centre) and C, D (M₂ as centre).
5. Join PA, PB and QC, QD.
Then PA, PB, QC and QD are tangents from the points P and Q respectively.
Justification : Join OA, OB, OC and OD ∠PAO = 90°
[Angle in a semi circle is 90°]
OA is the radius of circle.
PA is tangent to the given circle.
Similarly PB is the tangent to the given circle.
Again ∠OCQ = 90°
[Angle in a semi circle is 90°]
OC is the radius.
QC is the tangent to the given circle.
Similarly, QD is the tangent to the given circle.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm. which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction :
1. Draw a circle with centre O and radius 5 cm.
2. Construct radii OA and OB such that ∠AOB = 360° – (90° + 90° + 60°) = 120°.
3. Draw AL ⊥ OA at A and BM ⊥ OB at B. They intersect at P.
Then PA and PB Eire the required tangents inclined to each other at 60°.
Justification : PA ⊥ OA and PB ⊥ OB and ∠AOB = 120°.

⇒ ∠APB = 360° – 300°
⇒ ∠APB = 60°.
Therefore, tangents PA and PB are inclined at 60°.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction :
1. Draw a line segment AB = 8 cm.
2. A as centre and radius equal to 4 cm draw a circle.
3. B as centre and radius equal to 3 cm draw another circle.
4. Bisect line segment AB at M.

5. M is the centre and radius equal to AM draw the circle which intersects the previous circles at P, Q (A as centre) and R, S (B as centre).
6. Join BP, BQ, AR and AS.
7. Then, BP, BQ, AR and AS are the required tangents.
Justification : Join AP, RB
∠APB = 90° [Angle in a semi circle is 90°]
PB ⊥ AP.
Therefore, AP is radius and BP is tangent.
Similarly, AR, AS and BQ are tangents.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of Construction:
1. Construct ∆ABC such that BC = 8 cm, AB = 6 cm and ∠ABC = 90°.
2. Draw BD ⊥ AC.
3. ∠CDB = 90° So, BC is the diameter of circle passing through B, C, D.
4. Bisect line BC at O.
5. O as the centre, CO = OB as radius draw a circle which passes through B, C, D.

6. Join AO which intersects the circle at M.
7. M as centre and MA = OM as radius draw another circle which intersect the previous circle at P and B.
8. Join PA.
Then, AP and AB are the required tangents.
Justification:
∠ABC = 90° [By Construction]
AB ⊥ OB.
OB is radius and AB is tangent.
Similarly, AP is tangent.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction :
Take four points A, B, C and D on the given circle and join AB and CD.
1. Draw the perpendicular bisectors of non parallel chords AB and CD which intersect at O.
2. O is the centre of circle.
3. Take a point P outside the circle.

4. Join OP and bisected it at M.
5. M as the centre and radius PM = MO, draw another circle which intersect previous circle at R.
6. Join PR and PQ.
Then PR and PQ are the required tangents.
Justification : Join OR,
∠PRO = 90° [Angle in a semi circle is 90°]
PR ⊥ OR
OR is a radius and PR is a tangent.
Similarly, PQ is a tangent.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumference of the two circles.
Solution.
Circumference of the circle of radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle of radius 9 cm = 2π × 9 = 18π cm
Sum of the circumferences of two circles = 38π + 18π = 56π cm
Let R be the radius of the circle which has circumference equal to sum of the circumferences of the two circle.
2πR = 56π
⇒ R = \(\frac{56 \pi}{2 \pi}\)
⇒ R = 28 cm
Hence,radius of the circle = 28 cm.

Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution.
Area of circle of radius 8 cm = π × 8² = 64π cm2
Area of circle of radius 6 cm = π × 6² = 36π cm2
Sum of the areas of two circles = 64π + 36π = 100π cm²
Let R be the radius of the circle whose area is equal to the sum of the areas of the two circles.
⇒ πR² = 100π
⇒ R² = \(\frac{100 \pi}{\pi}\) = 100
⇒ R = √100 = 10 cm
Hence, radius of the circle = 10 cm.

Question 3.
Figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of the each of the five scoring regions.

Solution.
We have,
Radius of region representing Gold score (r₁) = \(\frac{21}{2}\) = 10.5 cm
Area of region representing Gold score = π × (10.5)²
= \(\frac{22}{7}\) × 10.5 × 10.5 = 346.5 cm²

Radius of region representing Red and Gold score areas (r₂) = 10.5 + 10.5 = 21 cm
Area of region representing Red score = πr₂² – πr₁²
= \(\frac{22}{7}\) × (21)² – 346.5
= 1386 – 346.5 = 1039.5 cm²

Radius of the region representing Blue, Red and Gold scoring areas (r₃) = (21 + 10.5) cm = 31.5 cm
Area of region representing Blue scoring area = πr₃² – (Area of Gold + Area of Red)
= \(\frac{22}{7}\) × (31.5)² – (346.5 + 1039.5)
= 3118.5 – 1386 = 1732.5 cm²

Radius of the region representing Black, Blue, Red and Gold scoring areas (r₄) = (31.5 + 10.5) cm = 42 cm
Area of region representing Black scoring area = πr₄² – (Area of Gold + Area of Red + Area of Blue)
= \(\frac{22}{7}\) × (42)² – (346.5 + 1039.5 + 1732.5)
= 5544 – 3118.5 = 2425.5 cm²

Radius of the region representing White, Black, Blue, Red and Gold, scoring areas (r₅) = (42 + 10.5) cm = 52.5 cm.
Area of region representing White scoring area = πr₅² – [Area of Gold + Area of Red + Area of Blue + Area of Black]
= \(\frac{22}{7}\) × (52.5)2 – (346.5 + 1039.5 + 1732.5 + 2425.5)
= 8662.5 – 5544 = 3118.5 cm²

Hence, areas of Gold, Red, Blue, Black and White scoring regions are 346.5 cm², 1039.5 cm², 1732.5 cm², 2425.5 cm², and 3118.5 cm² respectively.

Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km/hour ?
Solution.
We have,
Speed of car = 66 km/hour
Distance travelled by car in 1 minute = \(\frac{66}{100}\) × 1000 × 100 = 110000 cm
Distance travelled by car in 10 minutes = 110000 × 10 = 1100000 cm
Radius of car’s wheel = \(\frac{80}{2}\) = 40 cm
Circumference of wheel = 2πr
= 2 × \(\frac{22}{7}\) × 40 = \(\frac{1760}{7}\) cm
Distance covered by the wheel of a car in 1 revolution = \(\frac{1760}{7}\) cm
Number of revolutions made by the wheel of a car in 10 minutes = \(\frac{\text { Distance travelled by car in } 10 \text { minutes }}{\text { Distance travelled by car in1 revolution }}\)
= \(\frac{1100000}{\frac{1760}{7}}\)
= \(\frac{1100000 \times 7}{1760}\)
= 4375
Hence, number of revolutions made by wheel of a car = 4375.

Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units
Solution.
Correct option is (A)
Justification. Let the radius of circle is r units.
Acoording to questions
πr2 = 2πr
\(\frac{\pi r^2}{\pi r}\) = 2
r = 2.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 11 Constructions Ex 11.1

In each of the following, give the justification of the construction also:

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :
1. Draw a line segment AB = 7.6 cm.
2. Draw any ray AX, making an acute angle with AB.
3. Along AX mark (5 + 8) = 13 points A₁, A₂,
A₃ ……………………. A₁₃ such that AA₁ = A₁A₂ = A₂A₃ = …………………….. A₁₂A₁₃

4. Join A₁₃B.
5. Through A₅, draw A₅C || A₁₃B meeting AB at C.
6. Then AC : BC = 5 : 8.
7. On measuring
AC = 2.9 cm
CB = 4.7 cm
Justification : In ∆ABA₁₃, A₅C || A₁₃B.
\(\frac{\mathrm{AA}_5}{\mathrm{~A}_5 \mathrm{~A}_{13}}=\frac{\mathrm{AC}}{\mathrm{BC}}\) [By Basic proportionality theorem]
But \(\frac{\mathrm{AA}_5}{\mathbf{A}_5 \mathbf{A}_{13}}=\frac{5}{8}\) [By construction]
∴ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{8}\)
AC : BC = 5 : 8.

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the coi responding sides of the first triangle.
Solution:
Steps of construction :
1. Draw a line segment BC = 5 cm.
2. With B as a centre and radius 4 cm draw an arc.
3. With C as a centre and radius 6 cm draw another arc to intersect the first arc. at A.

4. Join BA and CA to get ∆ABC.
5. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
6. Along BX mark 3 points B₁, B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.
7. Join B₃C.
8. From B₂ draw B₂C’ || B₃C meeting BC at C’.
9. From C’ draw C’A’ || CA meeting BA at A’.
Then A’BC’ is the required triangle, each of whose side is \(\frac{2}{3}\) of corresponding Sides of ∆ABC.
Justification : Since, A’C’ || AC
Therefore, ∆A’BC’ ~ ∆ABC
\(\frac{\mathrm{AB}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{2}{3}\)

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle 7 whose sides are \(\frac{7}{5}\) of the corresponding sides of the 5 first triangle.
Solution:
Steps of Construction :
1. Draw a line segment BC = 6 cm.
2. With B as a centre and radius 5 cm draw an arc.
3. With C as a centre and radius 7 cm draw another arc to intersect the first arc at A.
4. Join BA and CA to get ∆ABC.

5. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
6. Along BX mark 7 points B₁, B₂, B₃, …….., B₇ such that BB₁ = B₁B₂ = B₂B₃ = ………….. = B₆B₇.
7. Join B₅C.
8. From B₇ draw B₇C’ || B₅₆C intersecting the produced line segment BC at C’.
9. From C’ draw CA’ || AC intersecting the produced line segment BA at A’.
Then, ∆A’BC’ is the required triangle each of whose side is \(\frac{7}{5}\) of the corresponding sides of the ∆ABC.
Justification : Since, AC || A’C’
Therefore, ∆A’BC’ ~ ∆ABC
⇒ \(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{7}{5}\)

Question 4.
Construct an Isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :
1. Let ABC be the isosceles triangle with base AB = 8 cm and altitude CD = 4 cm.
2. Draw a line segment AB = 8 cm.
3. Draw the perpendicular bisector of AB intersecting AB at D.
4. With D as a centre and radius 4 cm draw an arc intersecting the perpendicular bisector at C.
5. Join AC and BC to get the isosceles triangle ABC.

6. Produce AB to B’ AB’ = 1\(\frac{1}{2}\) of 8 = \(\frac{3}{2}\) × 8 = 12 cm.
7. Through B’ draw B’C’ || BC intersecting the produced line segment AC to C’.
Then AB’C’ is the required triangle each of whose side is 1\(\frac{1}{2}\) times of the corresponding sides of ∆ABC.
Justification : Since, BC || B’C’
Therefore, ∆C’AB’ ~ ∆CAB
⇒ \(\frac{\mathrm{AC}^{\prime}}{\mathrm{AC}}=\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{3}{2}\).

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction:
1. Draw a line segment BC = 6 cm.
2. At B, draw ∠CBY = 60°.
3. From B draw an arc AB = 5 cm meeting BY at A. Join AC. Thus ∆ABC obtained.
4. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
5. Along BX mark four points B₁, B₂, B₃, B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
6. Join B₄C.

7. From B₃ draw B₃C’ || B₄C meeting BC at C.
8. From C’ draw C’A’ || CA, meeting AB at A’. Then A’BC’ is the required triangle, each of whose side is \(\frac{3}{4}\) of corresponding sides of AABC.
Justification: Since, A’C || AC
Therefore, ∆A’BC ~ ∆ABC
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{3}{4}\).

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Solution:
Steps of Construction :
1. Draw line segment BC = 7 cm.
In ∆ABC, ∠B = 45°, ∠A = 105°,
∠C = 180° – (45° + 105°) = 30°
2. At B draw an angle ∠B = 45° and at C draw an angle ∠C *= 30° intersecting each other at A to get ∆ABC.
3. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
Along BX mark four points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

4. Join B₃C.
5. From B₄ draw B₄C’ || B₃C intersecting the produced line segment BC at C’.
6. From C’ draw C’A’ || CA intersecting the produced line segment BA at A’.
Then A’BC’ is the required triangle, each of whose side is \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Justification : Since, AC || A’C’
Therefore, ∆A’BC’ ~ ∆ABC
\(\frac{\mathrm{AB}^{\mathrm{B}}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{4}{3}\).

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
1. Let ABC be right triangle in which ∠B = 90°, BC = 4 cm and AB = 3 cm.
2. Draw BC = 4 cm
3. Draw ∠CBY = 90° at B.
4. Cut AB = 3 cm from BY. Join AC to get ∆ABC.
5. Draw any ray CX making an acute angle with BC on the side opposite to the vertex A.
6. Along CX mark five points C₁, C₂, C₃, C₄ and C₅ such that CC₁ = C₁C₂ = C₂C₃ = C₃C₄ = C₄C₄₅.
Join C₃B.

7. From C₅ draw C₅B’ || C₃B intersecting the produced line segment CB at B’.
8. From B’ draw B’A’ || BA intersecting the produced line segment CA at A’.
Then, A’CB’ is the required triangle.
Justification : Since, BA || B’A’
Therefore, ∆A’CB’ ~ ∆ACB
\(\frac{\mathrm{A}^{\prime} \mathrm{C}}{\mathrm{AC}}=\frac{\mathrm{B}^{\prime} \mathrm{C}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}=\frac{5}{3}\).

Haryana State Board HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60°
(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\ {cosec} 30^{\circ}}\)
(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ {cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
(v) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)
Solution :
(i) sin 60° cos 30° + sin 30° cos 60°
= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
= \(\frac{3}{4}+\frac{1}{4}\) = 1
Hence, sin 60° cos 30° + sin 30° cos 60° = 1.

(ii)2 tan² 45° + cos² 30° – sin² 60°
= 2 × (1)² + \(\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2\)
= 2 × 1 + \(\frac{3}{4}-\frac{3}{4}\) = 2
Hence, 2 tan² 45° + cos² 30° – sin² 60° = 2.

(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\ {cosec} 30^{\circ}}\)

Hence, \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\ {cosec} 30^{\circ}}\) = \(\frac{1}{8}\) (3√2 – √6).

(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ {cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)

Hence, \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ {cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\) = \(\frac{43-24 \sqrt{3}}{11}\)

(v) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)

Question 2.
Choose the correct option and justify your choice :
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\) =
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°

(ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) =
(a) tan 90°
(b) 1
(c) sin 45°
(d) 0

(iii) sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\) =
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°

Solution:
(i) (a)

∵

(ii) (d)
∵ \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-(1)^2}{1+(1)^2}\)
= \(\frac{1-1}{1+1}=\frac{0}{2}\) = 0

(iii) (a) When A = 0°
L.H.S. = sin 2A = sin 2 × 0°
= sin 0° = 0
R.H.S. = 2 sin A = 2 × sin 0°
= 2 × 0 = 0
In other option we shall find that L.H.S. ≠ R.H.S.

(iv) (c)
∵ \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\)
= \(\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}\)
= √3 = tan 60°.

Question 3.
If tan(A + B) = √3 and tan(A – B) = \(\frac{1}{\sqrt{3}}\); 0° < (A + B) ≤ 90°, A > B, find A and B.
Solution :
We have
tan(A + B) = √3
tan(A + B) = tan 60°
A + B = 60° ……………..(1)
and tan(A – B) = \(\frac{1}{\sqrt{3}}\)
tan(A – B) = tan 30°
A – B = 30° ……………….(2)
Adding equation (1) and (2), we get
A + B = 60°
A – B = 30°
2A = 90°
A = \(\frac{90^{\circ}}{2}\)
A = 45°
Substituting the value of A in equation (1), we get
45° + B = 60°
B = 60° – 45° = 15°
Hence, A = 45° and B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin(A + B) = sin A + sin B.
(ii) The value of a sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution :
(i) False. Taking A= 30°, B = 60°
LHS = sin (A + B)
= sin (30° + 60°)
= sin 90° = 1
RHS = sin A + sin B
= sin 30° + sin 60°
= \(\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\)
∴ LHS ≠ RHS
(ii) True.
(iii) False. Because value of cos θ decreases as θ increases.
(iv) False. Because sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
i.e., sin 30° ≠ cos 30°
(v) True. Because cot 0° = ∞ (not defined)

Haryana State Board HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C
Solution :
(i) In a right’triangle ABC we have ∠B = 90°, AB = 24 cm and BC = 7 cm.
By Pythagoras theorem, we have
AC² = AB² + BC²
⇒ AC² = 242 + 72
⇒ AC² = 576 + 49
⇒ AC² = 625
⇒ AC = \(\sqrt{625}\) = 25

∴ sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25}\)
cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}\)
Here, sin A = \(\frac{7}{25}\), cos A = \(\frac{24}{25}\)

(ii) For trigonometric ratio of ∠C, We have base = BC = 7 cm
Perpendicular = AB = 24 cm
Hypotenuse = AC = 25 cm
sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}\)
and cos C = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25}\)
Hence, sin C = \(\frac{24}{25}\), cos C = \(\frac{7}{25}\)

Question 2.
In figure, find tan P – cot R.

Solution:
We have, PQ = 12 cm, PR = 13 cm.
In right ∆PQR, ∠Q = 90°
By Pythagoras theorem, we have PR² = QR² + PQ²
⇒ (13)² = QR² + (12)²
⇒ QR² = (13)² – (12)²
⇒ QR² = (13 + 12) (13 – 12)
⇒ QR² = 25
⇒ QR = √25 = 5
cot R = \(\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}\)
For trigonometric ratio of ∠P, we have
Base = PQ = 12 cm
Perpendicular = QR = 5 cm
Hypotenuse = PR = 13 cm
∴ tan P = \(\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}\)
Now, tan P – cot R = \(\frac{5}{12}-\frac{5}{12}\) = 0
Hence, tan P – cot R = 0.

Question 3.
If sin A = \(\frac{3}{4}\) Calculate cos A and tan A.
Solution :
Consider a right triangle ABC in which ∠B = 90°
Then, sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}\)
Let BC = 3k and AC = 4k, where k is a positive integer.
By Pythagoras theorem, we have

AC² = AB² + BC²
(4k)² = AB² + (3k)²
AB² = (4k)² – (3k)²
AB² = 16k² – 9k²
AB² = 7k²
AB = √7 k
∴ cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{7} k}{4 k}=\frac{\sqrt{7}}{4}\)
and tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}}\)
Hence, cos A = \(\frac{\sqrt{7}}{4}\), tan A = \(\frac{3}{\sqrt{7}}\).

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution :

we have,
15 cot A = 8
cot A = \(\frac{8}{15}\)
Now consider a right 15 triangle ABC, in which ∠B = 90°
cot A = \(\frac{8}{15}\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{8}{15}\)
Let AB = 8k and BC = 15k,where k is a positive integer.
By Pythagoras theorem, we have
AC² = AB² + BC²
AC² = (8k)² + (15k)²
AC² = 64k² + 225k²
AC = 289k²
AC = \(\sqrt{289 k^2}\) = 17k.
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}:=\frac{15 k}{17 k}=\frac{15}{17}\)
and sec A = \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{17 k}{8 k}=\frac{17}{8}\)
Hence, sin A = \(\frac{15}{17}\) and sec A = \(\frac{17}{8}\).

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios.
Solution:
Consider a right triangle ABC in which ∠B = 90°, ∠CAB = 9, sec θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}\)
Let AC = 13k and AB = 12k, where k is a positive integer.

By Pythagoras theorem, we have
AC² = AB² + BC²
(13k)² = (12k)² + BC²
BC² = (13k)² – (12k)²
BC² = (13k + 12k) (13k – 12k)
BC² = 25k × k = 25k²
BC = \(\sqrt{25 k^2}\) = 5k

∴ sin θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 k}{13 k}=\frac{5}{13}\)
cos θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 k}{13 k}=\frac{12}{13}\)
tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5 k}{12 k}=\frac{5}{12}\)
cot θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12 k}{5 k}=\frac{12}{5}\)
cosec θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13 k}{5 k}=\frac{13}{5}\)

Hence, sin θ = \(\frac{5}{13}\),
cos θ = \(\frac{12}{13}\),
tan θ = \(\frac{5}{12}\),
cot θ = \(\frac{12}{5}\),
cosec θ = \(\frac{13}{5}\).

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B; then show that ∠A = ∠B.
Solution :
Let us consider right triangle ABC in which ∠C = 90° and cos A = cos B.
Draw CD ⊥ AB.
In right triangle ADC, ∠ADC = 90°.
cos A = \(\frac{\mathrm{AD}}{\overline{\mathrm{AC}}}\) ………………….(1)
In right ∆BDC,
∠BDC = 90°
cos B = \(\frac{\mathrm{BD}}{\overline{\mathrm{BC}}}\) ………………….(2)

According to question,
cos A = cos B
\(\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{BC}}\)
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}\) [each = \(\frac{C D}{C D}\)]
∆ADC ~ ∆BDC [By SSS similarity criterion]
[corresponding ∠s of two similar ∆s are equal]
Hence Proved.

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate:
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot² θ
Solution :

(i) Consider a triangle ABC in which ∠B = 90°,
∠ACB = θ and cot θ = \(\frac{7}{8}\)
⇒ \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{7}{8}\)
Let BC = 7k and AB = 8k, where k is a positive integer,
By Pythagoras theorem, we have
AC² = BC² + AB²
AC² = (7k)² + (8k)²
AC² = 49k² + 64k²
AC² = 113k²
AC = √113 k
∴ sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{8 k}{\sqrt{113} k}=\frac{8}{\sqrt{113}}\)

(ii) cot² θ = (cot θ)²
= \(\left(\frac{\mathrm{BC}}{\mathrm{AB}}\right)^2=\left(\frac{7 k}{8 k}\right)^2\)
= \(\left(\frac{7}{8}\right)^2=\frac{49}{64}\)

Hence, (i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}\)
(ii) cot² θ = \(\frac{49}{64}\)

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\) = cos² A – sin² A or not.
Solution :
Consider a triangle ABC in which ∠B = 90° and 3 cot A = 4
⇒ cot A = \(\frac{4}{3}\)
⇒ \(\frac{A B}{B C}=\frac{4}{3}\)
Let AB = 4k and BC = 3k,where k is a positive integer.

By pythagoras theorem, we have
AC² = AB² + BC²
⇒ AC² = (4k)² + (3k)²
⇒ AC² = 16k² + 9k²
⇒ AC² = 25k²
⇒ AC = \(\sqrt{25 k^2}\)
⇒ AC = 5k

Hence, \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos² A – sin² A.

Question 9.
In triangle ABC, right angled at B, if tan A = find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
In right triangle ABC, ∠B = 90° and tan A = \(\frac{1}{\sqrt{3}}\)
\(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}\)
Let BC = k and AB = √3k,where k is positive integer.

By pythagoras theorem, we have
AC² = AB² + BC²
AC² =(√3k)² + k²
AC² = 3k² + k² = 4k²
AC = \(\sqrt{4 k^2}\) = 2k
∴ sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}\)
and cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\)
Now, for trigonometric ratio of ∠C, we have
Base = BC = k
Perpendicular = AB = √3 k
Hypotenuse = AC = 2k
sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\)
cosC = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}\)

(i) sin A cos C + cos A sin C = \(\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\)
= \(\frac{1}{4}+\frac{3}{4}=\frac{4}{4}\) = 1

(ii)cos A cos C – sin A sinC= \(\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\) = 0

Hence, (i) sin A cos C + cos A sin C = 1
(ii) cos A cos C – sin A sin C = 0.

Question 10.
In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
In right ∆PQR, ∠Q = 90°, PQ = 5 cm and PR + QR = 25 cm

⇒ PR = 25 – QR
By Pythagoras theorem, we have
PR² = PQ² + QR²
(25 – QR)² = (5)² + QR²
⇒ 625 + QR² – 50 QR = 25 + QR²
⇒ QR² – 50QR – QR² = 25 – 625
⇒ – 50 QR = – 600
⇒ QR = \(\frac{-600}{-50}\) = 12 cm
and PR = 25 – 12 = 13 cm
∴ sin P = \(\frac{\mathrm{QR}}{\mathrm{PR}}=\frac{12}{13}\)
cos P = \(\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{5}{13}\)
tan P = \(\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{12}{5}\)
Hence, sin P = \(\frac{12}{13}\), cos P = \(\frac{5}{13}\) and tan P = \(\frac{12}{5}\).

Question 11.
State whether the following are true or false. Justify your Answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac{12}{5}\) for some value of ∠A.
(iii) cos A is the abbreviation used, for the cosecant of ∠A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution :
(i) False. Because tan 60° = √3 > 1
(ii) True. Because value of sec A is always ≥ 1.
(iii) False. Because cos A is abbreviation used for cosine A.
(iv) False. Because cot is meaningless without an angle.
(v) False. Because sin θ > 1.