HBSE 9th Class Maths Important Questions Chapter 12 Heron’s Formula

Haryana State Board HBSE 9th Class Maths Important Questions Chapter 12 Heron’s Formula Important Questions and Answers.

Haryana Board 9th Class Maths Important Questions Chapter 12 Heron’s Formula

Question 1.
Each side of an equilateral triangle measures 8 cm. Find :
(i) area of the triangle
(ii) height of the triangle.
Solution:
Each side of an equilateral triangle (a) = 8 cm, we know that
(i) Area of the equilateral triangle = $$\frac{\sqrt{3}}{4} a^2$$
$$\frac{\sqrt{3}}{4} \times(8)^2$$
= 16$$\sqrt{3}$$ cm2

(ii) Again, area of the triangle = $$\frac{1}{2}$$ × base × height
⇒ 16$$\sqrt{3}$$ = $$\frac{1}{2}$$ × 8 × height
⇒ 16$$\sqrt{3}$$ = 4 × height
⇒ height = $$\frac{16 \sqrt{3}}{4}$$
⇒ height = 4$$\sqrt{3}$$ cm
Hence, (i) area of the triangle = 16$$\sqrt{3}$$ cm2, (ii) height = 4$$\sqrt{3}$$ cm.

Question 2.
If perimeter of an equilateral triangle is equal to its area. Find the area of the equilateral triangle.
Solution:
Let a be the side of an equilateral triangle, then
perimeter of triangle = 3a …..(i)
and area of the triangle = $$\frac{\sqrt{3}}{4} a^2$$ …..(2)
According to question,
Perimeter of triangle = Area of triangle
⇒ 3a = $$\frac{\sqrt{3}}{4} a^2$$ [using (i) and (ii)]
$$\frac{3 \times 4}{\sqrt{3}}=\frac{a^2}{a}$$
4$$\sqrt{3}$$ = a [∵ 3 = $$\sqrt{3}$$ × $$\sqrt{3}$$]
a= 4$$\sqrt{3}$$ units
Putting the value of a in (ii), we get
Area of the equilateral triangle
= $$\frac{\sqrt{3}}{4} \times(4 \sqrt{3})^2$$
= $$\frac{\sqrt{3}}{4} \times 48$$
= 12$$\sqrt{3}$$ square units
Hence,
Area of equilateral triangle = 12$$\sqrt{3}$$ square units.

Question 3.
Find the percentage increase in the area of an equilateral triangle if its each side is doubled.
Solution:
Let the side of an equilateral triangle be a units, then
area of the triangle = $$\frac{\sqrt{3}}{4} a^2$$
Side of new triangle when side is doubled = 2a units

Hence, percentage increase in area of the triangle = 300%.

Question 4.
In a quadrilateral ABCD, diagonal AC = 36 cm and the lengths of the perpendicular from B and D to AC are 18 cm and 15 cm respectively. Find the area of the quadrilateral.
Solution:
Diagonal AC = 36 cm, BE = 18 cm and DF = 15 cm.

Area of the quadrilateral ABCD = $$\frac{1}{2}$$ × diagonal × (sum of perpendiculars)
= $$\frac{1}{2}$$ × AC × (BE + DF)
= $$\frac{1}{2}$$ × 36 × (18 + 15)
= 18 × 33
= 594 cm2
Hence, area of quadrilateral ABCD = 594 cm2.

Question 5.
The lengths of two adjacent sides of a parallelogram are respectively 48 cm and 36 cm. One of its diagonal is 44 cm. Find the area of the parallelogram.
Solution:
We know that diagonal of a parallelogram divides the parallelogram into two triangles of equal areas

∴ Area of parallelogram ABCD = 2 × Area of ΔABC
Sides of ΔABC are
a = 44 cm, b = 36 cm and c = 48 cm
s = $$\frac{44+36+48}{2}$$
⇒ s = $$\frac{128}{2}$$
⇒ s = 64 cm
By Heron’s formula, we have
Area of ΔABC = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{64(64-44)(64-36)(64-48)}$$
= $$\sqrt{64 \times 20 \times 28 \times 16}$$
= $$\sqrt{2 \times 2 \times 16 \times 2 \times 2 \times 5 \times 2 \times 2 \times 7 \times 16}$$
= 2 × 2 × 2 × 16$$\sqrt{35}$$
= 128$$\sqrt{35}$$
area of parallelogram ABCD = 2 × Area of ΔABC
= 2 × 128$$\sqrt{35}$$
= 256$$\sqrt{35}$$ cm2
Hence,area of parallelogram
ABCD = 256$$\sqrt{35}$$ cm2.

Question 6.
Find the area of a parallelogram whose one diagonal is 7.2 cm and the perpendicular distance of this diagonal from an opposite vertex is 3.4 cm.
Solution:
Since we know that diagonal of a parallelogram divides it into two triangles of equal area.
∴ Area of parallelogram ABCD = 2 × Area of ΔACD
Area of ΔACD = $$\frac{1}{2}$$ × base × height

= $$\frac{1}{2}$$ × AC × DM
= $$\frac{1}{2}$$ × 7.2 × 3.4
= 12.24 cm2
Area of parallelogram ABCD = 2 × Area of ΔACD
= 2 × 12.24
= 24.48 cm.
Hence,area of parallelogram ABCD = 24.48 cm2.

Question 7.
The area of a trapezium field is 780 m2. The perpendicular distance between the two parallel sides is 24 m. If one parallel side exceeds the other by 15 m, find the lengths of the parallel sides.
Solution:
Let one parallel side be x m, then
other parallel side = (x – 15) m
Area of trapezium field = $$\frac{1}{2}$$ × (sum of parallel sides) × distance between them
⇒ 780 = $$\frac{1}{2}$$(x + x – 15) × 24
⇒ 780 = (2x – 15) × 12
⇒ $$\frac{780}{12}$$ = 2x – 15
⇒ 65 = 2x – 15
⇒ 65 + 15 = 2x
⇒ 80 = 2x
⇒ x = $$\frac{80}{2}$$ = 40 m
Hence, lengths of parallel sides are 40 m and 40 – 15 = 25 m.

Question 8.
The cross-section of a canal is trapezium in shape. If the canal 18 m wide at the top, 12m wide at the bottom and the area of the cross-section is 135 m2, find its depth.
Solution:
Cross-section of a canal is the shape of a trapezium. Its parallel sides are 18 m and 12 m.
Area of cross-section = 135 m2
Let depth of cross-section be x m.
∴ Area of cross-section = $$\frac{1}{2}$$(sum of parallel sides) × distance between them
⇒ 135 = $$\frac{1}{2}$$(18 + 12) × x
⇒ 135 × 2 = 30 × x
⇒ $$\frac{135 \times 2}{30}$$ = x
⇒ 9 = x
⇒ x = 9 m
Hence, depth of cross section is 9 m.

Question 1.
The lengths of the sides of a triangle are 8 cm, 12 cm and 16 cm. Find the height corresponding to the longest side.
Solution:
Length of the sides of a triangle are a = 8 cm, b = 12 cm and c = 16 cm
s = $$\frac{a+b+c}{2}$$
s = $$\frac{8+12+16}{2}$$
s = $$\frac{36}{2}$$ = 18 cm
By Heron’s formula, we have
Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{18(18-8)(18-12)(18-16)}$$
= $$\sqrt{18 \times 10 \times 6 \times 2}$$
= $$\sqrt{3 \times 3 \times 2 \times 2 \times 5 \times 3 \times 2 \times 2}$$
= 2 × 2 × 3$$\sqrt{15}$$
= 12$$\sqrt{15}$$ cm2
Again,
area of the triangle = $$\frac{1}{2}$$ × base × height
12$$\sqrt{15}$$ = $$\frac{1}{2}$$ × 16 × height
[∵ Base = longest side of the triangle]
12$$\sqrt{15}$$ = 8 × height
⇒ height = $$\frac{12 \sqrt{15}}{8}$$
⇒ height = $$\frac{3 \sqrt{15}}{2}$$ cm
Hence, area of the triangle = 12$$\sqrt{15}$$ cm2
and height = $$\frac{3 \sqrt{15}}{2}$$ cm.

Question 2.
The height of an equilateral triangle measures 12 cm. Find the area of the triangle correct to 2 places of decimal. [Use $$\sqrt{3}$$ = 1.732]
Solution:
Let each side of an equilateral triangle be a cm.
Height of the triangle = 12 cm (given)
We know that,
Area of the equilateral triangle = $$\frac{\sqrt{3}}{4} a^2$$ …….(i)
Again, area of the triangle = $$\frac{1}{2}$$ × base × height
⇒ area of the triangle = $$\frac{1}{2}$$ × a × 12
⇒ area of the triangle = 6a
⇒ $$\frac{\sqrt{3}}{4} a^2=6 a$$ [using (i)]
⇒ $$\frac{a^2}{a}=\frac{6 \times 4}{\sqrt{3}}$$
⇒ a = 8$$\sqrt{3}$$ cm
Putting the value of a in (i), we get
area of the equilateral triangle = $$\frac{\sqrt{3}}{4} \times(8 \sqrt{3})^2$$
= $$\frac{\sqrt{3}}{4} \times 192$$
= 48$$\sqrt{3}$$
= 48 × 1.732
= 83.14 cm2.
Hence, area of the equilateral triangle = 83.14 cm2.

Question 3.
Base of an isosceles triangle is $$\frac{3}{2}$$ times each of the equal sides. If its perimeter is 28 cm, find the area and height of the triangle.
Solution:
Let each equal sides of an isosceles triangle be x cm, then according to question,
Base of isosceles triangle = $$\frac{3}{2}$$x
Perimeter of triangle = 28 cm (given)
⇒ x + x + $$\frac{3}{2}$$x = 28
⇒ $$\frac{2 x+2 x+3 x}{2}$$ = 28
⇒ 7x = 28 × 2
⇒ x = $$\frac{28 \times 2}{7}$$
⇒ x = 8 cm
∴ Equal sides are 8 cm and 8 cm and
base = $$\frac{3}{2}$$ × 8 = 12 cm
Here a = 8 cm, b = 8 cm and c = 12 cm
s = $$\frac{\text { Perimeter }}{2}=\frac{28}{2}$$
= 14 cm
By Heron’s formula, we have
Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
=$$\sqrt{14(14-8)(14-8)(14-12)}$$
= $$\sqrt{14 \times 6 \times 6 \times 2}$$
= $$\sqrt{2 \times 7 \times 6 \times 6 \times 2}$$
= 2 × 6$$\sqrt{7}$$
= 12$$\sqrt{7}$$ cm2
Again, Area of triangle = $$\frac{1}{2}$$ × base × height
12$$\sqrt{7}$$ = $$\frac{1}{2}$$ × 12 × height
12$$\sqrt{7}$$ = 6 × height height
height = $$\frac{12 \sqrt{7}}{6}$$
height = 2$$\sqrt{7}$$ cm
Hence,area of the triangle = 12$$\sqrt{7}$$ cm2
and height = 2$$\sqrt{7}$$ cm.

Question 4.
In the right triangle, the sides containing the right angle, one side exceeds the other by 4 cm. If area of the triangle is 96 cm2, find the perimeter of the triangle.
Solution:
Let the sides containing right angle be x cm and (x + 4) cm.

Then
Area of triangle = $$\frac{1}{2}$$ × base × height
⇒ 96 = $$\frac{1}{2}$$ × (x + 4) × x
⇒ 96 × 2 = x2 + 4x
⇒ 192 = x2 + 4x
⇒ x2 + 4x – 192 = 0
⇒ x2 + (16 – 12) – 192 = 0
⇒ x2 + 16x – 12x – 192 = 0
⇒ x(x + 16) – 12(x + 16) = 0
⇒ (x + 16) (x – 12) = 0
⇒ x + 16 = 0 or x – 12 = 0
⇒ x = – 16 or x = 12
Since side of a right triangle cannot be negative.
So, we neglect x = – 16 cm
∴ x = 12 cm
⇒ AB = 12 cm
and BC = 12 + 4 = 16 cm
In right triangle ABC, we have
AC2 = BC2 + AB2
(By Pythagoras theorem)
⇒ AC2 = 162 + 122
⇒ AC2 = 256 + 144
⇒ AC2 = 400
⇒ AC = $$\sqrt{400}$$
⇒ AC = + 20 cm
[Neglect x = -20]
⇒ AC = 20 cm
∴ Perimeter of the triangle
ABC = AB + BC + AC
= 12 + 16 + 20
= 48 cm
Hence,perimeter of the triangle = 48 cm.

Question 5.
The area of a right triangle is 150 cm2. If its altitude exceeds the base by 5 cm, calculate the perimeter of the triangle.
Solution:
Let base (BC) be x cm.
Altitude (AB) = (x + 5) cm

Area of the triangle = $$\frac{1}{2}$$ × base × height
⇒ 150 = $$\frac{1}{2}$$ × x × (x + 5)
⇒ 150 × 2 = x2 + 5x
⇒ 300 = x2 + 5x
⇒ x2 + 5x – 300 = 0
⇒ x2 + (20 – 15)x – 300 = 0
⇒ x2 + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
⇒ (x + 20) (x – 15) = 0
⇒ x + 20 = 0 or x – 15 = 0
⇒ x = – 20 or x = 15
Since, side of the right triangle cannot be negative.
So, we neglect x = – 20
⇒ x = 15 cm
⇒ BC = 15 cm
and AB = 15 + 5 = 20 cm
In right triangle ABC, we have
⇒ AC2 = BC2 + AB2
⇒ AC2 = 152 + 202
⇒ AC2 = 225 + 400
⇒ AC2 = 625
⇒ AC = $$\sqrt{625}$$
⇒ AC = ± 25
So,
Perimeter of the triangle = AB + BC + AC
= 20 + 15 + 25
= 60 cm
Hence,perimeter of the triangle = 60 cm.

Question 6.
A triangle and a rhombus have the same base and the same area. If the sides of the triangle are 30 cm, 32 cm and 34 cm and rhombus stands on the base 32 cm, find the height of the rhombus.
Solution:
Sides of the triangle are a = 30 cm, b = 32 cm and c = 34 cm
s = $$\frac{a+b+c}{2}$$
⇒ s = $$\frac{30+32+34}{2}$$
⇒ s = $$\frac{96}{2}$$
⇒ s = 48 cm
By Heron’s formula, we have
Area of the triangler = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{48(48-30)(48-32)(48-34)}$$
= $$\sqrt{48 \times 18 \times 16 \times 14}$$
= $$\sqrt{4 \times 4 \times 3 \times 3 \times 3 \times 2 \times 4 \times 4 \times 2 \times 7}$$
= 2 × 3 × 4 × 4$$\sqrt{21}$$
= 96$$\sqrt{21}$$ cm2
Since, rhombus has base 32 cm and rhombus and triangle have same base and same area.
∴ Area of rhombus = area of triangle
⇒ base × height = 96$$\sqrt{21}$$
⇒ 32 × height = 96$$\sqrt{21}$$
⇒ height = 96$$\frac{96 \sqrt{21}}{32}$$
⇒ height = 3$$\sqrt{21}$$ cm
Hence,
height of the rhombus = 3$$\sqrt{21}$$ cm.

Question 7.
In the given figure, sides of a ΔABC are 13 cm, 15 cm and 14 cm. PBC is a right angled triangle right angled at Pin which BP =9 cm. Find the area of shaded region.

Solution:
Sides of the ΔABC are
a = 13 cm, b = 15 cm and c = 14 cm
∴ s = $$\frac{1}{2}$$
⇒ s = $$\frac{42}{2}$$ = 21 cm
By Heron’s formula, we have
Area of the ΔABC = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{21(21-13)(21-15)(21-14)}$$
= $$\sqrt{21 \times 8 \times 6 \times 7}$$
= $$\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7}$$
= 2 × 2 × 3 × 7
= 84 cm2
In right ΔPBC, we have
BC2 = BP2 + CP2
⇒ 152 = 92 + CP2
⇒ 152 – 92 = CP2
⇒ (15 + 9) (15 – 9) = CP2
⇒ 24 × 6 = CP2
⇒ CP = $$\sqrt{24 \times 6}$$
⇒ CP = $$\sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 2}$$
⇒ CP = 2 × 2 × 3 = 12 cm
Area of right ΔPBC = $$\frac{1}{2}$$ × BP × PC
= $$\frac{1}{2}$$ × 9 × 12 = 54 cm2
Area of the shaded region = Area of ΔABC – Area of ΔPBC
= 84 – 54 = 30 cm2
∴ Area of the shaded region = 30 cm2.

Question 8.
Calculate the area of quadrilateral ABCD, in which ∠ABD = 90°,BCD is an equilateral triangle of side 20 cm and AD = 25 cm.
Solution:
BCD is an equilateral triangle of side 20 cm.

Area of equilateral ABCD = $$\frac{\sqrt{3}}{4} \times(20)^2$$
= 100$$\sqrt{3}$$ = 100 × 1.732
= 173.2 cm2
In right ΔABD, we have
(By Pythagoras theorem)
⇒ 252 = AB2 + 202
⇒ 252 – 202 = AB2
⇒ 625 – 400 = AB2
⇒ 225 = AB2
⇒ AB = 225
⇒ AB = 15 cm
Area of right ΔABD = $$\frac{1}{2}$$ × base × height
= $$\frac{1}{2}$$ × AB × BD
= $$\frac{1}{2}$$ × 15 × 20
= 150 cm2
Area of quadrilateral ABCD = 173.2 + 150
= 323.2 cm2
ABCD = 323.2.cm2.

Question 9.
Calculate the area of quadrilateral ABCD in which ∠A = 90°, AB = 36 cm, BC = 42 cm, CD = 39 cm and AD = 27 cm.
Solution:
In right ΔABD, we have
⇒ BD2 = 362 + 272
⇒ BD2 = 1296 + 729
⇒ BD2 = 2025
⇒ BD = $$\sqrt{2025}$$
⇒ BD = 45 cm

Area of right ΔABD = $$\frac{1}{2}$$ × base × height
= $$\frac{1}{2}$$ × AB × AD
= $$\frac{1}{2}$$ × 36 × 27
= 486 cm2
Sides of the ΔBCD are
a = 39 cm, b = 42 cm and c = 45 cm
s = $$\frac{a+b+c}{2}$$
= $$\frac{39+42+45}{2}$$
= $$\frac{126}{2}$$ = 63 cm
By Heron’s formula, we have.
Area of ΔBCD = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{63(63-39)(63-42)(63-45)}$$
= $$\sqrt{63 \times 24 \times 21 \times 18}$$
= $$\sqrt{3 \times 3 \times 7 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 2 \times 3 \times 3}$$
= 2 × 2 × 3 × 3 × 3 × 7
= 756 cm2
Area of quadrilateral ABCD = 486 + 756 = 1242 cm2
ABCD = 1242 cm2.

Question 10.
Two diagonals of a parallelogram are 12 cm and 16 cm. If its one side is 10 cm, find the area of a parallelogram.
Solution:
Let AC and BD are two diagonals of a parallelogram ABCD. We know that diagonals of a parallelogram bisect each other.

∴ AO = OC = $$\frac{16}{2}$$ = 8 cm
BO = OD = $$\frac{12}{2}$$ = 6 cm
Thus the sides of ΔAOB are
a = 8 cm, b = 6 cm, c = 10 cm
∴ s = $$\frac{a+b+c}{2}$$
= $$\frac{8+6+10}{2}$$
= $$\frac{24}{2}$$ = 12 cm
By Heron’s formula, we have
Area of ΔAOB = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{12(12-8)(12-6)(12-10)}$$
= $$\sqrt{12 \times 4 \times 6 \times 2}$$
= $$\sqrt{2 \times 2 \times 3 \times 2 \times 2 \times 2 \times 3 \times 2}$$
= 2 × 2 × 2 × 3
= 24 cm2
Since we know that diagonals of a parallelogram divide it into four triangles of equal areas.
∴ area of parallelogram ABCD = 4 × area of ΔAOB
= 4 × 24
= 96 cm2
Hence,area of parallelogram
ABCD = 96 cm2.

Question 11.
In the figure, AOBC is rhombus, three of whose vertices lie on the circle with centre O. If the area of the rhombus is 50$$\sqrt{3}$$ cm2, find the radius of the circle.

Solution:
Let the radius of the circle be x cm.
AO = OC = OB = x cm.
Since, we know that, in rhombus diagonals bisect each other at right angle.
∴ OD = CD = $$\frac{x}{2}$$ and AD = BD = $$\frac{A B}{2}$$
(By Pythagoras theorem)

Thus lengths of diagonals of a rhombus are
AB = $$\sqrt{3}$$x and OC = x
Area of rhombus = $$\frac{1}{2}$$ × Product of diagonals
⇒ 50$$\sqrt{3}$$ = $$\frac{1}{2}$$ × $$\sqrt{3}$$x × x
⇒ $$\frac{50 \sqrt{3} \times 2}{\sqrt{3}}$$ = x2
⇒ 100 = x2
⇒ x = $$\sqrt{100}$$ = 10 cm
Hence, radius of the circle is 10 cm.

Question 12.
AOBC is a rhombus, three of whose vertices lie on the circle with centre O. If the radius of the cricle is 12 cm, find the area of the rhombus.

Solution:
Join AB and OC.
We know that in a rhombus diagonals bisect each other at right angle.
∴ OD = CD = $$\frac{12}{2}$$ = 6 cm
AD = BD = $$\frac{1}{2}$$AB
(By Pythagoras theorem)
⇒ 122 = 62 + AD2
⇒ 122 – 62 = AD2
⇒ (12 + 6) (12 – 6) = AD2
⇒ 18 × 6 = AD2
⇒ AD = $$\sqrt{2 \times 3 \times 3 \times 2 \times 3}$$
⇒ AD = 2 × 3$$\sqrt{3}$$
⇒ AD = 6$$\sqrt{3}$$ cm
⇒ AB = 2 × AD
AB = 2 × 6$$\sqrt{3}$$
AB = 12$$\sqrt{3}$$ cm
Thus, diagonals of a rhombus are
OC = 12 cm
and AB = 12$$\sqrt{3}$$ cm
Area of the rhombus = $$\frac{1}{2}$$ × Product of diagonals
= $$\frac{1}{2}$$ × 12 × 12$$\sqrt{3}$$
= 72$$\sqrt{3}$$ cm2
Hence,area of the rhombus = 72$$\sqrt{3}$$ cm2.

Question 1.
If area of an isosceles triangle is 60 c2 and its each equal side is 13 cm, find :
(i) Base of the triangle
(ii) Height of the triangle.

Solution:
Let base of an isosceles triangle be 2x cm.
Here,
a= 13 cm, b = 13 cm and c = 20 cm

⇒ 602 = (13 + x)(13 – x)x2
[Squaring on both sides]
⇒ 3600 = (132 – x2)x2
[∵ (a + b)(a – b) = a2 – b2]
⇒ 3600 = (169 – x2)x2
⇒ 3600 = 169x2 – x4
⇒ x4 – 169x2 + 3600 = 0
Let x2 = y, we get
y2 – 169y + 3600 = 0
⇒ y2 – (144 + 25)y + 3600 = 0
⇒ y2 – 144y – 25y + 3600 = 0
⇒ y(y – 144) – 25(y – 144) = 0
⇒ (y – 144) (y – 25) = 0
⇒ y – 144 = 0 or y – 25 = 0
⇒ y = 144 or y = 25
⇒ x2 = 144 or x2 = 25 [∵ y = x2]
⇒ x = $$\sqrt{144}$$ or x = $$\sqrt{25}$$
[Taking square root on both sides]
⇒ x = ± 12 or x = ± 5.
Since base of an isosceles triangle cannot be negative.
∴ We neglect negative values of x.
∴ x = 12 or x = 5
⇒ BC = 2 × 12 or BC = 2 × 5
⇒ BC = 24 cm or BC = 10 cm
Since AD ⊥ BC and ABC is an isosceles triangle.
∴ BD = CD = x
There are two cases arise :
Case I: When x = 12,
In right triangle ABD, we have
[By Pythagoras theorem]
⇒ 132 = 122 + AD2
⇒ 132 – 122 = AD2
⇒ 169 – 144 = AD2
⇒ AD = $$\sqrt{25}$$ = 5 cm

Case II: When x = 5,
In right triangle ABD, we have
⇒ 132 = 52 + AD2
⇒ 132 – 52 = AD2
⇒ (13 + 5) (13 – 5) = AD2
⇒ 18 × 8 = AD2
$$\sqrt{18 \times 8}$$ = AD
$$\sqrt{3 \times 3 \times 2 \times 2 \times 2 \times 2}$$ = AD
3 × 2 × 2 = AD
Hence, (i) Base of the isosceles triangle = 24 cm or 10 cm.
(ii) Height of the isosceles triangle = 5 cm or 12 cm.

Question 2.
The perimeter of a right triangle is 40 cm and its hypotenuse is 17 cm. Calculate its area and verify the result by using Heron’s formula.
Solution:
Let base of a right traingle be x cm.

Perimeter of triangle = 40 cm (given)
⇒ AB + BC + AC = 40
⇒ AB + x + 17 = 40
⇒ AB = 40 – 17 – x
⇒ AB = (23 – x) cm
In right triangle ABC, we have
AC2 = BC2 + AB2
(By Pythagoras theorem)
⇒ 172 = x2 + (23 – x)2
⇒ 289 = 22 + 529 + x2 – 46x
⇒ 289 = 2x2 – 46x + 529
⇒ 2x2 – 46x + 529 – 289 = 0
⇒ 2x2 – 46x + 240 = 0
⇒ x2 – 23x + 120 = 0
⇒ x2 – (15 + 8)x + 120 = 0
⇒ x2 – 15x – 8x + 120 = 0
⇒ x(x – 15) – 8(x – 15) = 0
⇒ (x – 15) (x – 8) = 0
⇒ x – 15 = 0 or x – 8 = 0
x = 15 or x = 8
∴ BC = 15 cm or BC = 8 cm
AB = 23 – 15 cm or AB = 23 – 8
⇒ AB = 8 cm or AB = 15 cm
Area of triangle = $$\frac{1}{2}$$ × base × height
= $$\frac{1}{2}$$ × BC × AB
= $$\frac{1}{2}$$ × 15 × 8
= 60 cm2
Here a = 17, b = 15 and c = 8
and Perimeter of the triangle (2s) = 40 cm
⇒ s = $$\frac{40}{2}$$ = 20
By Heron’s formula
Again, Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{20(20-17)(20-15)(20-8)}$$
= $$\sqrt{20 \times 3 \times 5 \times 12}$$
= $$\sqrt{4 \times 5 \times 3 \times 5 \times 3 \times 4}$$
= 4 × 3 × 5 = 60 cm2
Hence, in both cases area of the triangle is same.
Verified

Question 3.
In an equilateral ΔABC, O is the point of concurrence of altitudes AD, BE and CF such that OD = 8 cm, OE = 10 cm and OF = 12 cm. Find the area of the triangle ABC.
Solution:
Let side of an equilateral ΔABC be x cm, then

Area of ΔABC = $$\frac{\sqrt{3}}{4} x^2$$ …(i)
Again, Area of ΔABC = Area of ΔBOC + Area of ΔAOC + Area of ΔAOB

Putting the value of x in (i), we get
Area of ΔABC = $$\frac{\sqrt{3}}{4} \times(20 \sqrt{3})^2$$
= $$\frac{\sqrt{3}}{4}$$ × 1200
= 300$$\sqrt{3}$$
= 300 × 1.732
= 519.6 cm2
Hence,
Area of ΔABC = 519.6 cm2.

Question 4.
A field is in the shape of a trapezium whose parallel sides are 32 m and 20 m and non-parallel sides are 10 m and 14 m. Find the area of the field.
Solution:
Through vertex C, draw CE || AD intersecting AB at E and draw CF ⊥ EB.

∵ AE || CD and AD || CE.
∴ AECD is a paralleogram.
CE = AD = 10 m,
AE = CD = 20 m
Then BE = AB – AE
⇒ BE = 32 – 20 = 12 m
Sides of ΔCEB are a = 10 m, b = 14 m and c = 12 m
∴ s = $$\frac{a+b+c}{2}$$
⇒ s = $$\frac{10+14+12}{2}$$
⇒ s = 18 m
By Heron’s formula, we have

Again, area of ΔCEB = $$\frac{1}{2}$$ × base × height
⇒ 24$$\sqrt{6}$$ = $$\frac{1}{2}$$ × 12 × CF
⇒ $$\frac{24 \sqrt{6} \times 2}{12}$$ = CF
⇒ 4$$\sqrt{6}$$ = CF
⇒ CF = 456 m
∴ Area of trapezium ABCD = $$\frac{1}{2}$$(sum of parallel sides) × distance between them
= $$\frac{1}{2}$$(32 + 20) × 4$$\sqrt{6}$$ m
= 104$$\sqrt{6}$$ m2.
Hence, area of trapezium = 104$$\sqrt{6}$$ m2

Question 5.
The area of a rhombus is 216 cm2. If its one diagonal is 24 cm, find :
(i) length of its other diagonal
(ii) perimeter of the rhombus.
Solution:
(i) Area of a rhombus = 216 cm2
Length of its one diagonal = 24 cm
Let the length of other diagonal be x cm.

Area of rhombus = $$\frac{1}{2}$$ × Product of diagonals
216 = $$\frac{1}{2}$$ × 24 × x
216 = 12 × x
x = $$\frac{216}{12}$$ = 18 cm.

(ii) We know that in a rhombus diagonals bisect each other at right angle.
∴ ∠AOB = 90°
and AO = OC = $$\frac{18}{12}$$ = 9 cm
OB = OD = $$\frac{24}{2}$$ = 12 cm

In right ΔAOB, we have
AB2 = AO2 + OB2
(By Pythagoras theorem)
⇒ AB2 = 92 + 122
⇒ AB2 = 81 + 144
⇒ AB2 = 225
⇒ AB = $$\sqrt{225}$$
⇒ AB = 15 cm
∴ Perimeter of the rhombus = 4 × side
= 4 × 15
= 60 cm
Hence (i) Length of other diagonal = 18 cm.
(ii) Perimeter of the rhombus = 60 cm.

Question 6.
The perimeter of a rhombus is 100 cm and one of its shorter diagonal is 30 cm long. Find the length of other diagonal and area of rhombus.
Solution:
Perimeter of a rhombus = 100 cm
Side of the rhombus = $$\frac{100}{4}$$ = 25 cm
Length of its one diagonal = 30 cm
Let length of its other diagonal = 2x cm

We know that in a rhombus diagonals bisect each other at right angle.
∴ ∠AOB = 90°
and AO = OC = $$\frac{2 x}{2}$$ = x cm
BO = OD = $$\frac{30}{2}$$ = 15 cm
In right ΔAOB, we have
AB2 = AO2 + OB2
(By Pythagoras theorem)
⇒ 252 = x2 + 152
⇒ x2 = 252 – 152
⇒ x2 = (25 + 15) (25 – 15)
⇒ x2 = 40 × 10
⇒ x2 = 400
⇒ x = $$\sqrt{400}$$ = 20 cm
So,the length of other diagonal = 2 × 20 × 40 cm
Area of the rhombus = $$\frac{1}{2}$$ × Product of diagonals
= $$\frac{1}{2}$$ × 30 × 40 = 600 cm2
Hence, length of other diagonal = 40 cm
and area of the rhombus = 600 cm2.

Question 7.
ABCD is a square with each side 16 cm. P is a point on DC such that, area of ΔAPD : area of trapezium ABCP = 7 : 25. Find the length of CP.
Solution:
Each side of the square = 16 cm
Let, CP = x cm, then
PD = (16 – x) cm

Area of right ΔAPD = $$\frac{1}{2}$$PD × AD
= $$\frac{1}{2}$$ × (16 – x) × 16
= 8(16 – x)
Area of the trapezium ABCP = 2 (sum of parallel sides) × distance between them
= $$\frac{1}{2}$$(16 + x) × 16 = 8(16 +x)
According to question,
Area of ΔAPD : Area of trapezium ABCP = 7 : 25
⇒ 8(16 – x) : 8(16 + x) = 7 : 25
⇒ $$\frac{8(16-x)}{8(16+x)}=\frac{7}{25}$$
⇒ $$\frac{16-x}{16+x}=\frac{7}{25}$$
⇒ 25(16 – x) = 7(16 + x)
⇒ 400 – 25x = 112 + 7x
⇒ 400 – 112 = 25x + 7x
⇒ 288 = 32x
⇒ x = $$\frac{288}{32}$$
⇒ x = 9 cm
Hence, length of CP = 9 cm.

Question 8.
An equilateral triangle is circumscribed and a square is inscribed in a circle of radius 7 cm. Prove that ratio of areas of square : equilateral triangle is 2 : 3$$\sqrt{3}$$.
Solution:
Let ΔABC be circumscribed and square PQRS be inscribed in a circle of radius 7 cm.

⇒ QS = 2 × 7
= 14 cm
Area of a square = $$\frac{1}{2}$$ × (diagonal)2
= $$\frac{1}{2}$$ × 142
= 98 cm2
We know that in equilateral triangle orthocentre is same as centroid.
∴ AO : OD = 2 : 1
⇒ $$\frac{A O}{O D}=\frac{2}{1}$$
⇒ AO = 2 × OD
⇒ AO = 2 × 7 = 14 cm
⇒ AD = AO + OD
⇒ AD = 14 + 7 = 21 cm
Let side of an equilateral triangle be x cm. We know that, in equilateral triangle perpendicular bisects the corresponding sides.
∴ BD = CD = $$\frac{x}{2}$$ cm
In right triangle ADB, we have
(By Pythagoras theorem)

Hence, area of the square : area of equilateral triangle = 2 : 3$$\sqrt{3}$$. Proved

Question 9.
In the given figure, ABCD is square with AC = 12 cm. A point Pon AB such that PQ ⊥ AC, PR ⊥ BD and Q, R are respectively mid points of AO and OB. Find the length of QR.

Solution:
We know that, in a square, diagonals are equal and bisect each other at 90°.
AO = OB = $$\frac{1}{2}$$ = 6 cm and
∠AOB = 90°
∠PQO = 90°, ∠PRO = 90°
[∵ PQ ⊥ AC, PR ⊥ BD]
In a quadrilateral PQOR, we have
∠PQO + ∠QOR + ∠PRO + ∠QPR = 360°
⇒ 90° + 90° + 90° + ZQPR = 360°
⇒ 270° + ∠QPR = 360°
⇒ ∠QPR = 360° – 270°
⇒ ∠QPR = 90°
Since, Q and R are mid points of AO and OB respectively.
∴ QO = OR = $$\frac{6}{2}$$ = 3 cm
Thus, in a quadrilateral PQOR, each angle is 90° and adjacent sides are equal.
∴ PQOR is a square of side 3 cm.
Area of the square PQOR = 3 × 3 = 9 cm2
Again, area of the square PQOR
= $$\frac{1}{2}$$ (diagonal)2
⇒ 9 = $$\frac{1}{2}$$QR2
⇒ 9 × 2 = QR2
QR = $$\sqrt{18}$$
QR = $$\sqrt{3 \times 3 \times 2}$$
QR = 3$$\sqrt{2}$$ cm
Hence, length of QR = 3$$\sqrt{2}$$ cm.

Question 10.
In the given figure, square PQRS is inside the square ABCD such that each side of PQRS can be extended to pass through the vertex of ABCD. Length of side of square ABCD is $$\sqrt{73}$$ cm. P and Q are points on AQ and BR so that AP = BQ = 3 cm. Find the length of PR.

Solution:
∠AQB + ∠AQR = 180°
(By linear pair axiom)
⇒ ∠AQB + 90° = 180°
⇒ ∠AQB = 180° – 90°
⇒ ∠AQB = 90°
In right ΔAQB, we have
AB2 = AQ2 + BQ2
⇒ ($$\sqrt{73}$$)2 = AQ2 + 32
⇒ 73 = AQ2 + 9
⇒ 73 – 9 = AQ2
⇒ 64 = AQ2
⇒ AQ = $$\sqrt{64}$$ = 8 cm
⇒ PQ = AQ – AP
⇒ PQ = 8 – 3 = 5 cm
Area of square PQRS = (5)2
= 25 cm2
Again,
area of square PQRS = $$\frac{1}{2}$$ × (diagonal)2
⇒ 25 = $$\frac{1}{2}$$ × PR2
⇒ 25 × 2 = PR2
⇒ PR2 = 50
⇒ PR = $$\sqrt{50}$$
⇒ PR = $$\sqrt{2 \times 5 \times 5}$$
⇒ PR = 5$$\sqrt{2}$$ cm
Hence,lenght of PR = 5$$\sqrt{2}$$ cm.

Question 11.
Triangle PQR inside the triangle ABC such that P, Q and R are the mid points of AH, BH and CH respectively and PQ = 13 cm, QR = 14 cm and PR = 15 cm. Prove that ratio of areas ΔPQR : ΔABC = 1 : 4.

Solution:
Sides of the ΔPQR are a = 13 cm, b = 14 cm, c = 15 cm
∴ s = $$\frac{a+b+c}{2}$$
⇒ s = $$\frac{13+14+15}{2}$$
⇒ s = $$\frac{42}{2}$$ = 21 cm
By Heron’s formula, we have
Area of ΔPQR = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{21(21-13)(21-14)(21-15)}$$
= $$\sqrt{21 \times 8 \times 7 \times 6}$$
= $$\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3}$$
= 2 × 2 × 3 × 7
= 84 cm2
In ΔABH, P and Q are the mid points of AH and BH respectively.
∴ PQ = $$\frac{1}{2}$$AB
(By mid point theorem)
⇒ 13 = $$\frac{1}{2}$$AB
⇒ AB = 26 cm
Similarly, in ΔBHC, Q and R are the mid points of BH and CH respectively.
∴ QR = $$\frac{1}{2}$$BC
⇒ 14 = $$\frac{1}{2}$$BC
⇒ BC = 28 cm
and in ΔAHC, P and R are the mid points of AH and CH respectively.
∴ PR = $$\frac{1}{2}$$AC
⇒ 15 = $$\frac{1}{2}$$AC
⇒ AC = 30 cm
Thus, the sides of ΔABC are a = 26 cm, b = 28 cm and c = 30 cm

Hence, area of ΔPQR : area of ΔABC = 1 : 4. Proved

Question 12.
An isosceles triangle ABC is circumscribed in a circle with centre ‘O’ such that AB = 9 cm, BC = 6 cm and AC = 9 cm. Find the area of shaded region.

Solution:
Draw OD ⊥ BC,OE ⊥ AC and OF ⊥ AB
Let radius of the circle be r сm.
Sides of the triangle ABC are a = 9 cm, b = 9 cm and c = 6 cm

Again,
area of ΔABC = area of ΔBOC + area of ΔAOC + area of ΔAOB
18$$\sqrt{2}$$ = $$\frac{1}{2}$$ BC × OD + $$\frac{1}{2}$$ AC × OE + $$\frac{1}{2}$$ AB × OF

Multiple Choice Questions

Choose the correct option in each of the following:

Question 1.
The sides of a triangle are 5 cm, 6 cm and 7 cm, then its semiperimeter is :
(a) 6 cm
(b) 7 cm
(c) 8 cm2
(d) 9 cm
(d) 9 cm

Question 2.
The base and hypotenuse of a right triangle are 8 cm and 10 cm long. Its area is :
(a) 40 cm2
(b) 24 cm2
(c) 30 cm2
(d) 35 cm2
(b) 24 cm2

Question 3.
The length of each side of an equilateral triangle having an area of 9$$\sqrt{3}$$ cm2 is:
(a) 36 cm
(b) 8 cm
(c) 6 cm
(d) 4 cm
(c) 6 cm

Question 4.
Length of median of an equilateral triangle is 3 cm. Each side of the triangle is :
(a) 2$$\sqrt{3}$$ cm
(b) 3$$\sqrt{3}$$ cm
(c) 4$$\sqrt{3}$$ cm
(d) 2$$\sqrt{6}$$ cm
(a) 2$$\sqrt{3}$$ cm

Question 5.
The perimeter of an equilateral triangle is 60 m. The area is :
(a) 10$$\sqrt{3}$$ m2
(b) 15$$\sqrt{3}$$ m2
(c) 20$$\sqrt{3}$$ m2
(d) 100$$\sqrt{3}$$ m2
(d) 100$$\sqrt{3}$$ m2

Question 6.
Area of an equilateral triangle is 16$$\sqrt{3}$$ cm2. Its perimeter is:
(a) 36 cm
(b) 30 cm
(c) 24 cm
(d) 12 cm
(c) 24 cm

Question 7.
The sides of a triangle are 13 cm, 14 cm and 15 cm. Its area is :
(a) 80 cm2
(b) 84 cm2
(c) 91 cm2
(d) 105 cm2
(b) 84 cm2

Question 8.
The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. The length of its longest altitude is :
(a) 16$$\sqrt{5}$$ cm
(b) 10$$\sqrt{5}$$ cm
(c) 24$$\sqrt{5}$$ cm
(d) 28 cm
(c) 24$$\sqrt{5}$$ cm

Question 9.
Area of an isosceles right triangle is 8 cm2. The length of its hypotenuse is :
(a) $$\sqrt{32}$$ cm
(b) $$\sqrt{16}$$ cm
(c) $$\sqrt{48}$$ cm
(d) $$\sqrt{24}$$ cm
(a) $$\sqrt{32}$$ cm
(a) $$\sqrt{3}$$ cm
(b) 2$$\sqrt{3}$$ cm
(c) 3$$\sqrt{2}$$ cm
(d) $$\sqrt{2}$$ cm
(b) 2$$\sqrt{3}$$ cm