HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.5

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 26 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.
Solution:
(i) The given sides of the triangle are 7 cm, 24 cm, 25 cm,
The longest side = 25cm.
The triangle is right angled if
(the longest side)2 = (sum of squares of other two sides) (by Pythagoras theorem)
(25)2 = (7)2 + (24)2
625 = 49 + 576
625 = 625
The given sides make a right triangle and length of hypotenuse = 25cm.

(ii) The given sides of the triangle are 3cm, 8 cm, 6 cm.
The longest side = 8 cm
(the longest side)2 = (sum of the squares of other two sides) (by Pythagoras theorem)
(8)2 = (3)2 + (6)2
64 = 9 + 36
But 64 â‰  45
The given sides do not make a right triangle.

(iii) The given sides of the triangle Eire 50 cm, 80 cm, 100 cm.
The longest side = 100 cm,
(the longest side)2 = (sum of the squares of other two sides) (by Pythagoras theorem)
(100)2 = (50)2 + (80)2
10000 = 2500 + 6400
But 10000 â‰  8900
âˆ´ The given sides do not mEike a right triangle.

(iv) The given sides of the triangle are 13 cm, 12 cm, 5 cm.
The longest side = 13 cm.
(the longest side)2 = (sum of the squares of other two sides) (by Pythagoras theorem)
(13)2 = (12)2 + (5)2
169 = 144 + 25
169 = 169
âˆ´ The given sides make a right triangle and length of hypotenuse = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM âŠ¥ QR. Show that PM2 = QM.MR.

Solution:
Given : A right âˆ†PQR in which âˆ P = 90Â° and PM âŠ¥ QR.
To Prove : PM2 = QM.MR
Proof : In right âˆ†QPR, PM âŠ¥ QR.
âˆ†RPM ~ âˆ†PQM [By theorem 6.7]
â‡’ $$\frac{\mathrm{MR}}{\mathrm{PM}}=\frac{\mathrm{PM}}{\mathrm{QM}}$$
[Corresponding sides of similar triangles sere proportional]
â‡’ PM2 = QM Ã— MR
Hence Proved.

Question 3.
In given figure, ABD is a triangle right angled at A and AC âŠ¥ BD. Show that
(i) AB2 = BC.BD
(ii) AC2 = BC.DC

Solution:
Given : A right triangle in which âˆ A = 90Â° sind AC âŠ¥ BD.
To Prove :
(i) AB2 = BC.BD
(ii) AC2 = BC.DC
Proof :
(i) In right âˆ†BAC, AC âŠ¥ BD.
âˆ†BAC ~ âˆ†BDA [By theorem 6.7]
â‡’ $$\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}$$
[Corresponding sides of similar triEingles are proportional]
â‡’ AB2 = BC.BD
Hence Proved.

(ii) Similsirly, âˆ†ABC~ âˆ†DAC [By theorem 6.7]
$$\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{AC}}$$
[Corresponding sides of similar triEingles Eire proportional]
â‡’ AC2 = BC.CD
Hence Proved.

(iii) Similarly, âˆ†DAC ~ âˆ†DBA [By theorem 6.7]
$$\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{BD}}$$
[Corresponding sides of similar triangles are proportional]
Hence Proved.

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
Given : âˆ†ABC is an isosceles triangle right angled at C.
i.e., AC = BC
To Prove : AB2 = 2AC2
Proof : In right âˆ†ACB

AB2 = BC2 + AC2 [By Pythagoras theorem]
â‡’ AB2 = AC2 + AC2 [âˆµ BC = AC]
â‡’ AB2 = 2AC2.
Hence Proved.

Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
Given: âˆ†ABC is an isosceles triangle in which AC = BC and AB2 = 2AC2
To Prove : âˆ†ABC is a right triangle.

Proof : In âˆ†ABC, AC = BC and AB2 = 2AC2
â‡’ AB2 = AC2 + AC2
â‡’ AB2 = BC2 + AC2 [v AC = BC]
â‡’ ABC is right angle triangle.
[By converse of Pythagoras theorem]
Hence Proved.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution :
Side of equilateral âˆ† = 2a.
And AD is its altitude. We know that altitude bisects its corresponding side.
i.e., BD = CD = a

AC2 = AD2 + CD2 [By Pythagoras theorem]
â‡’ (2a)2 = AD2 + a2
â‡’ 4a2 = AD2 + a2
â‡’ AD2 = 4a2 – a2 = 3a2
Hence, Altitude = âˆš3a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Given : ABCD is a rhombus

Proof: We know that diagonals of a rhombus bisect each other at right angle
âˆ´ âˆ AOB = âˆ BOC = âˆ COD = âˆ AOD = 90Â° and AO = CO, OD = BO.
In right âˆ†AOB, AB2 = OA2 + OB2 [By Pythagoras theorem]
AB2 = ($$\frac{1}{2}$$ AC)2 + ($$\frac{1}{2}$$ BD)2
[âˆµ OA = OC and OB = OD]
â‡’ AB2 = 7 AC2 + 7BD2
â‡’ 4AB2 = AC2 + BD2 ……………(1)
Similarly, we have
4BC2 = AC2 + BD2 ……………….(2)
4CD2 = AC2 + BD2 ……………….(3)
4AD2 = AC2 + BD2 ……………….(4)
Adding (1), (2), (3) and (4), we get
4(AB2 + BC2 + CD2 + AD2) = 4(AC2 + BD2)
â‡’ AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Hence Proved.

Question 8.
In given figure, O is a point in the interior of a triangle ABC, OD âŠ¥ BC, OE âŠ¥ AC and OF âŠ¥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Solution: Given: A âˆ†ABC in which OD âŠ¥BC, OE âŠ¥ AC, OF âŠ¥ AB.
To Prove :
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Construction : Join OB, OC and OA.

Proof :
(i) By Pythagoras theorem for each of the right triangles namely âˆ†OFA, âˆ†ODB, and âˆ†OEC, we get
OA2 = OF2 + AF2 …………….(1)
OB2 = OD2 + BD2 ……………(2)
OC2 = OE2 + CE2 ……………..(3)
Adding (1), (2) and (3), we get
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + CE2 = AF2 + BD2 + CE2
â‡’ OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Hence Proved.

(ii) In right triangles ODB and ODC, we have
OB2 = OD2 + BD2 …………….(4) [By Pythagoras theorem]
OC2 = OD2 + CD2 ……………..(5)
Subtracting (5) from (4), we get
OB2 – OC2 = BD2 – CD2 ………….(6)
Similarly OC2 – OA2 = CE2 – AE2 ……………(7)
and OA2 – OB2 = AF2 – BF2 ……………(8)
Adding (6), (7) and (8), we get
OB2 – OC2 + OC2 – OA2 + OA2 – OB2 = BD2 – CD2 + CE2 – AE2 + AF2 – BF2
â‡’ 0 = BD2 – CD2 + CE2 – AE2 + AF2 – BF2
â‡’ AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Hence Proved.

Question 9.
A ladder 10 m long reaches a window 8m above the ground. Find the distance of the foot of the ladder from the base of the wall.
Solution :
Let AC be ladder and AB be wall and BC be the distance of the ladder from wall.

âˆ´ AB = 8 m
AC = 10 m
In right âˆ†ABC, AC2 = BC2 + AB2
â‡’ 102 = BC2 + 82
â‡’ 100 = BC2 + 64
â‡’ BC2 = 100 – 64
â‡’ BC2 = 36
â‡’ BC = 6m.
Hence, distance of the foot of the ladder from base of the wall = 6m.

Question 10.
A guy wire attached to a vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Solution :
Let AC be guy wire and AB be vertical pole and BC be distance of the stake from the base of the pole.
âˆ´ AB = 18m, AC = 24m,
In right âˆ†ABC,
AC2 = BC2 + AB2 [By Pythagoras theorem]
â‡’ (24)2 = BC2 + (18)2
â‡’ 576 = BC2 + 324
â‡’ BC2 = 576 – 324

â‡’ BC2 = 252
â‡’ BC = âˆš252
â‡’ BC = 6âˆš7 m
Hence, distance of stake from the base of the pole = 6âˆš7 m.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1$$\frac{1}{2}$$hours?
Solution:
Let point B represents the position of airport. Then distance covered by first plane in the north direction (AB) = Speed Ã— Time
= 1000 Ã— 1$$\frac{1}{2}$$
= 1000 Ã— $$\frac{3}{2}$$
= 1500 km

and distance covered by second plane in the west direction (BC) = Speed Ã— Time
= 1200 Ã— $$\frac{1}{2}$$
= 1200 Ã— $$\frac{3}{2}$$ = 1800 km.

Now, in right âˆ†ABC,
AC2 = BC2 + AB2
[By Pythagoras theorem]
â‡’ AC2 = (1800)2 + (1500)2
â‡’ AC2 = 3240000 + 2250000
â‡’ AC2 = 5490000
â‡’ AC = $$\sqrt{5490000}$$
â‡’ AC = 300âˆš61 km
Hence, distance between two planes = 300âˆš61 km.

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
Let AB and CD be two poles, BD be distance between the feet of two poles and AC be distance between their tops.
âˆ´ AB = 11 m, BD = 12 m, CD = 6 m
âˆ B = âˆ D = 90Â°
CE || BD
âˆ AEC = 90Â°, CE = 12m, AE = 11 – 6 = 5 m
Now, in right âˆ†AEC,
âˆ´ AC2 = CE2 + AE2 [By Pythagoras theorem]

â‡’ AC2 = 122 + 52
â‡’ AC2 = 144 + 25
â‡’ AC2 = 169
â‡’ AC = âˆš169
â‡’ AC = 13 m.
Hence, distance between tops of two poles = 13 m.

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution :
Given : A triangle ABC in which âˆ C = 90Â°, D and E are points on the sides CA and CB respectively.
To Prove : AE2 + BD2 = AB2 + DE2
Proof: In right âˆ†ACE,
AE2 = AC2 + CE2 ………………(1)
[By Pythagoras theorem]
In right âˆ†DCB,
BD2 = BC2 + DC2 ………………(2)
In right âˆ†ABC,
AB2 = BC2 + AC2 ……………….(3)
In right âˆ†DCE,
DE2 = EC2 + CD2 ………………..(4)
Adding (1) and (2), we get

AE2 + BD2 = AC2 + BC2 + CE2 + CD2
AE2 + BD2 = AB2 + DE2 [Using (3) and (4)]
Hence Proved.

Question 14.
The perpendicular from A on side BC of a âˆ†ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB2 = 2AC2 + BC2.

Solution :
Given : A âˆ†ABC such that AD âŠ¥ BC and DB = 3CD
To Prove: 2AB2 = 2AC2 + BC2
Proof: DB = 3CD (given)
Now BC = DB + CD
â‡’ BC = 3CD + CD [. DB = 3CD]
â‡’ BC = 4CD
â‡’ CD = $$\frac{1}{4}$$BC
â‡’ and BD = 3CD = $$\frac{3}{4}$$BC
AB2 = BD2 + AD2 ………………..(1)
AC2 = CD2 + AD2 …………………(2)
Subtracting (2) from (1), we get
â‡’ AB2 – AC2 = ($$\frac{3}{4}$$BC)2 – ($$\frac{1}{4}$$BC)2
[âˆµ BD = $$\frac{3}{4}$$BC and CD = $$\frac{1}{4}$$BC]
â‡’ AB2 – AC2 = $$\frac{9}{16}$$ BC2 – $$\frac{9}{16}$$ BC2
â‡’ AB2 – AC2 = $$\frac{8}{16}$$BC2
â‡’ AB2 – AC2 = $$\frac{1}{2}$$BC2
â‡’ 2AB2 – 2AC2 = BC2
â‡’ 2AB2 = 2AC2 + BC2
Hence Proved.

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = $$\frac{1}{3}$$BC. Prove that 9AD2 = 7AB2.
Solution :
Given : An equilateral âˆ†ABC such that BD = $$\frac{1}{3}$$BC.
Construction : Draw AE âŠ¥ BC and Join AD.
Proof: Let each side of equilateral triangle be x units.

BD = $$\frac{1}{3}$$BC
â‡’ BD = $$\frac{x}{3}$$
We know that in equi1ateral perpendicular bisects its corresponding opposite side
BE = EC = $$\frac{x}{2}$$
DE = BE – BD
DE = $$\frac{x}{2}$$ – $$\frac{x}{3}$$
DE = $$\frac{x}{6}$$
In right âˆ†AED, AD2 = AE2 + DE2 ……………..(1)
[By Pythagoras theorem]
In right âˆ†AEB, AB2 = BE2 + AE2
[By Pythagoras theorem]
AE2 = AB2 – BE2
Put AE2 = AB2 – BE2 in (1), we get
AD2 = 2 BE2 + DE2
AD2 = x2 – ($$\frac{x}{2}$$)2 + ($$\frac{x}{6}$$)2
â‡’ AD2 = x2 – $$\frac{x^2}{4}+\frac{x^2}{36}$$
â‡’ AD2 = $$\frac{36 x^2-9 x^2+x^2}{36}$$
â‡’ AD2 = $$\frac{28 x^2}{36}$$
â‡’ AD2 = $$\frac{7}{9}$$ x2
â‡’ AD2 = $$\frac{7}{9}$$ AB2
Hence Proved.

Question 16.
In an equilateral triangle prove that three times the square of one side is equal to four times the square of one of its aftitudes.
Solution :
Given : A triangle ABC in which AB = BC = CA and AD âŠ¥ BC.
AB = AC (Given)

â‡’ BD = CD (CPCT)
â‡’ AB2 = AD2 + ($$\frac{1}{2}$$AB)2
[âˆµ BD = CD
âˆ´ BD = $$\frac{1}{2}$$AB]
â‡’ AB2 = 4AD2 + $$\frac{1}{4}$$AB2
â‡’ 4AB2 = 4AD2 + AB2
Hence Proved.

Question 17.
Tick the correct answer and justify. In âˆ†ABC, AB = 6âˆš3 cm, AC = 12 cm and BC = 6 cm. The angle B is :
(A) 120Â°
(B) 60Â°
(C) 90Â°
(D) 45Â°
Solution :
In âˆ†ABC,
AC2 = 122
â‡’ AC2 = 144
BC2 + AB2 = (6)2 + (6âˆš3)2
â‡’ BC2 + AB2 = 36 + 108
â‡’ BC2 + AB2 = 144
âˆ´ AC2 = BC2 + AB2

By converse of Pythagoras theorem,
âˆ B = 90Â°