# NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 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, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) 7 cm, 24 cm,-25 cm
(7)2 + (24)2 = 49 + 576 = 625 = (25)2 = 25
∴ The given sides make a right angled triangle with hypotenuse 25 cm

(ii) 3 cm, 8 cm, 6 cm(8)2 = 64
(3)2 + (6)2 = 9 + 36 = 45
64 ≠ 45
The square of larger side is not equal to the sum of squares of other two sides.
∴ The given triangle is not a right angled.

(iii) 50 cm, 80 cm, 100 cm
(100)2= 10000
(80)2 + (50)2 = 6400 + 2500
= 8900
The square of larger side is not equal to the sum of squares of other two sides.
∴The given triangle is not a right angled.

(iv) 13 cm, 12 cm, 5 cm
(13)2 = 169
(12)2 + (5)2= 144 + 25 = 169
= (13)2 = 13
Sides make a right angled triangle with hypotenuse 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Solution:
We have PQR is a right triangle and PM ⊥ QR.

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

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:

Given: In ∆ABC, ∠C = 90° and AC = BC
To Prove: AB2 = 2AC2
Proof: In ∆ABC,
AB2= BC2 + AC2
AB2 = AC2 + AC2 [Pythagoras theorem]
= 2AC2

Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2 , Prove that ABC is a right triangle.
Solution:
Given that ABC is an isosceles triangle with AC = BC and given that AB² = 2AC²
Now we have AB² = 2AC²
AB² = AC² + AC²
But AC= BC (Given)
AB² = AC² + BC²
Hence by Pythagoras theorem ∆ABC is a right triangle where AB is the hypotenuse of ∆ABC.

Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:
Given: In ∆ABC, AB = BC = AC = 2a

We have to find length of AD
In ∆ABC,
AB = BC = AC = 2a
BD = $$\frac { 1 }{ 2 }$$ x 2 a = a
⇒ AD2 = AB2 – BD2= (2a)2 – (a)2 = 4a2– a2= 3a2

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.
Solution:
Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.
To Prove: AB2+ BC2+ CD2+ DA2 = AC2+ BD2

Question 8.
In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Solution:
(i) Given: ∆ABC, O is any point inside it,
OD, OE and OF are perpendiculars to BC, CA and AB respectively.
To Prove:

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.
Solution:
By Pythagoras theorem

Hence the distance of the foot of the ladder from base of the wall in 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m 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 AB i.e., the vertical pole of height 18 m and AC be the guy wire of 24 m long. BC is the distance from the vertical pole to where the wire will be staked.
By Pythagoras theorem

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1$$\frac { 1 }{ 2 }$$ hours?
Solution:

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:
We have two poles.
We have
BC = 12 m
AB = 11 – 6
AB = 5 m

By Pythagoras theorem in right triangle ABC
AC² = AB² + BC²
AC² = (12)² +(5)²
AC² = 144 + 25
AC² = 169
AC = 13 m
Hence the distance between the tops is 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: ∆ABC is a right angled at C D and E are the points on the side CA and CB.
To Prove: AE² + BD² = AB² + DE²
Proof : ∆ACE is right angled at C
AE² = AC² + CE²… (i)
(Pythagoras theorem)

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

Solution:
We have
BD = 3CD
∴ BC = BD + DC
⇒ BC = 3CD + CD
BC = 4CD
⇒ CD = $$\frac { 1 }{ 4 }$$BC … (i)
And BD = 3CD
⇒ BD = $$\frac { 3 }{ 4 }$$BC …(ii)
Since ∆ABD is a right triangle, right angled at
Similarly, ∆ACD is right angled at D.
AC² = AD² + CD² …(iv)
Substracting (iv) from (iii)
We get

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:

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 altitudes.
Solution:
Let ABC be an equailateral traingle of each side x. AD be its altitude.
So, AB = BC = CA = x
and BD = DC = $$\frac { 1 }{ 2 }$$ BC = $$\frac { x }{ 2 }$$
In right triangle ADC in which ∠D = 90°
DC = base and AC = hypotenuse.
Apply Pythagorus theorem, we get

So, we get length of each side is x an length of altitude is $$\frac{\sqrt{3} x}{2}/latex] Then, three times the square of each side = 3 x (x)² = 3x² … (i) and four times, the square of its altitudes = 4 x [latex]\frac { 3 }{ 4 }$$x² = 3x² … (ii)
It shows that equations (i) and (ii) are same. Hence times the square of one side an equilateral triangle is equal to four times the square of its altitude.

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = 6$$\sqrt{3}$$ cm, AC = 12 cm and BC = 6 cm. The angle B is:
(a) 120°
(b) 60°
(c) 90°
(d) 45
Solution:
(12)= (6$$\sqrt{3}$$)² + (6)²