NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5.

• Triangles Class 10 Ex 6.1
• Triangles Class 10 Ex 6.2
• Triangles Class 10 Ex 6.3
• Triangles Class 10 Ex 6.4
• Triangles Class 10 Ex 6.5
• Triangles Class 10 Ex 6.6

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 6
Chapter Name	Triangles
Exercise	Ex 6.5
Number of Questions Solved	17
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths 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, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:

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 X MR.
Solution:
In a right triangle, perpendicular drawn from right angle to hypotenuse divides the triangle into two similar triangles

Question 3.
In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
Solution:

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

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², Prove that ABC is a right triangle.
Solution:

Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:

As all the altitudes of an equilateral triangle are equal hence, each of the altitudes of ∆ABC is of length .

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:
The figure given below shows a rhombus ABCD in which AB = BC = CD = DA. The diagonals AC and BD bisect each other at O.
In ∆AOB, ∠AOB = 90°

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:

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:

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:

Question 11.
An airplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another airplane 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 hours?
Solution:

Question 12.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:

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:

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:

Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = 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:

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

 

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