NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Get Free NCERT Solutions for Class 10 Maths Chapter 6 Ex 6.5 PDF. Triangles Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 6.5 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 6 Maths Class 10 Triangles Exercise 6.5 provided in NCERT TextBook.

• 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
• Extra Questions for Class 10 Maths Triangles

You can also download the free PDF of Ex 6.5 Class 10 Triangles NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

6. Triangles
6.1 Introduction
6.2 Similar Figures
6.3 Similarity Of Triangles
6.4 Criteria For Similarity Of Triangles
6.5 Areas Of Similar Triangles
6.6 Pythagoras Theorem
6.7 Summary

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 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 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:

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:

Download NCERT Solutions For Class 10 Maths Chapter 6 Triangles PDF

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:

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:

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 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 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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