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 PM ^{2} = 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) AB ^{2} = BC.BD
(ii) AC^{2} = BC.DC
(iii) AD^{2} = BD.CD**

**Solution:**

**Question 4.**

**ABC is an isosceles triangle right angled at C. Prove that AB ^{2} = 2AC^{2}.**

**Solution:**

**Question 5.**

**ABC is an isosceles triangle with AC = BC. If AB ^{2} = 2AC^{2}, 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)** OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

**(ii)** AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

**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 AE ^{2} + BD^{2} = AB^{2} + DE^{2}.**

**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 2AB ^{2 }= 2AC^{2} + BC^{2}.**

**Solution:**

**Question 15.**

**In an equilateral triangle ABC, D is a point on side BC, such that BD = BC. Prove that 9AD ^{2} = 7AB^{2}.**

**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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