NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5.

- The Triangle and its Properties Class 7 Ex 6.1
- The Triangle and its Properties Class 7 Ex 6.2
- The Triangle and its Properties Class 7 Ex 6.3
- The Triangle and its Properties Class 7 Ex 6.4
- The Triangle and its Properties Class 7 MCQ

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
The Triangle and its Properties |

Exercise |
Ex 6.5 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

**Question 1.**

PQR is a triangle right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

**Solution:**

QR^{2} = 10^{2} + 24^{2} By Pythagoras Property

⇒ = 100 + 576 = 676

⇒ QR = 26 cm.

**Question 2.**

ABC is a triangle right-angled at C. If AB – 25 cm and AC = 7 cm, find BC.

**Solution:**

AC^{2} + BC^{2} = AB^{2} By Pythagoras Property

⇒ 7^{2} + BC^{2} = 25^{2}

⇒ 49 + BC^{2} = 625

⇒ BC^{2} = 625 – 49

⇒ BC^{2} = 576

⇒ BC = 24 cm.

**Question 3.**

A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

**Solution:**

Let the distance of the foot of the ladder from the wall be a m. Then,

Hence, the distance of the foot of the ladder from the wall is 9 m.

**Question 4.**

Which of the following can be the sides of a right triangle ?

- 2.5 cm, 6.5 cm, 6 cm.
- 2 cm, 2 cm, 5 cm.
- 1.5 cm, 2 cm, 2.5 cm.

In the case of right-angled triangles, identify the right angles.

**Solution:**

1. 2.5 cm, 6.5 cm, 6 cm We see that

(2.5)^{2} + 6^{2} = 6.25 + 36 = 42.25 = (6.5)^{2}

Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 2.5 cm and 6 cm is a right angle.

2. 2 cm, 2 cm, 5 cm

∵ 2 + 2 = 4 \(\ngtr\) 5

∴ The given lengths cannot be the sides of a triangle

The sum of the lengths of any two sides of a triangle is greater than the third side

3. 1.5 cm, 2 cm, 2.5 cm We find that

1.5^{2} + 2^{2} = 2.25 + 4 = 6.25 = 2.5^{2}

Therefore, the given lengths can be the sides of a right triangle.

Also, the angle between the lengths 1.5 cm and 2 cm is a right angle.

**Question 5.**

A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

**Solution:**

AC = CD Given

In right angled triangle DBC, DC^{2} = BC^{2} + BD^{2}

by Pythagoras Property = 5^{2} + 12^{2} = 25 + 144 = 169

⇒ DC = 13 ⇒ AC = 13

⇒ AB = AC + BC = 13 + 5 = 18

Therefore, the original height of the tree = 18 m.

**Question 6.**

Angles Q and R of a ∆ PQR are 25° and 65°. Write which of the following is true:

**(i)** PQ^{2} + QR^{2} = RP^{2}

**(ii)** PQ^{2} + RP^{2} = QR^{2}

**(iii)** RP^{2} + QR^{2} = PQ^{2}

**Solution:**

**(ii)** PQ^{2} + RP^{2} = QR^{2} is true.

**Question 7.**

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

**Solution:**

In right-angled triangle DAB, AB^{2} + AD^{2} = BD^{2}

⇒ 40^{2} + AD^{2} = 41^{2} ⇒ AD^{2} = 41^{2} – 40^{2}

⇒ AD^{2} = 1681 – 1600

⇒ AD^{2} = 81 ⇒ AD = 9

∴ Perimeter of the rectangle = 2(AB + AD) = 2(40 + 9) = 2(49) = 98 cm

Hence, the perimeter of the rectangle is 98 cm.

**Question 8.**

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

**Solution:**

Let ABCD be a rhombus whose diagonals BD and AC are of lengths 16 cm and 30 cm respectively.

Let the diagonals BD and AC intersect each other at O.

Since the diagonals of a rhombus bisect each other at right angles. Therefore

BO = OD = 8 cm,

AO = OC = 15 cm,

∠AOB = ∠BOC

= ∠COD = ∠DOA = 90°

In right-angled triangle AOB.

AB^{2} = OA^{2} + OB^{2}

By Pythagoras Property

⇒ AB^{2} = 15^{2} + 8^{2}

⇒ AB^{2} = 225 + 64

⇒ AB^{2} = 289

⇒ AB = 17cm

Therefore, perimeter of the rhombus ABCD = 4 side = 4 AB = 4 × 17 cm = 68 cm

Hence, the perimeter of the rhombus is 68 cm.

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