# NCERT Solutions class 7 Maths Chapter-6 Exercise 6.5

NCERT Solutions Class-7 Maths chapter-6 Triangle and its properties Exercise-6.5 is prepared by academic team of pw all the questions of NCERT text book are solved step by step with proper and detail solutions explaining each and every questions . For More and additional questions of **CBSE class 7 maths **you can go to class 7 maths sections. **NCERT class 7 Maths Solutions** is the best way to enhanced your mathematics skill. And pw practice worksheet & question bank will help you a lot .

## NCERT Solutions class 7 Maths Chapter-6 Triangle and its properties

### Triangle and its properties Exercise-6.5

### Solutions of Chapter Triangle and its properties Exercise-6.5

**Question 1:**

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

**Answer:**

It is given in the question that,

PQ = 10 cm

Also,

PR = 24 cm

We have to find QR

We know that,

According to Pythagoras theorem,

In a right angled triangle:

(Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2}

By applying Pythagoras theorem in the given question, we get:

(QR)2 = (PQ)^{2 }+ (PR^{)2}

(QR)2 = (10)^{2 }+ (24)^{2}

(QR)2 = 100 + 576

(QR)2 = 676

Hence,

**Question 2:**

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

**Answer:**

It is given in the question that,

AB = 25 cm

Also,

AC = 7 cm

We have to find BC

We know that,

According to Pythagoras theorem,

In a right angled triangle:

(Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2}

By applying Pythagoras theorem in ∆ABC, we get:

(AB)^{2} = (AC)^{2 }+ (BC)^{2}

(BC)^{2} = (AB)^{2} - (AC)^{2}

(BC)^{2} = (25)^{2} – (7)^{2}

(BC)^{2 }= 625 – 49

(BC)^{2} = 576

Hence,

**Question 3:**

A 15 m long ladder reached a window 12 m high form the ground on placing it against a wall

at a distance a. Find the distance of the foot of the ladder from the wall.

**Answer:**

Hence,

Option (ii) is true

**Question 7:**

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

**Answer:**

Here,

It is given in the question that,

Length of a rectangle = 40 cm

Diagonal of a rectangle = 41 cm

We know that,

Diagonal of a rectangle divides it into two right angled triangles

By using Pythagoras theorem, we get

(Diagonal)^{2} = (Length)^{2} + (Breadth)^{2}

(41)^{2 }= (40)^{2} + b^{2}

b^{2} = (41)^{2} – (40)^{2}

b^{2} = 1681 – 1600

b^{2} = 81

b = 9 cm

Thus,

Perimeter of rectangle = 2 (l + b)

= 2 (40 + 9)

= 2 (49)

= 98 cm

**Question 8:**

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

**Answer:**

Let us assume a rhombus ABCD having diagonals AC and BD

Diagonals of this rhombus ABCD intersects each other at a point O

We know that,

Diagonals of rhombus bisects each other at 90^{o}

**NCERT CLASS 7 MATHEMATICS SOLUTIONS **

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