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

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its
Properties Ex 6.5 are the part of NCERT Solutions for Class 7 Maths. Here you
can find the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its
Properties Ex 6.5.

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5**

**Ex
6.5 Class 7 Maths Question 1.**

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

In right-angled triangle PQR, we have

QR

^{2}= PQ

^{2}+ PR

^{2}(Using Pythagoras property)

= (10)

^{2}+ (24)

^{2}

= 100 + 576 = 676∴ QR = √676 = 26 cm

Thus, the required length of QR is 26 cm.

**Ex
6.5 Class 7 Maths Question 2.**

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

BC

^{2}+ AC

^{2}= AB

^{2}(Using Pythagoras property)

⇒ BC

^{2}+ (7)

^{2}= (25)

^{2}

⇒ BC^{2} + 49 = 625

⇒ BC^{2} = 625 – 49

⇒ BC^{2} = 576

∴ BC = √576 = 24 cm

Thus, the required length of BC is 24 cm.

**Ex 6.5 Class 7 Maths 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:**

Here, the ladder forms a right-angled triangle.

∴ a

^{2}+ (12)

^{2}= (15)

^{2}(Using Pythagoras property)

⇒ a

^{2 }+ 144 = 225

⇒ a

^{2}= 225 – 144

⇒ a

^{2}= 81

∴ a = √81 = 9 m

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

**Ex 6.5 Class 7 Maths Question 4.**

Which of the following can be the
sides of a right triangle?(i) 2.5 cm, 6.5 cm, 6 cm

(ii) 2 cm, 2 cm, 5 cm

(iii) 1.5 cm, 2 cm, 2.5 cm

**Solution:****
**(i) Given sides are: 2.5 cm, 6.5
cm, 6 cm

Square of the longer side = (6.5)

^{2}= 42.25 cm

Sum of the squares of the other two sides

= (2.5)

^{2}+ (6)

^{2}

=
6.25 + 36

= 42.25 cm

Since, the square of the longer side in a
triangle is equal to the sum of the squares of the other two sides.

∴ The given sides form a right
triangle.

(ii) Given sides are: 2 cm, 2 cm, 5 cm

Square of the longer side = (5)^{2} = 25 cm

Sum of the squares of the other two sides

= (2)^{2} + (2)^{2} = 4 + 4 = 8 cm

Since 25 cm ≠ 8 cm

∴ The given sides do not form a right triangle.

(iii) Given sides are: 1.5 cm, 2 cm, 2.5 cm

Square of the longer side = (2.5)^{2} = 6.25 cm

Sum of the squares of the other two sides

= (1.5)^{2} + (2)^{2} = 2.25 + 4 = 6.25 cm

Since 6.25 cm = 6.25 cm

Since the square of the longer side in a triangle is equal to the sum of the
squares of the other two sides.

∴ The given sides form a right triangle.

**Ex
6.5 Class 7 Maths 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:**

Let
AB be the original height of the tree which is broken at C and touching the
ground at D such that AC = 5 m and AD = 12 m

In right-angled triangle ∆CAD,

AD^{2} + AC^{2} = CD^{2} (Using Pythagoras property)

⇒ (12)^{2} + (5)^{2} = CD^{2}

⇒ 144 + 25 = CD^{2}

⇒ 169 = CD^{2}

∴ CD = √169 = 13 m

But BC = CD = 13 cm

The tree, AB = AC + CB

AB = 5 m + 13 m

∴ AB = 18 m

Thus, the original height of the tree is 18 m.

**Ex 6.5 Class 7 Maths 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:**

We know that

∠P + ∠Q + ∠R = 180° (Angle sum property)

∠P + 25° + 65° = 180°

∠P + 90° = 180°

∠P = 180° – 90° = 90°

Therefore, ∆PQR is a right triangle, right angled at P.

(i) False

∴ PQ

^{2}+ QR

^{2}≠ RP

^{2}(Using Pythagoras property)

(ii) True

∴ PQ

^{2}+ RP

^{2}= QP

^{2}(Using Pythagoras property)

(iii) False

∴ RP

^{2}+ QR

^{2}≠ PQ

^{2}(Using Pythagoras property)

**Ex
6.5 Class 7 Maths Question 7.**

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

**Solution:
**It is given that, Length AB = 40 cm

Diagonal AC = 41 cm

In right-angled triangle ABC, we have

AB

^{2}+ BC

^{2 }= AC

^{2}(Using Pythagoras property)

⇒ (40)

^{2}+ BC

^{2}= (41)

^{2}

⇒ 1600 + BC

^{2}= 1681

⇒ BC

^{2}= 1681 – 1600

⇒ BC

^{2}= 81

∴ BC = √81 = 9 cm

∴ The required perimeter of the rectangle = 2(AB + BC)

= 2(40 + 9) cm

= 98 cm

**Ex
6.5 Class 7 Maths Question 8.**

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

**Solution:
**Let ABCD be a rhombus whose diagonals intersect each other at O such that
AC = 16 cm and BD = 30 cm

Since, the diagonals of a rhombus bisect each other at 90°.

Therefore, OA = OC = 8 cm and OB = OD = 15 cm

In right-angled ∆OAB,

AB

^{2}= OA

^{2}+ OB

^{2}(Using Pythagoras property)

= (8)

^{2}+ (15)

^{2}= 64 + 225

= 289

∴ AB = √289 = 17 cm

Since AB = BC = CD = DA (Using property of a rhombus)

∴ Required perimeter of the rhombus

= 4 × side = 4 × 17 = 68 cm

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