**The Triangle and Its Properties ****Exercise 6.5 ****NCERT Solution for Class 7 Maths Chapter 6**

**The Triangle and Its Properties ****Exercise 6.1**

**The Triangle and Its Properties ****Exercise 6.2**

**The Triangle and Its Properties ****Exercise 6.3**

**The Triangle and Its Properties ****Exercise 6.4**

**Exercise 6.5**

**Q.1:- PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR. **

**Solution 1:-**

Given :- PQ = 10 cm; PR = 24 cm

For finding QR :- we need to use Pythagoras theorem

Lets QR = x cm;

In the Right Angle Triangle RPQ,

(Hypotenuse)² = (Base)² + (Perpendicular)² [ by using the Pythagoras theorem]

x² = (10)² + (24)²

⇒ x² = 100 + 576

⇒ x² = 676

⇒ x = √676

⇒ x = 26

Hence, The length of QR = 26 cm.

**Q.2:- ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.**

**Solution 2:-**

Given :- AB = 25 cm; AC = 7 cm.

For finding BC :- we need to use Pythagoras theorem

Lets, BC = x cm

In the Right Angle Triangle ACB,

(Hypotenuse)² = (Base)² + (Perpendicular)² [ by using the Pythagoras theorem]

(25)² = (7)² + (x)²

⇒ x² = (25)² – (7)²

⇒ x² = 625 – 49

⇒ x = √576

⇒ x = 24

Hence, The length of BC = 24 cm.

**Q.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 3:- **

Given:- AC = 15 m, AB = 12 m, CB = a ;

Lets, the distance of the foot of the ladder from the wall = a meter

In the Right Angle Triangle CBA,

(Hypotenuse)² = (Base)² + (Perpendicular)² [ by using the Pythagoras theorem]

(15)² = (a)² + (12)²

⇒ a² = (15)² – (12)²

⇒ a² = 225 – 144

⇒ a = √81

⇒ a = 9

Hence, The length of CB = 9 m.

So, the distance of the foot of the ladder from the wall = 9 meter.

**Q.4:- Which of the following can be the sides of a right triangle?**

**(i) 2.5cm, 6.5cm, 6cm. **

** (ii) 2cm, 2cm, 5cm. **

** (iii) 1.5cm, 2cm, 2.5cm.**

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

** Solution 4 (i):-**

Given:- AC = 6.5 cm, AB = 6 cm, CB = 2.5 cm ;

For check :- Which can be the sides of a right triangle?

Using the Pythagoras theorem:-

(Hypotenuse)² = (Base)² + (Perpendicular)²

(6.5)² = (2.5)² + (6)²

L.H.S = 6.5 x 6.5 = 42.25;

R.H.S = 2.5 x 2.5 + 6 x 6 = 6.25 + 36 = 42.25 ;

Since, L.H.S = R.H.S

Therefore, the given triangle right angle and given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 6.5 cm, which is at B.

**Solution 4(ii):- **

Given:- AC = 2 cm, 2 cm, 5 cm ;

For check :- Which can be the sides of a right triangle?

Using the Pythagoras theorem:-

(Hypotenuse)² = (Base)² + (Perpendicular)²

(2)² = (2)² + (5)²

L.H.S = 2 x 2 = 4;

R.H.S = 2 x 2 + 5 x 5 = 4 + 25 = 29 ;

Since, L.H.S ≠ R.H.S

Therefore, the given triangle are not right angle and given sides are not of the right angled triangle.

**Solution 4(iii):- **

Given:- AC = 2 cm, 2 cm, 5 cm ;

For check :- Which can be the sides of a right triangle?

Using the Pythagoras theorem:-

(Hypotenuse)² = (Base)² + (Perpendicular)²

(2.5)² = (1.5)² + (2)²

L.H.S = 2.5 x 2.5 = 6.25;

R.H.S = 1.5 x 1.5 + 2 x 2 = 2.25 + 4 = 6.25 ;

Since, L.H.S = R.H.S

Therefore, the given triangle right angle and given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 2.5 cm, which is at Q.

**Q.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 5:-**

**Q.6:- Angles Q and R of a ∆PQR are 25º and 65º.
Write which of the following is true: **

** (i) PQ2 + QR2 = RP2 **

** (ii) PQ2 + RP2 = QR2 **

** (iii) RP2 + QR2 = PQ2 **

**Q.7:- Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.**

**Solution 7:- **

**Q.8:- The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. **

**Solution 8:-**

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