Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 6 The Triangles 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:

In the right APQR, using the Pythagoras property, we have:

QR² = PR² + PQ²

or QR² = 24² + 10²

or QR² = 576 + 100

or QR² = 676 = 26²

⇒ QR = 26

Thus,QR = 26 cm

Question 2.

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

Solution:

In the right A, using the Pythagoras property, we have

AC² + BC² = AB²

or 7² + x² = 25²

or 49 + x² = 625

or x² = 625 – 49 = 576

or x² = 24²

⇒ x = 24

Thus, 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:

The foot of ladder are ‘a’ metres from the wall. Using Pythagoras property, we have

a² + 12² = 15²

or a² + 144 = 225

or a² = 225 – 144 = 81

or a² = 9²

or a = 9m

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

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.

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

Solution:

(i) 2.5 cm, 6.5 cm, 6 cm The longest side is 6.5 cm.

Now, (2.5)² + (6)² = 6.25 + 36

= 42.25 = (6.5)²

∴ The given lengths can be the sides of a right triangle.

Obviously, the right angle is the angle between the sides 2.5 cm and 6 cm.

(ii) 2 cm, 2 cm, 5 cm

The longest side is 5 cm.

∴ 2² + 2² = 4 + 4 = 8

But 8 ≠ 5²

∴ The given lengths cannot be the sides of a right triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm The longest side is 2.5 cm

Now (1.5)² + (2)² = 2.25 + 4 = 6.25

Also (2.5)² =6.25

∴ (1.5)² + (2)² = (2.5)²

Thus, the given length can be sides of a right triangle.

Obviously, the right angle is the angle between the sides 1.5 cm and 2 cm.

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 the tree BC is broken at the point C, such that CD = CA.

Now, ABC is a right A,

∴ Using the Pythagoras property, we have:

AB² + BC² = AC²

or 12² + 5² = AC²

or 144 + 25 = AC²

or AC² = 169 = 13²

or AC = 13 m

Now, the height of the tree = BD

= BC + CD

= BC + AC

[∵ AC and CD are same]

= 5 m +13 m = 18 m

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

Question 6.

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

(i) PQ² + QR² = RP²

(ii) PQ² + RP² = QR²

(iii) RP² + QR² = PQ²

Solution:

In ∆PQR,

∠P + ∠Q + ∠R = 180°

∴ ∠P + 25° + 65° = 180°

or ∠P + 90° = 180°

or ∠P = 180° – 90° = 90°

So, ∆QPR is a right-angled triangle, having its right angle at P.

Now, Hypotenuse = The side opposite to P = QR

Using the Pythagoras property,

QR² = PQ² + RP²

The relation (ii), i.e. PQ² + RP² = QR² is true.

Question 7.

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

Solution:

Length of the rectangle = 40 cm

Let the breadth be x cm

In right ∆BAD, we have

BA² + AD² = BD²

or 40² + x² = 41²

or x² = 41² – 40²

= 1681 – 1600 = 81

or x² = 92 or x = 9

∴ Breadth = 9 cm

Now, Perimeter = 2(length + breadth)

= 2(40 cm + 9 cm)

= 2(49 cm) = 98 cm

Thus, 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 the given figure is a rhombus, such that AC and BD are its diagonal, and AC = 30 cm; BD = 16 cm.

Since, the diagonals of a rhombus bisect each other at right angles (Here they bisect each other at O).

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.

And, OA= OC = \(\frac { 1 }{ 2 }\)AC = \(\frac { 1 }{ 2 }\) x 30

= 15 cm

And, BO = OD = \(\frac { 1 }{ 2 }\)BD = \(\frac { 1 }{ 2 }\) x 16

= 8 cm

Now, in right-angled ∆AOB, we have

AB² = AO² + BO² = 15² + 8²

= 225 + 64 = 289

or AB² = 17²

or AB = 17 cm

Similarly, in right ABOC,BC =17 cm

in right ∆COD, CD = 17 cm

in right ∆AOD, AD = 17 cm

Now, Perimeter of the rhombus

= AB + BC + CD + AD

= 17 cm + 17 cm + 17 cm + 17 cm

= 68 cm.