Gujarat BoardĀ 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:

QRĀ² = PRĀ² + PQĀ²

or QRĀ² = 24Ā² + 10Ā²

or QRĀ² = 576 + 100

or QRĀ² = 676 = 26Ā²

ā 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

ACĀ² + BCĀ² = ABĀ²

or 7Ā² + xĀ² = 25Ā²

or 49 + xĀ² = 625

or xĀ² = 625 – 49 = 576

or xĀ² = 24Ā²

ā 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 āaā. Find the distance of the foot of the ladder from the wall.

Solution:

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

aĀ² + 12Ā² = 15Ā²

or aĀ² + 144 = 225

or aĀ² = 225 – 144 = 81

or aĀ² = 9Ā²

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)Ā² + (6)Ā² = 6.25 + 36

= 42.25 = (6.5)Ā²

ā“ 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.

ā“ 2Ā² + 2Ā² = 4 + 4 = 8

But 8 ā 5Ā²

ā“ 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)Ā² + (2)Ā² = 2.25 + 4 = 6.25

Also (2.5)Ā² =6.25

ā“ (1.5)Ā² + (2)Ā² = (2.5)Ā²

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,

ā“ Using the Pythagoras property, we have:

ABĀ² + BCĀ² = ACĀ²

or 12Ā² + 5Ā² = ACĀ²

or 144 + 25 = ACĀ²

or ACĀ² = 169 = 13Ā²

or AC = 13 m

Now, the height of the tree = BD

= BC + CD

= BC + AC

[āµ 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 āPQR are 25Ā° and 65Ā°. Write which of the following is true:

(i) PQĀ² + QRĀ² = RPĀ²

(ii) PQĀ² + RPĀ² = QRĀ²

(iii) RPĀ² + QRĀ² = PQĀ²

Solution:

In āPQR,

ā P + ā Q + ā R = 180Ā°

ā“ ā P + 25Ā° + 65Ā° = 180Ā°

or ā P + 90Ā° = 180Ā°

or ā P = 180Ā° – 90Ā° = 90Ā°

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

Now, Hypotenuse = The side opposite to P = QR

Using the Pythagoras property,

QRĀ² = PQĀ² + RPĀ²

The relation (ii), i.e. PQĀ² + RPĀ² = QRĀ² 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 āBAD, we have

BAĀ² + ADĀ² = BDĀ²

or 40Ā² + xĀ² = 41Ā²

or xĀ² = 41Ā² – 40Ā²

= 1681 – 1600 = 81

or xĀ² = 92 or x = 9

ā“ 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).

ā“ ā AOB = ā BOC = ā COD = ā DOA = 90Ā°.

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 āAOB, we have

ABĀ² = AOĀ² + BOĀ² = 15Ā² + 8Ā²

= 225 + 64 = 289

or ABĀ² = 17Ā²

or AB = 17 cm

Similarly, in right ABOC,BC =17 cm

in right āCOD, CD = 17 cm

in right āAOD, AD = 17 cm

Now, Perimeter of the rhombus

= AB + BC + CD + AD

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

= 68 cm.