# GSEB Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5

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 āaā. Find the distance of the foot of the ladder from the wall.

Solution:
The foot of ladder are āaā 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
[āµ 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 āPQR 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 āPQR,

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, āQPR 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

or 40Ā² + xĀ² = 41Ā²
or xĀ² = 41Ā² – 40Ā²
= 1681 – 1600 = 81
or xĀ² = 92 or x = 9
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 āAOB, 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 āCOD, CD = 17 cm
in right āAOD, AD = 17 cm
Now, Perimeter of the rhombus
= AB + BC + CD + AD
= 17 cm + 17 cm + 17 cm + 17 cm
= 68 cm.