KSEEB Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5
KSEEB Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5
Karnataka State Syllabus 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 = 10cm and PR = 24 cm, find QR.
Solution:
![KSEEB Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 1](https://i0.wp.com/www.kseebsolutions.com/wp-content/uploads/2020/10/KSEEB-Solutions-for-Class-7-Maths-Chapter-6-The-Triangles-and-Its-Properties-Ex-6.5-1.png?resize=144%2C148&ssl=1)
PQR is a right angled triangle. ∠RPQ = 90°
∴ QR2 = PQ2 + PR2 (By pythagoras property)
= 102 + 242
= 100 + 576
∴ QR =
Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Solution:
ABC is a right-angled triangle.
∠ACB = 90°
∴ AB2 = AC2 + BC2 (By pythagoras property)
252 = 72 + BC2
BC2 = 252 – 72 = 625 – 49
∴ BC = √576 = 24
∴ BC = 24 cms
Question 3.
A 15m 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 distance of the foot of the ladder from the wall is ‘a’ m.
Then
a2 + 122 = 122 (pythagoras property)
a2 = 152 – 122
a2 = 225 – 144
a2 = 81
∴ a =
∴ The distance of the foot of the ladder from the wall be 9m.
Question 4.
Which of the following can be the sides of a right triangle?
Solution:
In the case of right-angled triangles, identify the right angles.
i) 2.5 cm, 6.5 cm, 6 cm
Solution:
The greatest side is hypotenuse = 6.5 cm
(2.5)2 + (6)2 = (6.5)2 (According to Pythagoras property)
6.25 + 36 = 42.25
42.25 = 42.25
∴ This is a right-angled triangle. The given length can be the sides of a right-angled triangle. The right angle between the lengths of 2.5 cm and 6 cm.
ii) 2 cm, 2 cm, 5 cm
The greatest side is 5 cm
(2)2 + (2)2 = (5)2 (According to pythagoras property)
4 + 4 ≠ 25
∴ It is not the lengths of right angled triangle.
iii) 1.5 cm, 2 cm, 2.5 cms
The greatest side is 2.5 cms.
According to pythagoras property
(2.5)2 = (1.5)2 + (2)2
6.25 = 2.25 + 4
6.25 = 6.25
It is a right angled trianges, the given lengths are sides of the right angled triangles.
∴ The right angle between the lengths of 1.5 cms and 2 cms.
Question 5.
A Tree is broken at a height of 5m from the ground and its top touches the ground at a distance of 12 m from the base tree. Find the original height of the tree.
Solution:
According to the given statement the sketch we get is right angled triangle
RQ2 = PR2 + PQ2
= 52 + 122
= 25 + 144 = 169
∴ RQ = = 13m
∴ The original height of the tree is PR + RS = 5 + 13 = 18 m
Question 6.
Angles Q and R of a ∆ PQR are 25° and 65°.
Write which of the following is true :
Solution:
∠p = 90°
∴ According to Pythagoras property
ii) PQ2 + RP2 = QR2
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:
PQRS is a rectangle.
According to the figure, PQS is a right-angled triangle.
∴ QS2 = PQ2 + PS2
412 = 402 + PS2
∴ PS2 = 422 + 402
= 1681 – 1600
PS = = 9 cms
The perimeter of the rectangle = 2 (PQ + PS)
= 2 (40 + 9) = 2 (49)
= 98 cms
Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:
PQRS is a rhombus. The diagonals are PQ & RS.
PQ = 30 cms RS = 16 cms
According to fig In Rhombus the diagonals bisect
Each other. ∴ PQ & RS bisect at O.
PO = OQ SO = OR
Considering the ∆ POR
PR2 = PO2 + OR2
= 152 + 82
= 225 + 64 = 289
∴ PR = √289 = 17cms
∴ The perimeter of the rhombus = 4 × side PR
= 4 × 17 = 68 cms