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

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## 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: PQR is a right angled triangle. ∠RPQ = 90°
∴ QR2 = PQ2 + PR2 (By pythagoras property)
= 102 + 242
= 100 + 576
∴ QR = $\sqrt{676}$ = 26 cms

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 = $$\sqrt{576}$$ = 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 = $\sqrt{81}=9 \mathrm{cms}$
∴ 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 = $\sqrt{169}$ = 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 = $\sqrt{81}$ = 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 = $$\sqrt{289}$$ = 17cms
∴ The perimeter of the rhombus = 4 × side PR
= 4 × 17 = 68 cms