Students can Download Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## 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°

∴ QR^{2} = PQ^{2} + PR^{2} (By pythagoras property)

= 10^{2} + 24^{2}

= 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°

∴ AB^{2} = AC^{2} + BC^{2} (By pythagoras property)

25^{2} = 7^{2} + BC^{2}

BC^{2} = 25^{2} – 7^{2 }= 625 – 49

∴ BC = = 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

a^{2} + 12^{2} = 12^{2} (pythagoras property)

a^{2} = 15^{2} – 12^{2}

a^{2} = 225 – 144

a^{2} = 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

RQ^{2} = PR^{2} + PQ^{2}

= 5^{2} + 12^{2}

= 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) PQ^{2} + RP^{2} = QR^{2}

PQ^{2} + RP^{2} = QR^{2} 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.

∴ QS^{2} = PQ^{2} + PS^{2}

41^{2} = 40^{2} + PS^{2}

∴ PS^{2} = 42^{2} + 40^{2}

= 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

PO = OQ = \(\frac{30}{2}\) = 15 cms

SO = OR = \(\frac{16}{2}\) = 8 cms

Considering the ∆ POR

PR^{2} = PO^{2} + OR^{2}

= 15^{2} + 8^{2}

= 225 + 64 = 289

∴ PR = = 17cms

∴ The perimeter of the rhombus = 4 × side PR

= 4 × 17 = 68 cms