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Answer :

In right angled triangle PQR, we have

\(QR^2 = PQ^2 + PR^2 \) [From Pythagoras property]

\(QR^2 = 10^2 + 24^2 \)

\(QR^2 = 100 + 576 = 676 \)

\(\therefore QR = \sqrt{676} = 26 cm \)
The, the required length of QR = 26 cm.

- Q.2 ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
- Q.3 A 15 m 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.
- Q.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
- Q.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.
- Q.6 Angles Q and R of a APQR are 25° and 65°. Write which of the following is true. (i) \( PQ^2 + QR^2 = RP^2 \) (ii)\( PQ^2 + RP^2 = QR^2 \) (iii)\( RP^2 + QR^2 = PQ^2 \)
- Q.7 Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
- Q.8 The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

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