EXERCISE 6.5
1) PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Ans: By Pythagoras theorem we get,
QR 2 = PQ 2 + PR 2
= 102 + 242 ⇒ 100 + 576
QR 2 = 676 ⇒ QR = √676 ⇒ QR = 26
∴ the length of the hypotenuse QR = 26 cm.
2) ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Ans: By Pythagoras theorem
AB 2 = AC 2 + BC 2
25 2 = 7 2 + BC 2 ⇒ 25− 7 = BC 2
625 – 79 = BC 2 ⇒ BC = 576
BC = √576 ⇒ BC = 24 cm
Hence, the length of BC = 24 cm.
3) A 15 m long ladder reached a window 12 cm high from the ground placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Ans: By Pythagoras property
152 = a 2 + 122
225 = a 2 + 144 ⇒ a 2 = 225 − 144
a 2 = 81 ⇒ a = √81 ⇒ a = 9
Hence, the distance of the foot of the ladder from the wall is 9 cm.
4) Which of the following can be the sides of a right triangle?
2.5 cm , 6.5 cm , 6 cm.
Ans:
i) Let a = 2.5 cm . b=6.5 cm , c = 6 cm
Let us assume the largest value to be the hypotenuse
Then by Pythagoras theorem,
b 2 = a 2 + c 2
6.52 = 2.52 + 62 ⇒ 42.25 = 6.25 + 36
∴ 42.25 = 42.25
Since, the sum of squares of two sides of triangle is equal to the square of the third side.
∴ The given triangle is a right-angled triangle. The angle opposite to the largest side is a right angle.
ii) 2 cm , 2 cm , 5 cm
Ans: Let a = 2 cm , b = 2 cm , c = 5 cm
Let us assume the largest value to be the hypotenuse
Then by Pythagoras theorem,
c 2 = a 2 + b 2
52 = 22 + 22
25 = 4 + 4
25 = 8
Since, the sum of square of two sides of triangle is not equal to the square of the third side.
∴ The given triangle is not a right angled triangle.
iii) 1.5 cm , 2 cm , 2.5 cm
Ans: Let a = 1.5 cm , b = 2 cm , c = 2.5 cm
Let us assume the largest value to be the hypotenuse
Then by Pythagoras theorem,
c 2 = a 2 + b 2
2.52 = 1.52 + 22
6.25 = 2.25 + 4
6.25 = 6.25
Since, the sum of square of two sides of the triangle is equal to the square of third side,
∴ The given triangle is right-angled triangle. Right angle lies on the opposite side of the larger side, 2.5 cm.
5) A tree is broken at a height of 5 cm from the ground and its top touches the ground at a distance of 12 cm from the base of the tree. Find the original height of the tree.
Ans: Let ABC be the triangle and let A be the point where the tree is broken at a height of 5 cm from the ground.
By Pythagoras theorem,
AC 2 = AB 2 + BC 2
= 52 + 122 ⇒ 25 + 144
AC 2 = 169 ⇒ AC = √169
AC = 13 m
∴ the original height of the tree = AB + BC ⇒ 5 + 13 = 18 m
6) Angles Q and R of a ΔPQR are 25ᵒ and 65ᵒ. Write which of the following is true.
PQ2 + QR2 = RP2
PQ2 + RP2 = QR2
RP2 + QR2 = PQ2
Ans: ∠Q = 25ᵒ , ∠R = 65ᵒ
By angle sum property of triangle
∠P + ∠Q + ∠R = 180ᵒ
∠P + 25ᵒ + 65ᵒ = 180ᵒ
∠P + 90ᵒ = 180ᵒ
∠P = 180ᵒ − 90ᵒ
∠P = 90ᵒ
We know that the sides opposite to the right angle is hypotenuse. Hence, QR is the hypotenuse.
∴ QR 2 = PQ 2 + RP 2 is true.
7) Find the perimeter of the rectangle whose length is 40 cm and diagonal is 41 cm.
Ans: Let ABCD be the rectangle
AB = 40 cm BD = 41 cm
According to Pythagoras theorem,
BD 2 = AB 2 + AD 2
412 = 402 + AD 2
AD 2 = 412 − 402 = 1681 – 1600
AD 2 = 81 ⇒ AD = √81 ⇒ AD = 9
The perimeter of the rectangle = 2 ( length + breadth )
= 2 ( 40 + 9 ) ⇒ 2 (49) ⇒ 98 cm
∴ The perimeter of the rectangle is 98 cm.
8) The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Ans: Let ABCD be the rhombus. Its diagonals intersect at O. All the sides of the rhombus are equal and its diagonals bisect each other at 90ᵒ.
In Δ AOB , AO = 15 cm , BO = 8 cm
By Pythagoras theorem,
AB 2 = AO 2 + BO 2
= 152 + 82 ⇒ 225 + 64
AB 2 = 289 ⇒ AB = √289
AB = 17 cm
Hence the length of the side of the rhombus is 17 cm
Perimeter of rhombus = sum of all four sides
= AB + BC + CD + AD
= 17 + 17 + 17 + 17 ⇒ 68 cm
∴ Perimeter of rhombus is 68 cm.