Class 7, Maths, Chapter 6, Exercise 6.5 Solutions
Q.1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
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, 2cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Ans:
(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm
Let the largest value is the hypotenuse side
i.e. b = 6.5 cm. by Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
⟹ (b)2 = (a)2 + (c)2
⟹ (6.5)2 = (2.5)2 + (6)2
⟹ (6.5 x 6.5 ) = (2.5 x 2.5 ) + (6 x 6)
⟹ 42.25 = 6.25 + 36
⟹ 42.25 = 42.25
⟹ LHS = RHS
The sum of square of two side of triangle is equal to the square of third side,
∴The given triangle is right-angled triangle.
Right angle lies on the opposite of the greater side 6.5 cm.
(ii) 2 cm, 2 cm, 5 cm.
Let the largest value is the hypotenuse side i.e. c = 5 cm.
by Pythagoras theorem,(Hypotenuse)2 = (Base)2 + (Perpendicular)2
⟹ (c)2 = (a)2 + (b)2
⟹ (5)2 = (2)2 + (2)2
⟹ (5 x 5 ) = (2 x 2 ) + (2 x 2)
⟹ 25 = 4 + 4
⟹ 25 ≠ 8
⟹ LHS ≠ RHS
The sum of square of two side of triangle is not equal to the square of third side,
∴ The given triangle is not right-angled triangle.
(iii) 1.5 cm, 2cm, 2.5 cm.
Let the largest value is the hypotenuse side i.e. c = 2.5 cm.
by Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
⟹ (c)2 = (a)2 + (b)2
⟹ (2.5)2 = (1.5)2 + (2)2
⟹ (2.5 x 2.5 ) = (1.5 x 1.5 ) + (2 x 2)
⟹ 6.25 = 2.25 + 4
⟹ 6.25 = 6.25
⟹ LHS = RHS
The sum of square of two side of triangle is equal to the square of third side,
∴The given triangle is right-angled triangle. Right angle lies on the opposite of the greater side 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.
Ans:
Let TCB is the tree and after broken at point C, the top of Tree ‘T’ touches the ground at ‘A’ from the base of tree at a distance of AB = 12 m.
The, ΔABC is a right angled triangle in which AB = 12 m, BC = 5m and right angled at B.
Q.6. Angles Q and R of a ∆PQR are 25º and 65º. Write which of the following is true:
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
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.
NCERT Solutions For Class 7 Maths, Chapter 6, The Triangle and its Properties (All Exercises)
Class 7, Maths, Chapter 6, The Triangle and Its Properties
Class 7, Maths, Chapter 6, The Triangle and Its Properties, Exercise 6.1
Class 7, Maths, Chapter 6, The Triangle and Its Properties, Exercise 6.2
Class 7, Maths, Chapter 6, The Triangle and Its Properties, Exercise 6.3
Class 7, Maths, Chapter 6, The Triangle and Its Properties, Exercise 6.4
Class 7, Maths, Chapter 6, The Triangle and Its Properties, Exercise 6.5 ← You are here