NCERT Solutions for Class 7 Maths Exercise 6.5 Chapter 6 The Triangle and its Properties in simple PDF are given here. This exercise of NCERT Solutions for Class 7 Maths Chapter 6 contains topics related to the right-angled triangle and Pythagoras property. In the right-angled triangle, the side opposite to right angle is called the hypotenuse; the other two sides are known as the legs of the right angled-triangle. We suggest students learn more about these topics by solving the questions of NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties with the help of solutions provided here.
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1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Solution:-
Let us draw a rough sketch of right-angled triangle
By the rule of Pythagoras Theorem,
Pythagoras theorem states that for any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of square on the legs.
In the above figure RQ is the hypotenuse,
QR2 = PQ2 + PR2
QR2 = 102 + 242
QR2 = 100 + 576
QR2 = 676
QR = √676
QR = 26 cm
Hence, 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.
Solution:-
Let us draw a rough sketch of right-angled triangle
By the rule of Pythagoras Theorem,
Pythagoras theorem states that for any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of square on the legs.
In the above figure RQ is the hypotenuse,
AB2 = AC2 + BC2
252 = 72 + BC2
625 = 49 + BC2
By transposing 49 from RHS to LHS it becomes – 49
BC2 = 625 – 49
BC2 = 576
BC = √576
BC = 24 cm
Hence, the length of the BC = 24 cm.
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.
Solution:-
By the rule of Pythagoras Theorem,
Pythagoras theorem states that for any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of square on the legs.
In the above figure RQ is the hypotenuse,
152 = 122 + a2
225 = 144 + a2
By transposing 144 from RHS to LHS it becomes – 144
a2 = 225 – 144
a2 = 81
a = √81
a = 9 m
Hence, the length of a = 9 m.
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.
Solution:-
(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm
Let us assume the largest value is the hypotenuse side i.e. b = 6.5 cm.
Then, by Pythagoras theorem,
b2 = a2 + c2
6.52 = 2.52 + 62
42.25 = 6.25 + 36
42.25 = 42.25
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) Let a = 2 cm, b = 2 cm, c = 5 cm
Let us assume the largest value is the hypotenuse side i.e. c = 5 cm.
Then, by Pythagoras theorem,
c2 = a2 + b2
52 = 22 + 22
25 = 4 + 4
25 ≠8
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) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm
Let us assume the largest value is the hypotenuse side i.e. b = 2.5 cm.
Then, by Pythagoras theorem,
b2 = a2 + c2
2.52 = 1.52 + 22
6.25 = 2.25 + 4
6.25 = 6.25
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.
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.
Solution:-
Let ABC is the triangle and B is the point where tree is broken at the height 5 m from the ground.
Tree top touches the ground at a distance of AC = 12 m from the base of the tree,
By observing the figure we came to conclude that right angle triangle is formed at A.
From the rule of Pythagoras theorem,
BC2 = AB2 + AC2
BC2 = 52 + 122
BC2 = 25 + 144
BC2 = 169
BC = √169
BC = 13 m
Then, 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:
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP22 + QR2 = PQ2
Solution:-
Given that ∠B = 35o, ∠C = 55o
Then, ∠A =?
We know that sum of the three interior angles of triangle is equal to 180o.
= ∠PQR + ∠QRP + ∠RPQ = 180o
= 25o + 65o + ∠RPQ = 180o
= 90o + ∠RPQ = 180o
= ∠RPQ = 180 – 90
= ∠RPQ = 90o
Also, we know that side opposite to the right angle is the hypotenuse.
∴ QR2 = PQ2 + PR2
Hence, (ii) is true
7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution:-
Let ABCD be the rectangular plot.
Then, AB = 40 cm and AC = 41 cm
BC =?
According to Pythagoras theorem,
From right angle triangle ABC, we have:
= AC2 = AB2 + BC2
= 412 = 402 + BC2
= BC2 = 412 – 402
= BC2 = 1681 – 1600
= BC2 = 81
= BC = √81
= BC = 9 cm
Hence, the perimeter of the rectangle plot = 2 (length + breadth)
Where, length = 40 cm, breadth = 9 cm
Then,
= 2(40 + 9)
= 2 × 49
= 98 cm
8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:-
Let PQRS be a rhombus, all sides of rhombus has equal length and its diagonal PR and SQ are intersecting each other at a point O. Diagonals in rhombus bisect each other at 90o.
So, PO = (PR/2)
= 16/2
= 8 cm
And, SO = (SQ/2)
= 30/2
= 15 cm
Then, consider the triangle POS and apply the Pythagoras theorem,
PS2 = PO2 + SO2
PS2 = 82Â + 152
PS2 = 64 + 225
PS2 = 289
PS = √289
PS = 17 cm
Hence, the length of side of rhombus is 17 cm
Now,
Perimeter of rhombus = 4 × side of the rhombus
= 4 × 17
= 68 cm
∴ Perimeter of rhombus is 68 cm.