OP Malhotra Class 9 Maths Solutions Chapter 17 Circle: Circumference and Area Chapter Test

Students often turn to S Chand Class 9 Maths Solutions ICSE Chapter 17 Circle: Circumference and Area Chapter Test to clarify doubts and improve problem-solving skills.

S Chand Class 9 ICSE Maths Solutions Chapter 17 Circle: Circumference and Area Chapter Test

Question 1.
A cycle wheel makes 1000 revolutions in moving 440 m. What is the diameter of the wheel?
(a) 7 cm
(b) 14 cm
(c) 28 cm
(d) 21 cm
Solution:
Number of revolutions = 1000
Distance travelled = 440 m
∴ Perimeter of the wheel = \(\frac{440 \times 100}{1000}\) = 44 cm
and diameter = \(\frac{\text { Perimeter }}{\pi}\)
= \(\frac{44 \times 7}{22}\) = 14 cm

Question 2.
The area of a semi-circular fiels is 308 sq m; then taking π = \(\frac { 22 }{ 7 }\), the length of the railing to surround it has to be
(a) 88 cm
(b) 80 cm
(c) 44 cm
(d) 72 cm
Solution:
Area of a semi-circular field = 308 sq m

Perimeter = \(\frac { 1 }{ 2 }\) × 2πr + 2r = πr + 2r = \(\frac { 22 }{ 7 }\) × 14 + 2 × 14
= 44 + 28 = 72 cm

Question 3.
A wire of length of 36 cm is bent in the form of a semi-circle. What is the radius of the semi-circle?
(a) 9 cm
(b) 8 cm
(c) 7 cm
(d) 6 cm
Solution:
Length of wire = 36 cm
∴ Circumference of semi-circle = 36 cm
Let radius = r
∴ πr + 2r = 36
⇒ \(\frac { 22 }{ 7 }\)r + 2r = 36 ⇒ \(\frac { 36 }{ 7 }\)r = 36
⇒ r = \(\frac{36 \times 7}{36}\) = 7 cm

Question 4.
A wire is in the form of a circle of radius 42 cm. If it is bent into a square, then what is the side of the square?
(a) 66 cm
(b) 42 cm
(c) 36 cm
(d) 33 cm
Solution:
Radius of a circular wire = 42 cm
∴ Its perimeter = 2 πr
=2 × \(\frac { 22 }{ 7 }\) × 42 cm = 264 cm
∴ Perimeter of square = 264 cm
Side = \(\frac{\text { Perimeter }}{4}\) = \(\frac { 264 }{ 4 }\) = 66 cm

Question 5.
A metal wire when bent in the form of a square encloses an area 484 cm². If the same wire is bent in the form of a circle, then its area is \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
(a) 616 cm²
(b) 5040 cm²
(c) 1232 cm²
(d) 2464 cm²
Solution:
Area of square metal wire = 484 cm²
∴ Side = \(\sqrt{\text { Area }}\) = \(\sqrt{484}\) cm = 22 cm
∴ Perimeter = 4 × Side = 4 × 22 = 88 cm
∴ Perimeter of circle = 88 cm
∴ Radius of circle = \(\frac{\text { Perimeter }}{2 \pi}\)
= \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
and area = πr² = \(\frac { 22 }{ 7 }\) × 14 × 14 cm²
= 616 cm²

Question 6.
In the figure, the area enclosed between the two co-centric circles is 770 cm². If the radius of the outer circle is 21 cm, the radius of the inner circle is

(a) 14 cm
(b) 22 cm
(c) 12 cm
(d) 10.5 cm
Solution:
In the figure, enclosed area between the circles = 770 cm²
Radius of outer circle (R) = 21 cm
Let radius of inner circle (r) = r
Area of circles = π[R² – r²]

Question 7.
A circle circumscribes a rectangle with side 16 cm and 12 cm. What is the area of the circle?

(a) 48 π sq cm
(b) 50 π sq cm
(c) 100 π sq cm
(d) 200 π sq cm
Solution:
A circle is circumscribed by a rectangle with sides 16 cm and 12 cm

∴ Diagonal AC = \(\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}\)
= \(\sqrt{16^2+12^2}\) = \(\sqrt{256+144}\) cm
= \(\sqrt{400}\) = 20 cm
∴Diameter of circle = 20 cm
and radius = \(\frac { 20 }{ 2 }\) = 10 cm
Area of circle = πr² = π × 10 × 10 = 100π cm²

Question 8.
A person rides a bicycle round a circular path of radius 50 m. The radius of the wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 hour. What is the number of revolutions of the wheel in 15 min?
(a) 20
(b) 25
(c) 30
(d) 35
Solution:
Radius of circular path (R) = 50 m and radius of wheel of a bicycle = 50 cm Distance travelled by wheel of a cycle = 1 h Now circumference of path = 2πR
= \(\frac{2 \times 22}{7}\) × 50 m = \(\frac{2200}{7}\)m
and circumference of wheel
= \(\frac{2 \times 22}{7}\) × 50 cm = \(\frac{2200}{7}\)cm
∴Number of revolutions = \(\frac{2200 \times 100 \times 7}{7 \times 2200}\) = 100
Number of revolution in 15 minutes
= \(\frac{100 \times 15}{60}\) = 25 (1 hr = 60 min.)

Question 9.
The figure consists of four small semi-circles of equal radii (each 42 cm) the perimeter of the shaded region is

(a) 524 cm
(b) 264 cm
(c) 396 cm
(d) 504 cm
Solution:
The given figure consists of 4 small semicircles
Radii of each small semi-circle = 42 cm and radius of big semi-circles = 42 × 2 = 84 cm

Perimeter of shaded portion
= 4 × πr + 2 × πR
= 4 × \(\frac{22}{7}\) × 42 + 2 × \(\frac{22}{7}\) × 84 cm
= 528 + 528 = 1056 cm

Question 10.
A rectangular cardboard is of dimensions 18 cm × 10 cm. From the four corners of the rectangle quarter circles of radius 4 cm are cut. What is the perimeter (approximate) of the remaining portion?
(a) 47.1 cm
(b) 49.1 cm
(c) 51.0 cm
(d) 53.0 cm
Solutiion:
Radius of each quadrant = 4 cm
Length of cardboard = 18 cm
and width = 10 cm

Now circumference of four quadrants
=4 × \(\frac{1}{2}\) × πr
= 2 × \(\frac{22}{7}\) × 4 = \(\frac{176}{7}\) = 25.1 cm
Perimeter of remaining portion
= 2 (18 + 10) – 4 × 8 = 56 – 32 = 24 cm
Total perimeter = 25.1 + 24 = 49.1 cm (b)