Ch-Test of Circle: Circumference and Area Class 9 OP Malhotra ICSE Maths Solutions 2026-27. We Provide Step by Step Solutions / Answer of Circle: Circumference and Area OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Circle: Circumference and Area Class 9 OP Malhotra Ch-Test ICSE Maths Solutions

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-17	Circle: Circumference and Area
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Circle: Circumference and Area

Circle: Circumference and Area Class 9 OP Malhotra ICSE Maths Solutions 2026-27

Que-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

Sol: Number of revolutions = 1000
Distance travelled = 440 m
∴ Perimeter of the wheel = (440×100)/1000 = 44 cm
and diameter = Perimeter/π
= (44×7)/22 = 14 cm

Que-2: The area of a semi-circular fields is 308 sq m; then taking π = 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

Sol: Area of a semi-circular field = 308 sq m
radius (r) = √{(Area×2)/π}
= √{(308×2×7)/22}
= √(14×14)
= 14 cm
Perimeter = (1/2) × 2πr + 2r = πr + 2r
= (22/7) × 14 + 2 × 14
= 44 + 28 = 72 cm

Que-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

Sol:  Length of wire = 36 cm
∴ Circumference of semi-circle = 36 cm
Let radius = r
∴ πr + 2r = 36
⇒ (22/7r) + 2r = 36 ⇒ (36/7r) = 36
⇒ r = (36×7)/36 = 7 cm

Que-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

Sol: Radius of a circular wire = 42 cm
∴ Its perimeter = 2 πr
=2 × (22/7) × 42 cm = 264 cm
∴ Perimeter of square = 264 cm
Side = Perimeter/4 = 264/4 = 66 cm

Que-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

(a) 616 cm²
(b) 5040 cm²
(c) 1232 cm²
(d) 2464 cm²

Sol: Area of square metal wire = 484 cm²
∴ Side = √Area = √484 cm = 22 cm
∴ Perimeter = 4 × Side = 4 × 22 = 88 cm
∴ Perimeter of circle = 88 cm
∴ Radius of circle = Perimeter 2π
= (88×7)/(2×22) = 14 cm
and area = πr² = (22/7) × 14 × 14 cm²
= 616 cm²

Que-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

Sol: 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]
= (22/7) (21²-r²)
770 = (22/7) (441-r²)
(770×7)/22 = 441 – r²
245 = 441 – r²
r² = 441 – 245
r = √196
r = 14

Que-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

Sol: A circle is circumscribed by a rectangle with sides 16 cm and 12 cm
∴ Diagonal AC = √(AB²+BC²)
= √(16²+12²) = √(256+144) cm
= √400 = 20 cm
∴Diameter of circle = 20 cm
and radius = 202 = 10 cm
Area of circle = πr²
= π × 10 × 10 = 100π cm²

Que-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

Sol: 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
= (2×22)/7 × 50 m = 2200/7 m
and circumference of wheel
= (2×22)/7 × 50 cm = 2200/7 cm
∴Number of revolutions = (2200×100×7)/(7×2200) = 100
Number of revolution in 15 minutes
= (100×15)/60 = 25 (1 hr = 60 min.)

Que-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

Sol: 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 × (22/7) × 42 + 2 × (22/7) × 84 cm
= 528 + 528 = 1056 cm

Que-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

Sol: Radius of each quadrant = 4 cm
Length of cardboard = 18 cm
and width = 10 cm
Now circumference of four quadrants
= 4 × 1/2 × πr
= 2 × (22/7) × 4 = 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)

— : End of Circle: Circumference and Area Class-9 OP Malhotra Ch-Test ICSE Maths Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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