Circle or Semi Circle MCQ Quiz - Objective Question with Answer for Circle or Semi Circle - Download Free PDF
Last updated on May 15, 2025
Latest Circle or Semi Circle MCQ Objective Questions
Circle or Semi Circle Question 1:
What is the radius (in m) of a circular field whose area is equal to six times the area of a triangular field whose sides are 35 m, 53 m and 66 m?. (Take π = 22/7)
Answer (Detailed Solution Below)
Circle or Semi Circle Question 1 Detailed Solution
Given:
Area of circular field = 6 × area of triangular field
Formula used:
Area of circular field = πr2, where r is a radius of circle.
Area of triangle = √s (s - a) (s - b) (s - c),
where s = (a + b + c)/2
a, b and c are sides of triangle respectively.
Calculation:
a = 35 m, b = 53 m and c = 66 m
s = (35 + 53 + 66)/2
⇒ s = 77
Area of triangle = √s(s - a) (s - b) (s - c)
⇒ √77 (77 - 35) (77 - 53) (77 - 66)
⇒ √77 × 42 × 24 × 11
⇒ √7 × 11 × 2 × 3 × 7 × 2 × 3 × 2 × 2 × 11
⇒ 11 × 7 × 2 × 2 × 3
⇒ 924 m2
Now, 6 × Area of triangle = Area of circle
⇒ 924 × 6 = πr2
⇒ (924 × 6 × 7)/22 = r2
⇒ 1764 = r2
⇒ 42 = r
∴ Radius of circle is 42 m.
Circle or Semi Circle Question 2:
A bycle wheel makes 5000 revolutions in moving 11 km. How much cms will be the diameter of the wheel?
Answer (Detailed Solution Below)
Circle or Semi Circle Question 2 Detailed Solution
Given:
Number of revolutions = 5000
Distance covered = 11 km = 1100000 cm
Formula used:
Circumference of circle = 2πr
Calculation:
Circumference of circle = 1100000/5000
⇒ 220 cm
According to the question,
⇒ 2πr = 220 cm
⇒ 2 × (22/7) × r = 220 cm
⇒ r/7 = 5 cm
⇒ r = 35 cm
Diameter = 2r
⇒ 2 × 35
⇒ 70 cm
∴ The diameter of circle is 70 cm
Circle or Semi Circle Question 3:
The diameter of the driving wheel of a cart is 154 cm. Calculate the revolution per minute [RPM] of the wheel (approx) in order to keep a speed of 33 Kilo meter per hour.
Answer (Detailed Solution Below)
Circle or Semi Circle Question 3 Detailed Solution
Given:
Diameter of the wheel = 154 cm
Radius of the wheel = 154/2 = 77 cm
Speed = 33km/h
Formula used:
Circumference of a circle = 2πr
where r → radius of a circle
Distance covered = speed × time
Calculation:
Distance covered in one revolution = Circumference of wheel
⇒ 2πr = 2 × 22/7 × 77 = 484 cm
Speed = 33 km/hr = 33 × 100000/60 = 55000 cm/min
∴ Total Revolutions covered in one min
⇒ 55000/484 = 113.63 ≈ 114
Circle or Semi Circle Question 4:
A copper wire is bent in the shape of a square of area 81 cm2. If the same wire is bent from of a semicircle the radius (in cm) of the semicircle is \(\left(\text{Take} \ \pi = \frac{22}{7} \right)\)
Answer (Detailed Solution Below)
Circle or Semi Circle Question 4 Detailed Solution
Given:
Area of a square shape wire = 81 cm2
Formulas used:
Perimeter of a square with side 'a' = 4a
Area of square with side 'a' = a2
Perimeter of a semicircle = 2πr/2 + 2r = πr + 2r
Calculation:
Area of square = 81 cm2
⇒ a2 = 92
⇒ a = 9 cm
Perimeter of square = 4 × 9 = 36 cm
⇒ Perimeter of square = Perimeter of semicirlce
⇒ 36 cm = 22/7 × r + 2r
⇒ 36 = (22r + 14r)/7
⇒ 36 = 36r/7
⇒ 1 = r/7
⇒ r = 7 cm
∴ The radius of the semicircle is 7 cm.
Circle or Semi Circle Question 5:
If the area of a circle is 154 cm2, then the circumference of the circle is
Answer (Detailed Solution Below)
Circle or Semi Circle Question 5 Detailed Solution
Given:
Area of circle = 154 cm2
Formula used:
Area of circle = πr2
Circumference of circle = 2πr
Calculation:
154 = πr2
⇒ r2 = 154 × (7/22)
⇒ r2 = 7 × 7
⇒ r2 = 49
⇒ r = √49
⇒ r = 7 cm
Circumference = 2πr
⇒ Circumference = 2 × (22/7) × 7
⇒ Circumference = 44 cm
∴ The correct answer is option 2.
Top Circle or Semi Circle MCQ Objective Questions
Six chords of equal lengths are drawn inside a semicircle of diameter 14√2 cm. Find the area of the shaded region?
Answer (Detailed Solution Below)
Circle or Semi Circle Question 6 Detailed Solution
Download Solution PDFGiven:
Diameter of semicircle = 14√2 cm
Radius = 14√2/2 = 7√2 cm
Total no. of chords = 6
Concept:
Since the chords are equal in length, they will subtend equal angles at the centre. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.
Formula used:
Area of sector = (θ/360°) × πr2
Area of triangle = 1/2 × a × b × Sin θ
Calculation:
The angle subtended by each chord = 180°/no. of chord
⇒ 180°/6
⇒ 30°
Area of sector AOB = (30°/360°) × (22/7) × 7√2 × 7√2
⇒ (1/12) × 22 × 7 × 2
⇒ (77/3) cm2
Area of triangle AOB = 1/2 × a × b × Sin θ
⇒ 1/2 × 7√2 × 7√2 × Sin 30°
⇒ 1/2 × 7√2 × 7√2 × 1/2
⇒ 49/2 cm2
∴ Area of shaded region = 6 × (Area of sector AOB - Area of triangle AOB)
⇒ 6 × [(77/3) – (49/2)]
⇒ 6 × [(154 – 147)/6]
⇒ 7 cm2
∴ Area of shaded region is 7 cm2
The length of an arc of a circle is 4.5π cm and the area of the sector circumscribed by it is 27π cm2. What will be the diameter (in cm) of the circle?
Answer (Detailed Solution Below)
Circle or Semi Circle Question 7 Detailed Solution
Download Solution PDFGiven :
Length of an arc of a circle is 4.5π.
Area of the sector circumscribed by it is 27π cm2.
Formula Used :
Area of sector = θ/360 × πr2
Length of arc = θ/360 × 2πr
Calculation :
According to question,
⇒ 4.5π = θ/360 × 2πr
⇒ 4.5 = θ/360 × 2r -----------------(1)
⇒ 27π = θ/360 × πr2
⇒ 27 = θ/360 × r2 ---------------(2)
Doing equation (1) ÷ (2)
⇒ 4.5/27 = 2r/πr2
⇒ 4.5/27 = 2/r
⇒ r = (27 × 2)/4.5
⇒ Diameter = 2r = 24
∴ The correct answer is 24.
How many revolutions per minute a wheel of car will make to maintain the speed of 132 km per hour? If the radius of the wheel of car is 14 cm.
Answer (Detailed Solution Below)
Circle or Semi Circle Question 8 Detailed Solution
Download Solution PDFGiven:
Radius of the wheel of car = 14 cm
Speed of car = 132 km/hr
Formula Used:
Circumference of the wheel = \(2\pi r\)
1 km = 1000 m
1m = 100 cm
1hr = 60 mins.
Calculation:
Distance covered by the wheel in one minute = \(\frac{132 \times 1000 \times 100}{60}\) = 220000 cm.
Circumference of the wheel = \(2\pi r\) = \(2\times \frac{22}{7} \times 14\) = 88 cm
∴ Distance covered by wheel in one revolution = 88 cm
∴ The number of revolutions in one minute = \(\frac{220000}{88}\) = 2500.
∴ Therefore the correct answer is 2500.
In a circle with centre O, chords PR and QS meet at the point T, when produced, and PQ is a diameter. If \(\angle\)ROS = 42º, then the measure of \(\angle\)PTQ is
Answer (Detailed Solution Below)
Circle or Semi Circle Question 9 Detailed Solution
Download Solution PDFGiven:
∠ROS = 42º
Concept used:
The sum of the angles of a triangle = 180°
Exterior angle = Sum of opposite interior angles
Angle made by an arc at the centre = 2 × Angle made by the same arc at any point on the circumference of the circle
Calculation:
Join RQ and RS
According to the concept,
∠RQS = ∠ROS/2
⇒ ∠RQS = 42°/2 = 21° .....(1)
Here, PQ is a diameter.
So, ∠PRQ = 90° [∵ Angle in the semicircle = 90°]
In ΔRQT, ∠PRQ is an exterior angle
So, ∠PRQ = ∠RTQ + ∠TQR
⇒ 90° = ∠RTQ + 21° [∵ ∠TQR = ∠RQS = 21°]
⇒ ∠RTQ = 90° - 21° = 69°
⇒ ∠PTQ = 69°
∴ The measure of ∠PTQ is 69°
AB is a diameter of a circle with center O. A tangent is drawn at point A. C is a point on the circle such that BC produced meets the tangent at P. If ∠APC = 62º, then find the measure of the minor arc AC(i.e.∠ ABC).
Answer (Detailed Solution Below)
Circle or Semi Circle Question 10 Detailed Solution
Download Solution PDFGiven:
AB is a diameter of a circle with a center O
∠APC = 62º
Concept used:
The radius/diameter of a circle is always perpendicular to the tangent line.
Sum of all three angles of a triangle = 180°
Calculation:
Minor arc AC will create angle CBA
∠APC = 62º = ∠APB
∠BAP = 90° (diameter perpendicular to tangent)
In Δ APB,
∠APB + ∠BAP + ∠PBA = 180°
⇒ ∠PBA = 180° - (90° + 62°)
⇒ ∠PBA = 28°
∴ The measure of minor arc AC is 28°
Mistake PointsMeasure of the minor arc AC is asked,
∠ABC marks arc AC,
∴ ∠ABC is the correct angle to show a measure of arc AC
This is a previous year's question, and according to the commission, this is the correct answer.
The two sides holding the right-angle in a right-angled triangle are 3 cm and 4 cm long. The area of its circumcircle will be:
Answer (Detailed Solution Below)
Circle or Semi Circle Question 11 Detailed Solution
Download Solution PDFThe two sides holding the right angle in a right-angled triangle are 3 cm and 4 cm long,
⇒ Length of hypotenuse = (32 + 42)1/2 = 5 cm
⇒ Radius of circum-circle = 5/2 = 2.5 cm
∴ Area = 22/7 × (2.5)2 = 6.25π cm2An arc of length 23.1 cm subtends an 18° angle at the centre. What is the area of the circle? [Use \(π = \frac{22}{7}\)]
Answer (Detailed Solution Below)
Circle or Semi Circle Question 12 Detailed Solution
Download Solution PDFGiven:
Length of an arc = 23.1 cm
Angle subtended on center by arc = 18°
Formula used:
Length of an arc = (2 × π × R × θ)/360
Area of circle = π × R2
Where, R = radius
Calculation:
Length of an arc = (2 × π × R × θ)/360
⇒ 23.1 = (2 × 22 × R × 18)/(360 × 7)
⇒ 23.1 = (22 × R)/(10 × 7)
⇒ R = (2.1 × 70)/2 = 73.5 cm
Area of circle = π × R2
⇒ (22/7) × 73.5 × 73.5
⇒ 22 × 10.5 × 73.5
⇒ 16978.50 cm2
∴ The correct answer is 16978.50 cm2.
One-quarter of a circular pizza of diameter 28 cm was removed from the whole pizza. What is the perimeter (in cm) of the remaining pizza? (Take π = 22/7)
Answer (Detailed Solution Below)
Circle or Semi Circle Question 13 Detailed Solution
Download Solution PDFGiven:
Diameter of pizza = 28cm
Formula:
Circumference of circle = πd
Calculation:
Radius of pizza = 28/2 = 14cm
Total circumference of pizza = 22/7 × 28 = 88cm
Circumference of 3/4 of pizza = 88 × 3/4 = 66cm
∴ Perimeter of remaining pizza = 66 + 14 + 14 = 94cm
A circular play ground has a circular path with a certain width around it. If the difference between the circumference of the outer and inner circle is 144 cm, then find the approximate width of the path. (Take π = 22/7)
Answer (Detailed Solution Below)
Circle or Semi Circle Question 14 Detailed Solution
Download Solution PDFGiven:
A play ground has a circular path with a certain width around it.
Difference between the circumference of the outer and inner circle is 144 cm
Formula used:
Circumference of a circle = 2πr unit
where r → radius of the circle.
Calculation:
Let the inner radius and outer radius be r cm and R cm respectively.
The width of the path will be (R - r) cm
Difference between the circumference of the outer and inner circle = 144 cm
⇒ 2πR - 2πr = 144
⇒ 2π(R - r) = 144
⇒ R - r = (144 × 7)/44
⇒ R - r = 22.9 ≈ 23
∴ The width of the path is 23 cm.
The circumference of the two circles is 198 cm and 352 cm respectively. What is the difference between their radii?
Answer (Detailed Solution Below)
Circle or Semi Circle Question 15 Detailed Solution
Download Solution PDFGiven:
The circumference of the two circles is 198 cm and 352 cm respectively.
Concept used:
Circumference of the two circles = 2πr
Where, r = radius
Calculation:
Let the radius of two circle is r1 & r2
According to the question,
2πr2 - 2πr1 = 352 - 198
⇒ 2π(r2 - r1) = 154
⇒ π(r2 - r1) = 77
⇒ r2 - r1 = 77 × 7/22
⇒ r2 - r1 = 49/2
⇒ r2 - r1 = 24.5
∴ The required answer is 24.5 cm