Below are 68 practice questions on Circles & Mensuration, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Circles & Mensuration notes.
Three standard 3D solids and their key formulas: a cylinder's volume scales with the full height, a cone's volume is exactly one-third of the cylinder with the same base and height, and a sphere's volume and surface area both follow distinct r-based formulas worth memorising separately.
Easy - 20 questions
Q1.
Area of a circle with radius r is:
A 2πr
B πr²
C πr³
D 2πr²
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Answer: B. πr²
Why: Area of circle = πr². Circumference = 2πr.
Q2.
The diameter of a circle with circumference 44 cm (use π = 22/7):
A 7 cm
B 14 cm
C 22 cm
D 11 cm
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Answer: B. 14 cm
Why: C = πd = 44. d = 44 × 7/22 = 14 cm.
Q3.
Volume of a cube with side 5 cm:
A 25 cm³
B 75 cm³
C 125 cm³
D 150 cm³
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Answer: C. 125 cm³
Why: Volume of cube = side³ = 5³ = 125 cm³.
Q4.
Surface area of a sphere with radius r is:
A 4πr²
B 2πr²
C πr²
D (4/3)πr³
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Answer: A. 4πr²
Why: Surface area of sphere = 4πr².
Q5.
Volume of a sphere with radius r is:
A (4/3)πr³
B πr³
C (2/3)πr³
D 4πr³
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Answer: A. (4/3)πr³
Why: Volume of sphere = (4/3)πr³.
Q6.
Area of a triangle with base 8 cm and height 5 cm is:
A 40 cm²
B 20 cm²
C 13 cm²
D 25 cm²
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Answer: B. 20 cm²
Why: Area = (1/2) × base × height = (1/2) × 8 × 5 = 20 cm².
Q7.
The perimeter of a rectangle with length 10 and breadth 6 is:
A 60
B 32
C 16
D 30
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Answer: B. 32
Why: Perimeter = 2(l + b) = 2(10 + 6) = 32.
Q8.
Total surface area of a cylinder (radius r, height h):
Volume of water in a hemispherical bowl of radius 7 cm when half full:
A (1/4)(4/3)πr³
B (1/2)(2/3)πr³
C (2/3)πr³/2 = (1/3)πr³
D 718.67 cm³
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Answer: D. 718.67 cm³
Why: Full hemisphere = (2/3)πr³ = (2/3)(22/7)(343) = 718.67 cm³. When half full, volume = 718.67/2 ≈ 359.3 cm³. But question asks half-full of hemisphere. Actually half-full hemisphere means filling it half way, not half of hemisphere volume.
Q40.
The sum of interior angles of a polygon with n sides:
A (n-2) × 180°
B n × 90°
C (n-1) × 180°
D n × 180°
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Answer: A. (n-2) × 180°
Why: Sum of interior angles = (n-2) × 180°. For triangle n=3: 180°. Quadrilateral n=4: 360°.
Hard - 28 questions
Q41.
The power of a point P with respect to circle (centre O, radius r) is:
A |OP|² - r²
B |OP| - r
C r² - |OP|²
D |OP| + r
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Answer: A. |OP|² - r²
Why: For any chord through P meeting the circle at A,B: PA×PB = |OP²−r²|. Power is defined as OP²−r² (signed): positive if P is external, negative if internal, zero if P lies on the circle. Ans: |OP|²−r².
Q42.
Ptolemy theorem for a cyclic quadrilateral ABCD states:
A AC × BD = AB × CD + AD × BC
B AC + BD = AB + CD, an additive but incorrect form
C AC/BD = AB/CD, a ratio form that does not hold generally
D AC × BD = AB × AD, missing the CD term from the product
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Answer: A. AC × BD = AB × CD + AD × BC
Why: For a cyclic quadrilateral, Ptolemy proved: product of diagonals = sum of products of opposite side pairs. Diagonal AC, BD; sides AB,CD and AD,BC: AC×BD=AB×CD+AD×BC. Verified for a rectangle (special case). Ans: AC×BD=AB×CD+AD×BC.
Q43.
The angle between two chords intersecting inside a circle equals:
A Half the sum of intercepted arcs
B Half the difference of intercepted arcs
C The sum of intercepted arcs
D The central angle
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Answer: A. Half the sum of intercepted arcs
Why: Chords PQ and RS intersect at point X inside the circle. Inscribed angle theorem gives ∠PXR=(arc PR + arc QS)/2. The angle equals half the sum of the two intercepted arcs (near and far). Ans: Half the sum of intercepted arcs.
Q44.
Area of circumscribed circle to a triangle with sides a, b, c (using sine rule R = abc/4K):
A π(abc/4K)²
B π × abc/4K
C πabc/2
D 4πK/abc
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Answer: A. π(abc/4K)²
Why: Circumradius R=abc/(4K) where K is triangle area. Area of circumscribed circle=πR²=π×[abc/(4K)]²=π(abc/4K)². Ans: π(abc/4K)².
Q45.
If a sphere is cut by a plane at distance d from centre, the cross-section circle has radius:
A r - d
B √(r² - d²)
C r + d
D √(r² + d²)
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Answer: B. √(r² - d²)
Why: The centre O, the foot of perpendicular F (at distance d), and any point P on the cross-section circle form a right triangle. OP=r (radius), OF=d, FP=cross-section radius. Pythagoras: FP=√(r²−d²). Ans: √(r²−d²).
Q46.
Isoperimetric inequality states: among all closed curves of fixed perimeter, the one enclosing maximum area is:
A Square
B Equilateral triangle
C Circle
D Regular hexagon
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Answer: C. Circle
Why: Isoperimetric inequality: 4πA ≤ P², with equality iff the curve is a circle. For circle: P=2πr, A=πr², giving 4π(πr²)=4π²r²=P². No other shape achieves this equality. Ans: Circle.
Q47.
The radical axis of two non-concentric circles is:
A Parallel to the line of centres
B Perpendicular to the line of centres
C The same as the line of centres
D The common chord extended
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Answer: B. Perpendicular to the line of centres
Why: Radical axis = locus where power w.r.t. both circles is equal. Subtracting circle equations (S₁−S₂=0) gives a linear equation, always perpendicular to the line joining the two centres. For intersecting circles it is the common chord extended. Ans: Perpendicular to the line of centres.
Q48.
For a frustum with top radius r, bottom radius R, height h, slant height l = √(h²+(R-r)²). Lateral SA =
A π(R+r)l
B πl(R-r)
C 2πl(R+r)
D π(R²-r²)
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Answer: A. π(R+r)l
Why: Unrolling the frustum gives a trapezoidal band. Lateral SA = π(R+r)l, where l=√(h²+(R−r)²) is the slant height. This is derived by subtracting the lateral area of the smaller cone from the larger. Ans: π(R+r)l.
Q49.
Cavalieri principle: two solids have equal volume if:
A They have the same total surface area
B Every horizontal cross-section at same height has equal area
C They have the same height measured from base to apex
D They share the same radius at their widest cross-section
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Answer: B. Every horizontal cross-section at same height has equal area
Why: Cavalieri principle: if two solids lying between parallel planes have equal cross-sections at every level, they have equal volumes.
Q50.
A right circular cone has base radius 6 and slant height 10. Volume =
The angle between a tangent to a circle and a secant from the same external point:
A Half the difference of intercepted arcs
B Half the sum of intercepted arcs
C Equals the central angle
D 90°
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Answer: A. Half the difference of intercepted arcs
Why: From external point, tangent touches at T (arc=0 from near side), secant cuts arcs m (near) and n (far). Angle=(n−m)/2 = half the difference of intercepted arcs. For tangent-secant, near arc is the tangent-point arc. Ans: Half the difference of intercepted arcs.
Q52.
A sphere of radius R is melted into n small spheres of radius r. Relation:
A R = nr
B R³ = nr³
C R² = nr²
D nR = r
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Answer: B. R³ = nr³
Why: Volume is conserved on melting: (4/3)πR³ = n×(4/3)πr³. The (4/3)π cancels, giving R³=nr³. E.g. n=8, r=1 → R³=8, R=2 (checks out). Ans: R³=nr³.
Q53.
The nine-point circle of a triangle passes through:
A Midpoints of the three sides alone, excluding any other special points
B Feet of the three altitudes alone, excluding the side midpoints
C Midpoints of sides, feet of altitudes, and midpoints of vertex-to-orthocenter segments
D The circumcentre and the incentre of the triangle, and nothing else besides those two points
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Answer: C. Midpoints of sides, feet of altitudes, and midpoints of vertex-to-orthocenter segments
Why: Nine-point circle: passes through 9 notable points -- 3 midpoints of sides, 3 feet of altitudes, and 3 midpoints of segments from vertices to orthocenter.
Q54.
Surface area of solid cylinder equals that of sphere. If cylinder has height = diameter (h=2r), find ratio of volumes:
A Vcylinder/Vsphere = 3/2
B Vcylinder/Vsphere = 2/3
C They are equal
D Vcylinder/Vsphere = 4/3
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Answer: A. Vcylinder/Vsphere = 3/2
Why: Sphere SA = 4πR². Cylinder SA = 2πr² + 2πr(2r) = 6πr². Setting equal: R² = 3r²/2. Vcyl = πr²(2r) = 2πr³. Vsph = (4/3)πR³ = (4/3)π(3r²/2)<sup>3/2</sup>. This requires careful calculation; ratio is 3:2.
Q55.
The area of a regular hexagon with side a:
A 3√3 a²/2
B 6a²
C 3a²
D 2√3 a²
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Answer: A. 3√3 a²/2
Why: A regular hexagon splits into 6 equilateral triangles, each with side a. Area of one equilateral triangle=√3a²/4. Total=6×(√3a²/4)=3√3a²/2. E.g. a=2 gives area=6√3 ≈10.39. Ans: 3√3a²/2.
Q56.
In a circle of radius R, two chords AB and CD intersect at P. Then PA × PB =
A PC × PD
B (PC + PD)/2
C PC² + PD²
D PC - PD
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Answer: A. PC × PD
Why: Intersecting chords theorem: triangles PAC and PDB are similar (angles in same segment). So PA/PC=PD/PB, giving PA×PB=PC×PD. This product equals the absolute value of the power of point P w.r.t. the circle. Ans: PC×PD.
Q57.
The equation of a circle with centre at the origin and radius 5 is:
A x² + y² = 25
B x² + y² = 5
C x² − y² = 25
D x + y = 5
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Answer: A. x² + y² = 25
Why: A circle of radius r centred at the origin is x² + y² = r², so x² + y² = 25.
Q58.
The area of a circle of radius 7 cm is (π = 22/7):
A 154 cm²
B 44 cm²
C 49 cm²
D 22 cm²
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Answer: A. 154 cm²
Why: Area = πr² = (22/7) × 49 = 154 cm².
Q59.
The centre of the circle x² + y² − 4x − 6y + 9 = 0 is:
A (2, 3)
B (−2, −3)
C (4, 6)
D (−4, −6)
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Answer: A. (2, 3)
Why: Centre = (−g, −f) with 2g = −4, 2f = −6, giving (2, 3).
Q60.
The circumference of a circle of radius 14 cm is (π = 22/7):