Areas Related to Circles 2-Mark Questions Class 10: CBSE PYQs with Answers

10π = (θ/360) × 2π × 18

θ = 100°

Answer: 100°

Distance in one minute = 66 × 1000 / 60 = 1100 m = 110000 cm

Wheel circumference = ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 140 = 440 cm

Number of revolutions = (110000)/(440)$\frac{\text{110000}}{\text{440}}$ = 250

Answer: 250 revolutions

Angle = ((360)/(60)$\frac{\text{360}}{\text{60}}$) × 20 = 120°

Area = ((120)/(360)$\frac{\text{120}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14² = (616)/(3)$\frac{\text{616}}{\text{3}}$ cm²

Answer: (616)/(3)$\frac{\text{616}}{\text{3}}$ cm² (approximately 205.33 cm²)

The three corner sectors each have radius 14 m; their angles add to 180°.

The sectors do not overlap: each side is greater than 14 + 14 = 28 m.

Total area = ((180)/(360)$\frac{\text{180}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14² = 308 m²

Answer: 308 m²

Angle = ((360)/(60)$\frac{\text{360}}{\text{60}}$) × 5 = 30°

Area = ((30)/(360)$\frac{\text{30}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14² = (154)/(3)$\frac{\text{154}}{\text{3}}$ cm²

Answer: (154)/(3)$\frac{\text{154}}{\text{3}}$ cm² (approximately 51.33 cm²)

Arc length = ((60)/(360)$\frac{\text{60}}{\text{360}}$) × 2 × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 42 = 44 cm

Answer: 44 cm

Angle = ((360)/(60)$\frac{\text{360}}{\text{60}}$) × 45 = 270°

Area = ((270)/(360)$\frac{\text{270}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 21² = 1039.5 cm²

Answer: 1039.5 cm²

Angle swept in 5 minutes = ((5)/(60)$\frac{\text{5}}{\text{60}}$) × 360° = 30°.

Area = ((30)/(360)$\frac{\text{30}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 4.2 × 4.2 = 4.62 cm².

Answer: 4.62 cm²

Elapsed time = 35 minutes.

Angle swept = ((35)/(60)$\frac{\text{35}}{\text{60}}$) × 360° = 210°.

Area swept = ((210)/(360)$\frac{\text{210}}{\text{360}}$) × π × 6².

Using π = (22)/(7)$\frac{\text{22}}{\text{7}}$, area = 66 cm².

Answer: 66 cm²

Let the interior angles of ABCD be A, B, C and D.

A + B + C + D = 360°.

Total shaded area = [(A + B + C + D)/360] × π × 7².

= π × 49 = 154 cm², using π = 22/7.

Answer: 154 cm²

Radius r = (28)/(2)$\frac{\text{28}}{\text{2}}$ = 14 cm; arc length l = 22 cm.

Sector area = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × r × l.

= ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 14 × 22 = 154 cm².

Answer: 154 cm²

Radius r = (20)/(2)$\frac{\text{20}}{\text{2}}$ = 10 cm.

The central angle is 60°, so the triangle formed by the chord and radii is equilateral.

Sector area = ((60)/(360)$\frac{\text{60}}{\text{360}}$) × 3.14 × 10² = (157)/(3)$\frac{\text{157}}{\text{3}}$ cm².

Triangle area = ((√3)/(4)$\frac{\text{√3}}{\text{4}}$) × 10² = (173)/(4)$\frac{\text{173}}{\text{4}}$ cm², using √3 = 1.73.

Minor segment area = sector area − triangle area.

= (157)/(3)$\frac{\text{157}}{\text{3}}$ − (173)/(4)$\frac{\text{173}}{\text{4}}$ = (109)/(12)$\frac{\text{109}}{\text{12}}$ cm² ≈ 9.08 cm².

Answer: (109)/(12)$\frac{\text{109}}{\text{12}}$ cm² ≈ 9.08 cm²

Area = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14² − ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 14² = 154 − 98.

Answer: 56 cm²

Arc length = 75 − 2 × 21 = 33 cm.

Area of sector = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × r × arc length = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 21 × 33.

Answer: 346.5 cm²

Area of 90° sector = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14² = 154 cm².

Area of triangle AOB = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 14 × 14 = 98 cm².

Area of minor segment = 154 − 98 = 56 cm².

Answer: 56 cm²

22 = ((60)/(360)$\frac{\text{60}}{\text{360}}$) × 2 × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × r, so r = 21 cm.

Area = ((60)/(360)$\frac{\text{60}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 21² = 231 cm².

Answer: 231 cm²

Arc length = 27.2 − 2 × 5.6 = 16 cm.

Area of sector = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × radius × arc length = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 5.6 × 16 = 44.8 cm².

Answer: 44.8 cm²

The two radii have length 5√2 cm.

10² = (5√2)² + (5√2)², so the central angle is 90° by the converse of Pythagoras.

Sector area = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × 3.14 × (5√2)² = 39.25 cm².

Triangle area = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × (5√2)² = 25 cm².

Minor segment area = 39.25 − 25 = 14.25 cm².

Answer: 14.25 cm²

Let R be the radius of the required circle.

πR² = π × 40² + π × 9².

R² = 1600 + 81 = 1681, so R = 41 cm.

Diameter = 2R = 82 cm.

Answer: 82 cm

Area of the 90° sector = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × 3.14 × 10² = 78.5 cm².

The two radii are perpendicular, so the triangle area = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 10 × 10 = 50 cm².

Area of the minor segment = sector area − triangle area.

= 78.5 − 50 = 28.5 cm².

Answer: 28.5 cm²

The grazed parts are three sectors of radius 14 m at the triangle vertices.

The sectors do not overlap: each side is greater than 14 + 14 = 28 m.

The three sector angles add to 180°.

Total grazed area = ((180)/(360)$\frac{\text{180}}{\text{360}}$) × π × 14².

= ((1)/(2)$\frac{\text{1}}{\text{2}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 196 = 308 m².

Answer: 308 m²

Area of the minor sector = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × π × 5² = 25π/4 cm².

The radii are perpendicular, so triangle area = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 5 × 5 = (25)/(2)$\frac{\text{25}}{\text{2}}$ cm².

Minor segment area = (25π/4 − (25)/(2)$\frac{\text{25}}{\text{2}}$) cm².

Major segment area = circle area − minor segment area.

= 25π − (25π/4 − (25)/(2)$\frac{\text{25}}{\text{2}}$) = (75π/4 + (25)/(2)$\frac{\text{25}}{\text{2}}$) cm².

Answer: (75π/4 + (25)/(2)$\frac{\text{25}}{\text{2}}$) cm²

The figure shows a quadrant centred at A, with radius 5 cm.

Area = 10² − ((90)/(360)$\frac{\text{90}}{\text{360}}$) × 3.14 × 5² = 100 − 19.625 = 80.375 cm² ≈ 80.38 cm²

Answer: 80.375 cm² ≈ 80.38 cm²

Area of sector = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14² = 154 cm²

Area of triangle = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 14² = 98 cm²

Minor segment = 154 − 98 = 56 cm²

Answer: 56 cm²

The figure shows four circles centred at the four corners of a square of side 14 cm.

Shaded area = 14² − 4 × ((90)/(360)$\frac{\text{90}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 7² = 196 − 154 = 42 cm²

Answer: 42 cm²

Shaded area = area of quadrant OPRQ − area of triangle POS.

Area = ((1)/(4)$\frac{\text{1}}{\text{4}}$) × 3.14 × 10² − ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 10 × 6 = 78.5 − 30.

Answer: 48.5 cm²

Remaining area = πr² − ((1)/(4)$\frac{\text{1}}{\text{4}}$)πr² = ((3)/(4)$\frac{\text{3}}{\text{4}}$)πr².

Area = ((3)/(4)$\frac{\text{3}}{\text{4}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 70² = 11550 cm².

Answer: 11550 cm²

Arc length l = 80 − 2 × 15 = 50 cm.

Area = ((1)/(2)$\frac{\text{1}}{\text{2}}$)rl = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × 15 × 50.

Answer: 375 cm²

Since AB ∥ DC, ∠A = 180° − 60° = 120° and ∠C = 180° − 120° = 60°.

Shaded area = ((120)/(360)$\frac{\text{120}}{\text{360}}$)π × 3² + ((60)/(360)$\frac{\text{60}}{\text{360}}$)π × 6² = 3π + 6π.

Answer: 9π cm²

Each circle has radius (10)/(2)$\frac{\text{10}}{\text{2}}$ = 5 cm.

Shaded area = rectangle area − 2 × circle area = 20 × 10 − 2 × 3.14 × 5² = 43 cm².

Answer: 43 cm²

Area of sector = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × radius × arc length.

66 = ((1)/(2)$\frac{\text{1}}{\text{2}}$) × r × 22, so r = 6 cm.

Answer: 6 cm

The three sector angles are A, B and C, with A + B + C = 180°.

Total removed area = [(A + B + C)/360] × π × 14².

= ((1)/(2)$\frac{\text{1}}{\text{2}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 196 = 308 cm².

Answer: 308 cm²

The horizontal and vertical extent of the white region is 14 − 3 − 3 = 8 cm.

Let a be the side of its central square. Each semicircle has diameter a.

Total width = a/2 + a + a/2 = 2a, so a = 4 cm.

The semicircle radius is 2 cm.

Unshaded area = square area + four semicircle areas.

= 4² + 4 × ((1)/(2)$\frac{\text{1}}{\text{2}}$) × π × 2² = (16 + 8π) cm².

Answer: (16 + 8π) cm²

The figure shades the annular sector with central angle 60°.

Perimeter = ((60)/(360)$\frac{\text{60}}{\text{360}}$) × 2π × 3 + ((60)/(360)$\frac{\text{60}}{\text{360}}$) × 2π × 2 + 2(3 − 2) = 5π/3 + 2.

Using π = (22)/(7)$\frac{\text{22}}{\text{7}}$, perimeter = (152)/(21)$\frac{\text{152}}{\text{21}}$ cm.

Answer: (152)/(21)$\frac{\text{152}}{\text{21}}$ cm ≈ 7.24 cm

πr² = ((120)/(360)$\frac{\text{120}}{\text{360}}$)πR², so r²/R² = 1/3.

Taking positive square roots, r : R = 1 : √3.

Answer: 1 : √3

From the figure ∠AOB = 60°, so triangle AOB is equilateral and AB = 7 cm.

Length of minor arc AB = ((60)/(360)$\frac{\text{60}}{\text{360}}$) × 2 × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 7 = (22)/(3)$\frac{\text{22}}{\text{3}}$ cm.

Perimeter of shaded minor segment = arc AB + chord AB = (22)/(3)$\frac{\text{22}}{\text{3}}$ + 7 = (43)/(3)$\frac{\text{43}}{\text{3}}$ cm.

Answer: (43)/(3)$\frac{\text{43}}{\text{3}}$ cm ≈ 14.33 cm

AD = √(OA² + OD²) = √(8² + 6²) = 10 cm; CD = 8 − 6 = 2 cm.

Arc ABC = ((90)/(360)$\frac{\text{90}}{\text{360}}$) × 2 × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 8 = (88)/(7)$\frac{\text{88}}{\text{7}}$ cm.

Perimeter = arc ABC + AD + CD = (88)/(7)$\frac{\text{88}}{\text{7}}$ + 10 + 2 = (172)/(7)$\frac{\text{172}}{\text{7}}$ cm.

Answer: (172)/(7)$\frac{\text{172}}{\text{7}}$ cm ≈ 24.57 cm

Perimeter = remaining 270° arc + two radii = ((3)/(4)$\frac{\text{3}}{\text{4}}$) × 2πr + 2r.

With r = 10 and π = (22)/(7)$\frac{\text{22}}{\text{7}}$, perimeter = (330)/(7)$\frac{\text{330}}{\text{7}}$ + 20 = (470)/(7)$\frac{\text{470}}{\text{7}}$ cm.

Answer: (470)/(7)$\frac{\text{470}}{\text{7}}$ cm ≈ 67.14 cm

The figure shows the major sector: its arc subtends the reflex angle 360° − 90° = 270°.

Perimeter = ((270)/(360)$\frac{\text{270}}{\text{360}}$) × 2 × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 42 + 2 × 42 = 198 + 84.

Answer: 282 cm

Total shaded angle = 35° + 50° + 95° = 180°.

Area = ((180)/(360)$\frac{\text{180}}{\text{360}}$) × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 5² = (275)/(7)$\frac{\text{275}}{\text{7}}$ cm².

Answer: (275)/(7)$\frac{\text{275}}{\text{7}}$ cm² ≈ 39.29 cm²

((√3)/(4)$\frac{\text{√3}}{\text{4}}$)a² = 121√3, so a = 22 cm. Wire length = 3a = 66 cm.

2πr = 66 ⇒ r = 10.5 cm, using π = 22/7.

Area of circle = ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 10.5².

Answer: 346.5 cm²

Perimeter = two radii + arc length.

= 2r + ((90)/(360)$\frac{\text{90}}{\text{360}}$) × 2πr.

= 2 × 14 + ((1)/(4)$\frac{\text{1}}{\text{4}}$) × 2 × ((22)/(7)$\frac{\text{22}}{\text{7}}$) × 14.

= 28 + 22 = 50 cm.

Answer: 50 cm

Let R be the radius of the required circle.

πR² = π × 24² + π × 7².

R² = 576 + 49 = 625, so R = 25 cm.

Diameter = 2R = 50 cm.

Answer: 50 cm