HOTS Questions on Chapter 11 Class 10 Areas Related To Circles

HOTS Questions on Class 10 Maths Chapter 11 Areas Related to Circles help students understand sectors, segments, arcs, and shaded regions in a simple way. They improve knowledge of circle formulas, area calculations, and real life application based problems. This chapter builds a strong base for solving geometry questions and improving math skills. A downloadable PDF is also available for easy revision and practice. 

HOTS Questions Based on Sectors of a Circle

Question 1: A sector of a circle of radius 12 cm has an angle of 120°. Find the area of the sector. Also find the area of the triangle formed by the two radii and the chord. Hence find the area of the minor segment.

Answer:

Given: r = 12 cm, θ = 120°

Step 1: Area of sector

= (θ/360) × πr²

= (120/360) × (22/7) × 144

= (1/3) × (22/7) × 144

= 3168/21

= 150.86 cm²

Step 2: Area of triangle OAB

= (1/2) × r² × sin θ

= (1/2) × 144 × sin 120°

= (1/2) × 144 × (√3/2)

= 36√3

= 36 × 1.732

= 62.35 cm²

Step 3: Area of minor segment

= Area of sector - Area of triangle

= 150.86 - 62.35

= 88.51 cm²

Answer: Sector = 150.86 cm², Triangle = 62.35 cm², Segment = 88.51 cm²

Question 2: Two circular discs of the same radius 14 cm are placed such that each passes through the center of the other. Find the area of the overlapping region.

Answer:

Each circle passes through center of other.

Distance between centers = 14 cm = r

Each sector has θ = 60° (equilateral triangle)

Wait: central angle = 120° for this configuration.

In each circle:

The overlapping region = 2 segments

For each circle, the chord connecting

intersection points subtends 120° at center.

Area of one sector (120°):

= (120/360) × π × 14²

= (1/3) × (22/7) × 196

= (1/3) × 616

= 205.33 cm²

Area of triangle (equilateral, side = 14):

= (√3/4) × 14²

= (√3/4) × 196

= 49√3

= 84.87 cm²

Area of one segment:

= 205.33 - 84.87 = 120.46 cm²

Total overlapping area = 2 segments

= 2 × 120.46

= 240.92 cm²

Answer: Overlapping area = 240.92 cm²

Question 3: A student claims that if you double the radius of a sector while keeping the angle the same, the area becomes double. Is the student correct? Justify.

Answer:

The student is INCORRECT.

Area of sector = (θ/360) × πr²

If r is doubled, new radius = 2r

New area = (θ/360) × π(2r)²

= (θ/360) × π × 4r²

= 4 × [(θ/360) × πr²]

= 4 × original area

So area becomes FOUR TIMES (not double)

when radius is doubled.

This is because area depends on r²

(radius squared), not r.

Example:

r = 7, θ = 90°

Area₁ = (90/360) × (22/7) × 49 = 38.5 cm²

r = 14, θ = 90°

Area₂ = (90/360) × (22/7) × 196 = 154 cm²

154/38.5 = 4 (four times, not two times)

HOTS Questions Based on Segments of a Circle

Question 4: A chord of a circle of radius 10 cm subtends a right angle at the center. Find the areas of: a) The minor segment b) The major segment c) What percentage of the circle does the minor segment occupy?

Answer:

Given: r = 10 cm, θ = 90°

Step 1: Area of minor sector

= (90/360) × π × 100

= (1/4) × (22/7) × 100

= 2200/28

= 78.57 cm²

Step 2: Area of triangle OAB

= (1/2) × r × r (since θ = 90°)

= (1/2) × 10 × 10

= 50 cm²

Step 3: Area of minor segment

= 78.57 - 50

= 28.57 cm²

Step 4: Total circle area

= πr² = (22/7) × 100 = 314.28 cm²

Step 5: Area of major segment

= 314.28 - 28.57

= 285.71 cm²

Step 6: Percentage occupied by minor segment

= (28.57/314.28) × 100

= 9.09%

Answer: Minor = 28.57 cm², Major = 285.71 cm², Percentage = 9.09%

Question 5: In a circle of radius 21 cm, a chord AB makes an angle of 60° at the center O. Find the area of the minor segment cut off by chord AB.

Answer:

Given: r = 21 cm, θ = 60°

Step 1: Area of sector OAB

= (60/360) × (22/7) × 21²

= (1/6) × (22/7) × 441

= (1/6) × 1386

= 231 cm²

Step 2: Area of triangle OAB

Since OA = OB = 21 cm and θ = 60°:

Triangle OAB is equilateral (all sides = 21)

Area = (√3/4) × 21²

= (√3/4) × 441

= 110.25√3

≈ 110.25 × 1.732

≈ 190.95 cm²

Step 3: Area of minor segment

= 231 - 190.95

= 40.05 cm²

Answer: Area of minor segment = 40.05 cm²

Question 6: Compare the area of a minor segment when: Case 1: r = 14 cm, θ = 90° Case 2: r = 7 cm, θ = 180° (semicircle)

Which segment has greater area?

Answer:

Case 1: r = 14, θ = 90°

Sector area = (1/4) × (22/7) × 196 = 154 cm²

Triangle area = (1/2) × 14 × 14 = 98 cm²

Segment = 154 - 98 = 56 cm²

Case 2: r = 7, θ = 180° (semicircle)

Segment = semicircle (no triangle below diameter)

Area = (1/2) × (22/7) × 49

= (1/2) × 154

= 77 cm² 

Comparison:

Case 1 segment = 56 cm²

Case 2 segment = 77 cm²

Case 2 has greater area even with

smaller radius because θ = 180°.

Answer: Case 2 (semicircle) has greater area at 77 cm²

HOTS Questions Based on Shaded Regions

Question 7: A square ABCD of side 14 cm has four quadrant circles drawn with each vertex as center and radius 7 cm (half the side). Find the area of the shaded region inside the square but outside all quadrants.

Answer:

Step 1: Area of square

= 14² = 196 cm²

Step 2: Each quadrant has r = 7, angle = 90°

Area of one quadrant

= (90/360) × (22/7) × 7²

= (1/4) × (22/7) × 49

= (1/4) × 154

= 38.5 cm²

Step 3: Total area of 4 quadrants

= 4 × 38.5 = 154 cm²

Step 4: Shaded area (inside square, outside quadrants)

= 196 - 154

= 42 cm²

Answer: Shaded area = 42 cm²

Question 8: Find the area of the shaded region in a figure where an equilateral triangle of side 28 cm has a circle inscribed in it, and three sectors (each of 60°) are drawn at the three vertices with radius 14 cm.

Answer:

Step 1: Area of equilateral triangle

= (√3/4) × 28²

= (√3/4) × 784

= 196√3

= 196 × 1.732

= 339.48 cm²

Step 2: Each vertex sector has θ = 60°, r = 14

(Interior angle of equilateral triangle = 60°)

Area of one sector = (60/360) × (22/7) × 14²

= (1/6) × (22/7) × 196

= (1/6) × 616

= 102.67 cm²

Step 3: Total area of 3 sectors

= 3 × 102.67 = 308 cm²

Step 4: Shaded area = Triangle - 3 sectors

= 339.48 - 308

= 31.48 cm²

Answer: Shaded area = 31.48 cm²

Question 9: ABCD is a square of side 10 cm. Semicircles are drawn on each side as diameter, all pointing inward. Find the total shaded area inside the square but outside all semicircles.

Answer:

Step 1: Area of square = 10² = 100 cm²

Step 2: Semicircle diameter = 10 cm, r = 5 cm

Area of one semicircle = (1/2) × π × 25

= (1/2) × (22/7) × 25

= 275/7

= 39.28 cm²

Step 3: Total area of 4 semicircles

= 4 × 39.28 = 157.14 cm²

But semicircles overlap in the center!

Actual area covered = Total - Overlaps

For a square with inward semicircles on all 4 sides:

Covered area = π × r² = π × 25 = 78.57 cm²

(The four semicircles together cover a full circle area)

Step 4: Shaded area = Square - Circle

= 100 - 78.57

= 21.43 cm²

Answer: Shaded area = 21.43 cm²

HOTS Questions Based on Composite Figures

Question 10: A rectangular park is 40 m long and 30 m wide. Two semicircular flower beds are made on the two shorter sides (breadth). Find the cost of planting grass in the remaining rectangular area at ₹5 per m².

Answer:

Step 1: Rectangle area = 40 × 30 = 1200 m²

Step 2: Semicircle on breadth (30 m side)

Diameter = 30 m, r = 15 m

Area of one semicircle = (1/2) × π × 15²

= (1/2) × (22/7) × 225

= 2475/7

= 353.57 m²

Step 3: Two semicircles total = 2 × 353.57 = 707.14 m²

Step 4: Remaining grass area

= 1200 - 707.14

= 492.86 m²

Step 5: Cost = 492.86 × 5 = ₹2464.28

Answer: Cost = ₹2464.28

Question 11: In the figure, ABCD is a square of side 14 cm. A quadrant of a circle is drawn from B with radius BC = 14 cm. Also a semicircle is drawn on BC as diameter. Find the area of the shaded region.

Answer:

Step 1: Area of quadrant (center B, r = 14)

= (1/4) × (22/7) × 14²

= (1/4) × (22/7) × 196

= (1/4) × 616

= 154 cm²

Step 2: Area of semicircle on BC

Diameter = BC = 14 cm → r = 7 cm

= (1/2) × (22/7) × 7²

= (1/2) × (22/7) × 49

= (1/2) × 154

= 77 cm²

Step 3: Area of triangle ABC

= (1/2) × 14 × 14 = 98 cm²

Step 4: Shaded area

= Area of semicircle + (Area of quadrant - Area of triangle)

= 77 + (154 - 98)

= 77 + 56

= 133 cm²

Answer: Shaded area = 133 cm²

HOTS Questions Based on Real Life Applications

Question 12: A gardener wants to design a circular garden of radius 21 m. Inside the garden, he plans to grow roses in two sectors, each having a central angle of 45°. The remaining area will be grass. Find the area used for roses and the area of grass. Find the cost of growing roses at ₹10 per m² and grass at ₹5 per m².

Answer:

Step 1: Total circle area

= π × 21² = (22/7) × 441 = 1386 m²

Step 2: Each rose sector has θ = 45°, r = 21

Area of one sector = (45/360) × (22/7) × 441

= (1/8) × 1386

= 173.25 m²

Step 3: Total rose area = 2 × 173.25 = 346.5 m²

Step 4: Grass area = 1386 - 346.5 = 1039.5 m²

Step 5: Cost calculation

Rose cost = 346.5 × 10 = ₹3465

Grass cost = 1039.5 × 5 = ₹5197.50

Total cost = ₹3465 + ₹5197.50 = ₹8662.50

Answer: Rose area = 346.5 m², Grass area = 1039.5 m², Total cost = ₹8662.50

Question 13: A running track has two straight sections of 90 m each and two semicircular ends. The inner radius of the semicircular ends is 35 m and the track is 7 m wide throughout. Find: a) The total area of the track b) The distance a runner covers in one complete round on the inner edge

Answer:

Step 1: Track consists of:

- Two straight rectangular sections

- Two semicircular ring sections (annular sectors)

Inner radius (r) = 35 m

Outer radius (R) = 35 + 7 = 42 m

Straight length = 90 m

Step 2: Area of two straight sections

= 2 × (90 × 7) = 1260 m²

Step 3: Area of two semicircular rings

= 2 × (1/2) × π(R² - r²)

= π(R² - r²)

= (22/7) × (42² - 35²)

= (22/7) × (1764 - 1225)

= (22/7) × 539

= 22 × 77

= 1694 m²

Step 4: Total track area = 1260 + 1694 = 2954 m²

Step 5: Inner edge distance (one round):

= 2 straight sections + 2 semicircles

= 2 × 90 + 2 × π × r

= 180 + 2 × (22/7) × 35

= 180 + 220

= 400 m

Answer: Track area = 2954 m², One round on inner edge = 400 m

HOTS Question 14: An architect designs a window with a rectangular base of width 1.2 m and height 1.5 m, topped by a semicircle of the same width. Find the total area of glass needed for the window. If glass costs ₹450 per m², find the total cost.

Answer:

Step 1: Rectangle area

= 1.2 × 1.5 = 1.8 m²

Step 2: Semicircle on top

Diameter = 1.2 m, r = 0.6 m

Area = (1/2) × π × 0.6²

= (1/2) × (22/7) × 0.36

= (1/2) × 1.131

= 0.566 m²

Step 3: Total glass area

= 1.8 + 0.566

= 2.366 m²

Step 4: Total cost

= 2.366 × 450

= ₹1064.57

Answer: Glass area = 2.366 m², Cost = ₹1064.57

Quick Formula Reference

Area of sector = (θ/360) × πr²

Arc length = (θ/360) × 2πr

Area of minor segment = Sector - Triangle

Area of triangle in sector = (1/2) × r² × sin θ

Area of major segment = πr² - Minor segment

Area of annular ring = π(R² - r²)

Area of quadrant = (1/4) × πr²

Area of semicircle = (1/2) × πr²

Perimeter of sector = 2r + arc length

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