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Mensuration: Areas Related to Circles - Solve perimeter and area problems involving combinations of circular plane figures

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Combining regular polygons and circles: Area problems often involve subtraction or addition of plane figures. For instance, finding the area of a shaded region between a square and inscribed/circumscribed circles requires calculating the difference between their individual areas.

A circle inscribed inside a square showing the relationship between side and radius.
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Design analysis in circular figures: Complex patterns like flowers or gears can be broken down into identical sectors or segments. The total area is the sum of these congruent parts, often calculated as n×Area of one partn \times \text{Area of one part}, where nn is the number of symmetric regions.

A circle divided into six equal sectors with alternating shaded regions.
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The Perimeter of combined figures: When calculating the perimeter of a shaded region, identify the boundaries. The perimeter is the sum of all external and internal boundary lengths, which may include straight edges (sides of polygons) and curved edges (circular arcs).

A running track shape showing straight lengths and semi-circular ends.
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Quarter-circle subtractions: Problems frequently involve a square with quadrants removed from corners or semi-circles drawn on sides. The area of the remaining region is calculated by subtracting the sum of the areas of these circular components from the total area of the square.

📐Formulae

Area of a circle: A=πr2A = \pi r^2

Circumference of a circle: C=2πrC = 2 \pi r

Area of a sector with central angle θ\theta: A=θ360∘×πr2A = \frac{\theta}{360^\circ} \times \pi r^2

Length of an arc with central angle θ\theta: l=θ360∘×2πrl = \frac{\theta}{360^\circ} \times 2 \pi r

Area of a segment of a circle: A=θ360∘πr2−12r2sin⁡θA = \frac{\theta}{360^\circ} \pi r^2 - \frac{1}{2} r^2 \sin \theta

Area of an equilateral triangle: A=34a2A = \frac{\sqrt{3}}{4} a^2

Area of a square: A=s2A = s^2

Relationship between radius (rr) and side (aa) of an inscribed equilateral triangle: a=r3a = r \sqrt{3}

💡Examples

Problem 1:

A square ABCDABCD has a side of 1414 cm. From each corner of the square, a quadrant of a circle of radius 3.53.5 cm is cut and also a circle of diameter 77 cm is cut from the center. Find the area of the remaining (shaded) portion of the square.

Solution:

Step 1: Calculate the area of the square ABCDABCD. Areasquare=side2=14×14=196 cm2Area_{square} = side^2 = 14 \times 14 = 196 \text{ cm}^2 Step 2: Calculate the area of the four quadrants at the corners. Since 4×quadrant=1 full circle4 \times \text{quadrant} = 1 \text{ full circle}: Area4_quadrants=πr2=227×3.5×3.5=38.5 cm2Area_{4\_quadrants} = \pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5 \text{ cm}^2 Step 3: Calculate the area of the central circle. The diameter is 77 cm, so the radius r=3.5r = 3.5 cm. Areacenter_circle=πr2=227×3.5×3.5=38.5 cm2Area_{center\_circle} = \pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5 \text{ cm}^2 Step 4: Find the area of the remaining portion. Arearemaining=Areasquare−(Area4_quadrants+Areacenter_circle)Area_{remaining} = Area_{square} - (Area_{4\_quadrants} + Area_{center\_circle}) Arearemaining=196−(38.5+38.5)=196−77=119 cm2Area_{remaining} = 196 - (38.5 + 38.5) = 196 - 77 = 119 \text{ cm}^2

Explanation:

We use the subtraction method. The total area of the square is found first, then the areas of the parts 'removed' (the four quadrants and the central circle) are calculated and subtracted from the total.

Problem 2:

Find the area of the shaded region in a circle of radius 66 cm where a central angle of 60∘60^\circ forms a sector, and an equilateral triangle is formed by the radii and the chord joining their endpoints.

Solution:

Step 1: Area of the sector with θ=60∘\theta = 60^\circ and r=6r = 6 cm. Areasector=60360×π×62=16×36π=6π cm2Area_{sector} = \frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times 36\pi = 6\pi \text{ cm}^2 Step 2: Since the central angle is 60∘60^\circ and the two sides are radii (equal), the triangle is equilateral with side a=6a = 6 cm. Areatriangle=34×62=34×36=93 cm2Area_{triangle} = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \text{ cm}^2 Step 3: Area of the shaded segment. Areasegment=Areasector−Areatriangle=(6π−93) cm2Area_{segment} = Area_{sector} - Area_{triangle} = (6\pi - 9\sqrt{3}) \text{ cm}^2 Taking π≈3.14\pi \approx 3.14 and 3≈1.73\sqrt{3} \approx 1.73: Area≈(6×3.14)−(9×1.73)=18.84−15.57=3.27 cm2Area \approx (6 \times 3.14) - (9 \times 1.73) = 18.84 - 15.57 = 3.27 \text{ cm}^2

Explanation:

This problem requires calculating the area of a minor segment. We identify that a 60∘60^\circ sector with equal radii must contain an equilateral triangle, then subtract the triangle's area from the sector's area.

Problem 3:

A square lawn of side 2020 m has four semi-circular flower beds at each side. Find the total area of the flower beds. (Use π=3.14\pi = 3.14)

Square with semi-circles attached to all four sides.

Solution:

  1. Side of the square s=20s = 20 m.
  2. Since the flower beds are semi-circles on the sides, the diameter of each semi-circle is equal to the side of the square.
  3. Diameter d=20d = 20 m, so radius r=202=10r = \frac{20}{2} = 10 m.
  4. Total area of 4 semi-circles = 4×12πr2=2πr24 \times \frac{1}{2} \pi r^2 = 2 \pi r^2.
  5. Area = 2×3.14×(10)22 \times 3.14 \times (10)^2
  6. Area = 2×3.14×100=6282 \times 3.14 \times 100 = 628 m2m^2.

Explanation:

The problem asks for the area of the components added to the square. Since there are four semi-circles with the same diameter, they effectively form two full circles.

Problem 4:

In the given figure, ABCDABCD is a square of side 1010 cm. Semi-circles are drawn with each side of the square as diameter. Find the area of the shaded region (the four petals). (Use π=3.14\pi = 3.14)

Square with four internal overlapping semi-circles forming a petal pattern.

Solution:

  1. Area of Square ABCD=10×10=100ABCD = 10 \times 10 = 100 cm2cm^2.
  2. Let the unshaded regions be I, II, III, and IV.
  3. Area of (I + III) = Area of Square - Area of 2 semi-circles (on sides AD and BC)
  4. Area (I + III) = 100−(2×12π×52)=100−3.14×25=100−78.5=21.5100 - (2 \times \frac{1}{2} \pi \times 5^2) = 100 - 3.14 \times 25 = 100 - 78.5 = 21.5 cm2cm^2.
  5. Similarly, Area (II + IV) = 21.521.5 cm2cm^2.
  6. Area of shaded region = Area of Square - Area (I + II + III + IV)
  7. Area of shaded region = 100−(21.5+21.5)=100−43=57100 - (21.5 + 21.5) = 100 - 43 = 57 cm2cm^2.

Explanation:

To find the area of the overlapping petals, we first find the area of the 'empty' spaces by subtracting semi-circles from the square. Then, we subtract those empty spaces from the total square area.