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Areas Related to Circles - Areas of Combinations of Plane Figures

Grade 10CBSE

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

🔑Concepts

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Combinations of Plane Figures: Calculating areas of composite figures involves identifying individual standard shapes like circles, sectors, segments, squares, rectangles, and triangles. Visually, these problems often present a large 'outer' boundary containing 'inner' shapes that are either added to or subtracted from the total area.

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Subtraction Method for Shaded Regions: The most common strategy for finding the area of a shaded region is to calculate the area of the entire outer figure and subtract the area of the unshaded or 'cut-out' parts. Visually, imagine a 'doughnut' shape where the area is the outer circle minus the inner circle.

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Area of a Sector: A sector is the portion of a circle enclosed by two radii and an arc. Visually, it looks like a slice of pizza. The area is proportional to the central angle θ\theta subtended at the center.

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Area of a Segment: A segment is the region between a chord and the corresponding arc of a circle. Visually, it looks like a 'cap' on top of a circle. To find its area, you calculate the area of the sector and subtract the area of the triangle formed by the center and the two endpoints of the chord.

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Geometric Symmetry: Many complex figures are symmetrical, meaning you can calculate the area of one part and multiply it by the number of identical parts. For example, if a square has four identical quadrants at its corners, you calculate the area of one quadrant and multiply by 4.

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Inscribed and Circumscribed Figures: Figures are often drawn inside one another. A circle 'inscribed' in a square touches all four sides, meaning the circle's diameter equals the square's side length. Conversely, if a square is 'inscribed' in a circle, the square's diagonal equals the circle's diameter.

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Boundary vs. Area: In combined figures, the perimeter of the resulting shape is not simply the sum of the perimeters of the original shapes. Only the 'exposed' outer edges contribute to the final perimeter, while internal boundaries are ignored.

📐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.