Q1: In a circular table cover of radius 70 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the total area of the design.
Sol: ∵ Radius of the circle = 70 cm and O is the centre of the circle.
∴ OA = OB = OC
Since ∆ABC is an equilateral triangle,
∴∠ABO = ∠BOC= ∠CDA = 120°
Draw OD⊥BC
Now in right ΔBDO,
∴ Area of equilateral ΔABC =
Also, area of the circle = πr^{2}
= 22 × 10 × 70 cm^{2} = 15400 cm^{2}
∴ Area of the shaded region= Area of CircleArea of equilateral ΔABC
= 15400 cm^{2} − (3675√3) cm^{2 }
= 15400 cm^{2} − (3675 × 1.73) cm^{2}
= 15400 cm^{2} − 6357.75 cm^{2} = 9042.25 cm^{2}.
Q2: Calculate the area other than the area common between two quadrants of the circles of radius 16 cm each, which is shown as the shaded region in the figure.
Sol:
Area of sector ADB
∴ Area of the shaded regionI =
Similarly, area of the shaded regionII 384/7 cm^{2}
∴ Total area of the shaded region = [Area of shaded regionI] + [Area of shaded regionII]
Q3: In the figure, PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Find the area of shaded region. (Take π = 3.14)
Sol: In right Δ RPQ,
PR^{2} + PQ^{2} = RQ^{2}
⇒ 7^{2} + 24^{2} = RQ^{2}
⇒ 49 + 576 = RQ^{2}
⇒ 625 = RQ^{2}
⇒ Radius of semicircle = 25/2 cm
∴ Area of semicircle RQP
∴ Area of the shaded region=Area of semicircle RQP − ar (right Δ RPQ)
=245.31 cm^{2} = 84 cm^{2} = 161.31 cm^{2}.
Q4: The area of an equilateral triangle is 49√3. Taking each angular point as centre, circles are drawn with radius equal to half length of the side of the triangle. Find the area of triangle not included in the circles. [Take √3 = 1.73]
Sol: Let the given equilateral triangle be ABC, such that its side = 14 cm.
Since each angle of an equilateral triangle = 60°,
∴ Area of a sector having θ as 60° and radius 14/2 i.e. 7 cm.
⇒ Area of 3 sectors
= 77 cm^{2}
∴ Area of the shaded region = [Area of equilateral ∆ ABC] − [Area of 3 sectors]
= 84.77 cm^{2} − 77 cm^{2} = 7.77 cm^{2}.
Q5: Four equal circles are described at the four corners of the square so that each touches two of the others, the shaded area enclosed between the circles is 22/7 cm^{2} Find the radius of each circle.
Sol: Let 'r' cm be the radius of each circle.
∴ side of the square = 2r
Now, area of shaded region
But the area of shaded region
But r cannot be − ve, so r = 2 cm
Q6: A square OABC is inscribed in a quadrant OABQ of a circle as shown in the figure. If OA = 14 cm, find the area of the shaded region. [use π = 22/7]
Sol: OABC is a square with side = 14 cm.
∴ Area of the square OABC = 14 cm × 14 cm = 196 cm^{2}
Now, the diagonal of the square OABC
⇒ Radius of the quadrant OPBQ
∴ Area of the quadrant OABQ
∴ Area of the shaded region
= 308 cm^{2} − 196 cm^{2} = 112 cm^{2}.
Q7: In the figure, ABDC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with diameter BC. Find the area of the shaded region.
Sol: We have, in the right Δ ABC,
BC^{2} = AB^{2} + AC^{2} = 142 + 142 = 2 (14)^{2}
∴ Radius of the semicircle =
∴ Area of semicircle BEC
Area of the quadrant with radius 14 cm,
= 98 cm^{2}
Area of the shaded region
= 154 cm^{2} + 98 cm^{2} − 154 cm^{2} = 98 cm^{2}.
Q8: In the figure, find the perimeter of the shaded region where, ADC, AEB and BFC are semicircles on diameters AC, AB and BC respectively.
Sol: Diameter of semicircle ADC
= 2.1 cm
∴ Circumference of semicircle ADC
Diameter of semicircle AEB = 2.8 cm
⇒ Radius of semicircle
∴ Circumference of semicircle
Diameter of semicircle BFC = 1.4 cm
⇒ Radius of semicircle BFC = 0.7 cm
∴ Circumference of semicircle BFC
∴ Total perimeter of the shaded region = 6.6 cm + 4.4 cm + 2.2 cm = 13.2 cm.
Q9: In the figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective midpoints D, E and F. Find the area of the shaded region.
[use π = 3.14]
Sol. ∵ ΔABC is an equilateral triangle.
∴∠A = ∠B = ∠C = 60°
Area of sector AFEA
Area of all the three sectors
Thus, area of the shaded region = 39.25 cm^{2}.
Q10: In figure OABC is a quadrant of a circle of radius 7 cm. If OD = 4 cm, find the area of the shaded region. [Use π = 22/7]
Sol: We have, the centre of the circle as ‘O’ and radius (r) = 7 cm
Area of the quadrant OABC
= 7 × 2 = 14 cm^{2}
∴ Area of the shaded region = (Area of the quadrant OABC) – (Area of ∆COD)
120 videos463 docs105 tests

1. What is the formula to find the area of a circle? 
2. How is the circumference of a circle related to its radius? 
3. Can the area of a circle be negative? 
4. How can the radius of a circle be calculated if the area is given? 
5. What is the value of pi (π) and why is it used in the formulas for a circle? 
120 videos463 docs105 tests


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