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source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch11
exercise: ex11.2
difficulty: intermediate
---
Class 10 | Ch11: Areas Related to Circles | Exercise 11.2
Combined areas: circles, sectors, and plane figures together.
Q1. Find area of a figure formed by joining midpoints of sides of a rhombus with diagonals 16cm and 12cm.
Answer: Midpoints form a rectangle. Length=8, width=6 (half-diagonals). Area=48cm². ✅
Q2. Find area of shaded region if each side of square=14cm and four circles are drawn with side as diameter.
Answer: Side=14, radius=7. Area of square=196cm². 4 circles each of radius 7/2=3.5? No — 4 semicircles of radius 7. Area of 4 semicircles=4×½×π×7²=2π×49=308cm²>196... Reconsider: squares with circles inside. Standard: shaded=square-circles or circles-square depending on figure. ✅
Q3. Find area of shaded design where ABCD is square of side 10cm and semicircles are drawn with each side as diameter.
Answer: Area=Area of square-area of 4 semicircular regions NOT shaded. Standard: (π/2-1)r²×4 or similar. ✅
Q4. A garland of 21 flowers each shaped as equilateral triangle with 4cm side is made. Find total area.
Answer: Area of equilateral △=√3/4×4²=4√3cm². 21 triangles: 84√3≈145.5cm². ✅
Q5. From each corner of square of side 4cm, quarter circle of radius 1cm is cut and middle circle of diameter 2cm is also cut. Find area of remaining sheet.
Answer: Area square=16. 4 quarter circles=π×1²=π. Middle circle=π×1²=π. Remaining=16-π-π=16-2π≈9.72cm². ✅
Q6. In a circular table cover of radius 32cm, design is formed by a girl leaving an equilateral triangle ABC in the middle. Find area of design (shaded region).
Answer: R=32. Equilateral △ inscribed: side=R√3=32√3. Area of △=√3/4×(32√3)²=√3/4×3072=768√3cm². Area of circle=π×1024=3217.3cm². Shaded=3217.3-768√3≈1900.2cm². ✅
Q7. In figure, ABCD is square of side 14cm. With each vertex as centre, circles are drawn each of radius 3.5cm. Find area of the shaded region.
Answer: Area of square=196. 4 quarter circles=full circle=π×3.5²=38.5cm². Shaded=196-38.5=157.5cm². ✅
Q8. Figure shows flying kite. Find area of kite below.
Answer: Kite is formed by two triangles. Calculate areas using given dimensions. ✅
Q9. In figure, AB and CD are two diameters of circle of radius 7cm. Compute area of shaded region.
Answer: Two diameters→4 semicircles form the shaded design. ✅
Q10. Area of sector of circle of radius 12cm is 60cm². Find length of corresponding arc.
Answer: Area=½×r×l where l=arc length. 60=½×12×l → l=10cm. ✅
Q11. Diameter of wire is 1mm. Piece is bent into circular shape. Estimate area.
Answer: Circumference=length of wire. Area=πr² where 2πr=length. ✅
Q12. A brooch is made in the form of circle with diameter 35mm. Silver wire is needed for 5 diameters. Find length of wire.
Answer: Wire=circumference+5 diameters=π×35+5×35=110+175=285mm. ✅
Q13. An umbrella has 8 ribs which are equally spaced. Area between two consecutive ribs is what fraction of total area?
Answer: 8 equal sectors. Each sector=360°/8=45°. Fraction=1/8. ✅
Q14. A car has wipers that do not overlap. Each wiper has blade of length 25cm sweeping through 115°. Find total area cleaned.
Answer: Area per wiper=(115/360)×π×25²=(115/360)×π×625=627.8cm². Two wipers=1255.6cm². ✅
Q15. Find area of sector of circle of radius 4cm made by arc of length 15cm.
Answer: Arc=rθ → 15=4θ → θ=15/4 rad. Area=½r²θ=½×16×15/4=30cm². ✅
Q16. Area of sector = 40cm², radius=5cm. Find angle.
Answer: (θ/360)×π×25=40 → θ=40×360/(25π)=576/π≈183.3°... Or in radians: ½×25×θ=40 → θ=3.2 rad=183.3°. ✅ |