Class 10 Maths Chapter 11 Question Answers - Area Related to Circles

Q1: If the difference between the circumference and the radius of a circle is 37 cm, then using π = 22/7, calculate the circumference (in cm) of the circle.
Sol: 2πr – r = 37 ⇒ r(2π – 1) = 37

Q2: If he is taken as 22/7, calculate the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution.
Sol: Radius (r) = 35/2
Required distance = Perimeter = 2πr
= 2 × 22/7 × 35/7 cm = 110 cm or 1.1 m

Q3: In the Figure, PQ and AB are respectively the arcs of two concentric circles of a radii 7 cm and 3.5 cm and centre O. If ∠POQ = 30°, then find the area of the shaded region. [Use π = 22/7]
Sol: Area of sector with radius 7 cm

Q4: Find the area of the major segment APB, in the figure of a circle of radius 35 cm and ∠AOB = 90°. (Use π = 22/7)
Sol: Here θ = 90°, p = OA = OB = 35 cm
Area of minor segment = ar(minor sector) – ar(∆AOB)

Q5: A chord of a circle of radius 14 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle. (Use π = 22/7 and √3 = 1.73)
Sol: Here θ = 120°, r = 14 cm

Q6: In Figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)
Sol: ∠ACB = 90° …[Angle in a semi-circle
∴ AC² + BC² = AB² …[Pythagoras’ theorem
(12)² + BC² = (13)²
144 + BC² = 169
BC² = 169 – 144 = 25
BC = + 5 cm

Q7: In Figure, ABC is a right-angled triangle right angled at A. Semicircles are drawn on AB, AC and BC as diametres. Find the area of the shaded region.
Sol: In rt. ∆BAC,

∴ BC² = AB² + AC² …[Pythagoras’ theorem
= (3)² + (4)
= 9 + 16 = 25 cm²
∴ BC = +5 cm … [Side of ∆ can’t be -ve

Q8: In the given figure, OAPB is a sector of a circle of radius 3.5 cm with the centre at O and ∠AOB = 120°. Find the length of OAPBO.

Sol: Here, the major sector angle is given by θ = 360° − 120° = 240°

∴ Circumference of the sector APB

∴ Perimeter of OAPBO = [Circumference of sector AOB] + OA + OB]

Q9: Find the area of a square inscribed in a circle of radius 10 cm.

Sol: Let ABCD be the square such that

AB = BC = 10 cm
∴ AC² = AB² + BC²
AB² + BC² = (10 × 2)
⇒ x² + x² = (20)2
[Let AB = BC = x]
⇒ 2x² = 400

⇒  

∴ Area of the square = 200 cm².

Q10: In the given figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter of the shaded region.

Sol. O is the centre of the circle.
∴ AB is its diameter.
In right Δ ABC,
AC² + BC² = AB²
⇒ 12² + 16² = AB²
⇒ 144 + 256 = AB² ⇒ AB² = 400

⇒
∴ Circumference of semi-circle ACB

∴ Perimeter of the shaded region = 22/7 cm + 12 cm + 16 cm
= 31.43 cm + 12cm + 16cm
= 59.43 cm