Math Olympiad Test: Circles- 2 - Class 10 MCQ

Let O be the centre of a circle of radius 5.2 cm. Let OABO be the sector with perimeter 16.4 cm.

OA + O B + arc AB = 27.2
5.2 + 5 .2 + arc AB = 16.4
⇒ arc AB = 16.4 - 16.4 = 6 cm
Area of sector OACBO

Distance moved by wheel 
= 2 × π × r × 5000 = 11000 m

= 35 cm

PT² + OT² = OP²
⇒ PT² = (25)² -(7)²
⇒ PT = = 24 cm.

Distance moved by rope
= (1.1) × (60 + 28)
∴ No. of revolutions = 

In ΔABO,
∠y + 25° + 100° = 180°
⇒ y = 55°
∠C + ∠A = 180° [∵ ABCD is cycle]
⇒ ∠ = 80°
In ΔPBC,
∠x + 55° + 80° = 180°
⇒ ∠x = 45°

∵ EC = ED
∴ ∠ECD = ∠EDC = 62°
∴ ∠DEC = 180° - (2 × 62°) =56°

∴ ∠AED = ∠CED + ∠AEC
= 56° + 62° = 118°

∵ A, B, C and D are on the circumference of the circle
∴ ABCD is a cyclic quadrilateral, i.e.,
∠A + ∠C = 180°
⇒ ∠A = 180° - ∠C = 180° - 130° = 50°
∴ ∠DAB= 50°.

Let r be the radius
2πr = 22

Area of quadrant = 

Let the major arc be x cm
Length of minor arc = x/5
Circumference = 
⇒ 
⇒ 
⇒ x = 55 cm
Required area = 
=    288.75 cm².

Length of pendulum
= radius of the sector = x cm
Arc length = 8.8 cm
⇒ 

Angle described by the minute hand in 35 minutes = 
θ = 210º, r = 12 cm
Area of the sector = 

= 264 cm².

(area ΔOAC) + (area ΔOAB) + (area ΔOBC)
= area (ΔABC)



⇒ 8r + 6r + 10r = 48
⇒ 24r = 48 ⇒ r = 2

Area of the sector OACBO,

= 154 cm².
Area of ΔOAB = 

∴ Area of minor segment of circle
= 154 - 98 = 56 cm².

Area of the circular filed = 
πr² = 3850

r = 35 m
Circumference = 2πr = = 220 m
Cost of fencing = 220 × 8.50 = Rs. 1870

Area of the shaded region = Area of square ABCD - Area of two semicircles

= 196 - 154 = 42 cm²