Math Olympiad Test: Circles- 1 - Class 10 MCQ

For any external point P,
AP × BP = PD × PC
⇒ (AB + BP) × BP = 12 × PC
⇒ (8 +10) × 10 = 12 × PC
⇒ PC = 180/12 = 15 cm.

In DOPT,

OT² + PT² = OP²
⇒ PT = 

Diameter of the circumscribed circle
= diagonal of the square
= 10√2cm
Radius = 

Area of circumscribed circle = πr²


= 157 cm².

Let r be the radius of the circle
2πr - r = 37
⇒ r (2π - 1) = 37
⇒ 
⇒ 

⇒ r = 7 cm

Area = πr² = = 22 × 7 = 154 cm².

The tangent at any point of a circle and the radius through this point are perpendicular to each other as shown below:

In the figure above,
∠OPA +∠OPQ = 180^∘ ........(1) [Linear pair angle]
Also, ∠OPA = ∠OPQ ......(2)
∴ ∠OPA +∠OPQ = 180^∘
2 ×∠OPQ = 180^∘
∠OPQ = 90^∘
Hence, OP is perpendicular to PQ ⇒ OP is perpendicular to AB.

Radius of front of wheel = 40 cm

Circumference =
Distance moved by it in 800 revolutions

Circumference of real wheel = 2π × 1 = 2π m.
No. of revolutions = 

Area of square = 484
Side of square = = 22 cm
⇒ 2πr= 22 × 4

Area of circle = πr²

= 44 × 14
= 616 cm².

Let R be the radius of outer circle and r be the radius of inner circle
2πR = 503

and 2πr = 437

Width of the track = R - r = 

Area of the track = π(R² - r²)
= π(R + r) (R - r)

= 4935 m²

Let A be the area of each quadrant of the circle

Radius of circel = 14 cm

Area of 4 quadrants = 4 ×154 = 616 m² Area of square park = (100)² = 10000 m²
Area of remaining part = 10000 - 616
= 9384 m²

Area of the sector =