In the adjoining figure O is the centre of the circle. The radius OP bisects a rectangle ABCD, at right angle. DM = NC = 2cm and AR = SB = 1 cm and KS = 4cm and OP = 5 cm. What is the area of the rectangle?

$8 c m^{2}$

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$10 c m^{2}$

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$12 c m^{2}$

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None of these

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Solution

The correct option is B
$10 c m^{2}$

$(O S)^{2} = (O K)^{2} + (K S)^{2}$
25 = $O K^{2} + 16 \Rightarrow O K = 3$
and $(O N)^{2} = (O L)^{2} + (L N)^{2} 25 = (O L)^{2} + 9 \Rightarrow O L = 4 c m$
$∴$ KL = OL - OK = 1 cm
$∴$ Area of rectangle = 1 $\times$ 10 = $10 c m^{2}$

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In the adjoining figure O is the centre of the circle. The radius OP bisects a rectangle ABCD, at right angle. DM = NC = 2cm and AR = SB = 1 cm and KS = 4cm and OP = 5 cm. What is the area of the rectangle?

The correct option is B 10 cm2 (OS)2=(OK)2+(KS)2 25 =OK2+16⇒OK=3 and (ON)2=(OL)2+(LN)225=(OL)2+9⇒ OL=4 cm ∴ KL = OL - OK = 1 cm ∴ Area of rectangle = 1 × 10 = 10 cm2

The correct option is B
$10 c m^{2}$

$(O S)^{2} = (O K)^{2} + (K S)^{2}$
25 = $O K^{2} + 16 \Rightarrow O K = 3$
and $(O N)^{2} = (O L)^{2} + (L N)^{2} 25 = (O L)^{2} + 9 \Rightarrow O L = 4 c m$
$∴$ KL = OL - OK = 1 cm
$∴$ Area of rectangle = 1 $\times$ 10 = $10 c m^{2}$