In the circle given below,let OA = 1 unit,OB | vedclass

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(B) Let $PA = AQ = \lambda$.Since $PQ \perp OB$,by the property of intersecting chords in a circle,we have $OA \cdot AB = PA \cdot AQ$.Given $OA = 1$

Vedclass · Accepted Answer

$24 \sqrt{3}$

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(B) Let $PA = AQ = \lambda$.Since $PQ \perp OB$,by the property of intersecting chords in a circle,we have $OA \cdot AB = PA \cdot AQ$.Given $OA = 1$ and $OB = 13$,we have $AB = OB - OA = 13 - 1 = 12$.Substituting the values,$1 \cdot 12 = \lambda \cdot \lambda$.$\lambda^2 = 12 \Rightarrow \lambda = \sqrt{12} = 2 \sqrt{3}$.The area of $\Delta PQB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes PQ 	imes AB$.Since $PQ = PA + AQ = \lambda + \lambda = 2\lambda = 4 \sqrt{3}$,Area $= \frac{1}{2} 	imes (4 \sqrt{3}) 	imes 12 = 2 \sqrt{3} 	imes 12 = 24 \sqrt{3}$ square units.

In the circle given below,let $OA = 1$ unit,$OB = 13$ units and $PQ \perp OB$. Then,the area of the triangle $PQB$ (in square units) is

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