Question

In the figure given below, PB and QA are perpendiculars to the line segment AB. If PO = 4 cm, QO = 10 cm and the area of triangle POB = 233 cm², find the area of triangle QOA. [Give your answer correct to two decimal places]
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Answer 1

In △OAQ and △OBP, we have

∠A=∠B [Each equal to 90°]

∠AOQ=∠BOP

So, by AA-criterion of similarity, we have

△AOQ∼△BOP

$\bold{ \implies \: \frac{Area(△AOQ}{ \triangle \: BOP} = \frac{OQ {}^{2} }{OP {}^{2} } }$

$\bold{ \implies \: \frac{Area(△AOQ)}{ Area(\triangle \:BOP)} = \frac{10 {}^{2} }{4 {}^{2} } }$

$\bold{ \implies \: \frac{Area(△AOQ)}{ 233 {}^{2} } = \frac{10 {}^{2} }{4 {}^{2} } }$

$\small \bold{ ⇒ Area(△AOQ) = \frac{100}{16} \times 233 }$

=1456.25 cm²