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Question

In the given figure $$PQRS$$ is a cyclic quadrilateral $$PQ$$ and $$SR$$ produced meet at $$T$$. Find area of quadrilateral $$PQRS$$ if area of $$\triangle PTS = 27\ cm^{2}$$ (given Triangle TPS is similar to TRQ)
578694_43567ead2685499aa9e8464491d93735.png


Solution

$$\cfrac {\text{area of triangle}\,\, TPS}{\text{area of triangle}\,\, TRQ} = \dfrac {TP^{2}}{TR^{2}} [\because \triangle TPS \sim \triangle TRQ]$$
$$\cfrac {27}{\text{area of triangle}\,\, TRQ} = \dfrac {18^{2}}{6^{2}}$$
$$\cfrac {27}{\text{area of triangle}\,\, TRQ} = \dfrac {9}{1}$$
$$\text{area of triangle}\,\, TRQ = 3\ cm^{2}$$
$$\therefore \text{area of quadrilateral}\,\, PQRS = \text{area of triangle}\,\, TPS - \text{area of triangle}\,\, TRQ$$
$$= 27 - 3$$
$$= 24\ cm^{2}$$ 

Mathematics

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