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)

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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