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ICSE Class X Math 2016 Paper

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In the given figure PQRSP Q R S is a cyclic quadrilateral PQP Q and SRS R produced meet at TT.
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Question : 26 of 57
Marks: +1, -0
Find area of quadrilateral PQRS if area of △PTS=27cm2\triangle PTS = 27 \text{cm}^2.
Solution:  
 area of △TPS area of △TRQ=TP2TR2\frac{\text{ area of }\triangle TPS}{\text{ area of }\triangle TRQ} = \frac{TP^2}{TR^2}
[∵△TPS∼△TRQ][\because \triangle TPS \sim \triangle TRQ]
27 area of △TRQ=18262\frac{27}{\text{ area of }\triangle TRQ} = \frac{18^2}{6^2}
27 area of △TRQ=91\frac{27}{\text{ area of }\triangle TRQ} = \frac{9}{1}
 area of △TRQ=3cm2\text{ area of }\triangle TRQ = 3 \text{cm}^2
 area of quadrilateral PQRS\text{ area of quadrilateral } PQRS
= area of △TPS− area of △TQR= \text{ area of }\triangle TPS - \text{ area of }\triangle TQR
=27−3= 27 - 3
=24cm2.= 24 \text{cm}^2.
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