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ICSE Class X Math 2019 Paper

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Question : 16 of 32
Marks: +1, -0
In the given figure, ∠PQR=∠PST=90∘, PQ=5 cm\angle PQR = \angle PST = 90^{\circ},\ PQ=5\ \text{cm} and PS=2 cmPS=2\ \text{cm}.
(i) Prove that △PQR∼△PST\triangle PQR \sim \triangle PST.
(ii) Find Area of â–³PQR\triangle PQR : Area of quadrilateral SRQT.
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