In the given figure, \angle {\PQR}=\angle {\PST}=90^{\circ}, {\PQ}=5 {\cm} and {\PS}=2 {\cm}. (i) Prove that \triangle {\PQR} ∼ \triangle {\PST}. (ii) Find Area of \triangle P Q R : Area of quadrilateral SRQT.

ICSE Class X Math 2019 Solved Paper - Question 16

Question : 16 of 32

Marks: +1, -0

In the given figure,

∠ {\PQR}=∠ {\PST}=90^{∘}, {\PQ}=5 {\cm}

and

{\PS}=2 {\cm}

.
(i) Prove that

△ {\PQR} ∼ △ {\PST}

.
(ii) Find Area of

△ P Q R

: Area of quadrilateral SRQT.

Solution:

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