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(a) A vertical pole of height $10\text{ m}$ casts a shadow of $15\text{ m}$ on the ground and at the same time, a tower casts a shadow of $45\text{ m}$ on the ground. Find the height of the tower.
OR
(b) In the given figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Prove that $\Delta PQS \sim \Delta TQR$.
OR
(b) In the given figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Prove that $\Delta PQS \sim \Delta TQR$.
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(a)
$\because \Delta ABC \sim \Delta PQR$ [$\frac{1}{2}$ mark]
$\therefore \frac{10}{PQ} = \frac{15}{45} \Rightarrow PQ = 30$
Height of the tower $= 30\text{ m}$ [$1\frac{1}{2}$ marks]
OR
(b) In $\Delta PQR$, $\angle 1 = \angle 2 \Rightarrow PQ = PR$ [$\frac{1}{2}$ mark]
$\therefore$ In $\Delta PQS$ and $\Delta TQR$
$\frac{QR}{QS} = \frac{QT}{PR} \Rightarrow \frac{QR}{QS} = \frac{QT}{QP}$ [$\frac{1}{2}$ mark]
also $\angle Q = \angle Q$ (Common) [$\frac{1}{2}$ mark]
$\therefore \Delta PQS \sim \Delta TQR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]
$\because \Delta ABC \sim \Delta PQR$ [$\frac{1}{2}$ mark]
$\therefore \frac{10}{PQ} = \frac{15}{45} \Rightarrow PQ = 30$
Height of the tower $= 30\text{ m}$ [$1\frac{1}{2}$ marks]
OR
(b) In $\Delta PQR$, $\angle 1 = \angle 2 \Rightarrow PQ = PR$ [$\frac{1}{2}$ mark]
$\therefore$ In $\Delta PQS$ and $\Delta TQR$
$\frac{QR}{QS} = \frac{QT}{PR} \Rightarrow \frac{QR}{QS} = \frac{QT}{QP}$ [$\frac{1}{2}$ mark]
also $\angle Q = \angle Q$ (Common) [$\frac{1}{2}$ mark]
$\therefore \Delta PQS \sim \Delta TQR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]