Question : In a $\triangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $\triangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$, then:
Option 1: $\triangle ABC \cong \triangle PQR$
Option 2: $ar(\triangle ABC)\neq ar(\triangle PQR)$
Option 3: $ar(\triangle ABC) \leq ar(\triangle PQR)$
Option 4: $ar(\triangle ABC)=ar(\triangle PQR)$
Correct Answer: $ar(\triangle ABC)\neq ar(\triangle PQR)$
Solution :
Given: In a $\triangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $\triangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$
Since the sides are not equal, $\triangle ABC \cong \triangle PQR$ is wrong.
In $\triangle ABC$,
$AB=\sqrt{9^2-5^2}=\sqrt{56} =2\sqrt{14}$
Area of $\triangle ABC = \frac{1}{2}\times 5 \times 2\sqrt{14} =5\sqrt{14} =\sqrt{350}$ cm$^2$
In $\triangle PQR$,
$PQ=\sqrt{8^2-3^3}=\sqrt{55}$ cm
Area of $\triangle PQR = \frac{1}{2}\times 3 \times \sqrt{55} = \sqrt{247.5}$ cm$^2$
Hence, the correct answer is $ar(\triangle ABC)\neq ar(\triangle PQR)$.
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