Question : In a $ riangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $ riangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$, then:
Option 1: $ riangle ABC \cong riangle PQR$
Option 2: $ar( riangle ABC) eq ar( riangle PQR)$
Option 3: $ar( riangle ABC) \leq ar( riangle PQR)$
Option 4: $ar( riangle ABC)=ar( riangle PQR)$

Question

Question : In a $ riangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $ riangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$, then:

Option 1: $ riangle ABC \cong riangle PQR$

Option 2: $ar( riangle ABC) eq ar( riangle PQR)$

Option 3: $ar( riangle ABC) \leq ar( riangle PQR)$

Option 4: $ar( riangle ABC)=ar( riangle PQR)$

Team Careers360 · Accepted Answer

Correct Answer: $ar( riangle ABC) eq ar( riangle PQR)$

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

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Question : In a △ABC, if ∠A=90∘,AC=5 cm,BC=9 cm and in △PQR,∠P=90∘,PR=3 cm,QR=8 cm, then:Option 1: △ABC≅△PQROption 2: ar(△ABC)≠ar(△PQR)Option 3: ar(△ABC)≤ar(△PQR)Option 4: ar(△ABC)=ar(△PQR)

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