Question
In Figure. wo triangles are congruent by $\ce{RHS}.$
$i.$
$ii.$
$i.$
$ii.$
✓
Answer
True. Solution:
$i.$
$ii.$
In $\triangle\text{ABC}, \text{AC}=\sqrt{\text{AB}^{2}+\text{BC}^{2}} =\sqrt{\text{4}^{2}+\text{5}^{2}}=\sqrt{41\ \text{cm}} [$by Pythagoras theoram$]$
In $\triangle\text{PQR}, \text{PR}=\sqrt{\text{PQ}^{2}+\text{QR}^{2}} =\sqrt{\text{4}^{2}+\text{5}^{2}}=\sqrt{41\ \text{cm}} [$by Pythagoras theoram$]$
Now, in $\triangle\text{ABC}$ and $\triangle\text{PQR},$
$\text{AB}=\text{PQ}=4\ \text{cm}$
$\text{AC}=\text{PR}=\sqrt{41\ \text{cm}}$
$\angle\text{ABC}=\angle\text{PQR}=90^{\circ}$
By $\ce{RHS}$ congruence criterian, $\triangle\text{ABC}\cong\triangle\text{PQR}$
$i.$
$ii.$
In $\triangle\text{ABC}, \text{AC}=\sqrt{\text{AB}^{2}+\text{BC}^{2}} =\sqrt{\text{4}^{2}+\text{5}^{2}}=\sqrt{41\ \text{cm}} [$by Pythagoras theoram$]$
In $\triangle\text{PQR}, \text{PR}=\sqrt{\text{PQ}^{2}+\text{QR}^{2}} =\sqrt{\text{4}^{2}+\text{5}^{2}}=\sqrt{41\ \text{cm}} [$by Pythagoras theoram$]$
Now, in $\triangle\text{ABC}$ and $\triangle\text{PQR},$
$\text{AB}=\text{PQ}=4\ \text{cm}$
$\text{AC}=\text{PR}=\sqrt{41\ \text{cm}}$
$\angle\text{ABC}=\angle\text{PQR}=90^{\circ}$
By $\ce{RHS}$ congruence criterian, $\triangle\text{ABC}\cong\triangle\text{PQR}$
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