Question : In $\triangle P Q R, \angle P=90^{\circ}. {S}$ and ${T}$ are the mid points of sides ${PR}$ and ${PQ}$, respectively. What is the value of $\frac{\text{RQ}^2}{(\text{QS}^2 + \text{RT}^2)}$?
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{1}{2}$
Option 4: $\frac{2}{3}$
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Correct Answer: $\frac{4}{5}$
Solution :
$\angle$ P = 90°
Pythagoras theorem
Hypotenuse$^2$ = perpendicular$^2$ + base$^2$
Take the sides as shown in the figure
S and T are the midpoints of PR and PQ.
QR$^2$ = QP$^2$ + PR$^2$
⇒ QR$^2$ = 2$^2$ + 2$^2$
⇒ QR$^2$ = 4 + 4 = 8
⇒ QR = $2\sqrt{2}$
QS$^2$ = QP$^2$ + PS$^2$
⇒ QS$^2$ = 2$^2$ + 1$^2$
⇒ QS$^2$ = 4 + 1
⇒ QS = $\sqrt{5}$
RT$^2$ = PT$^2$ + PR$^2$
⇒ RT$^2$ = 1$^2$ + 2$^2$
⇒ RT$^2$ = 5
⇒ RT = $\sqrt{5}$
So, $\frac{\text{RQ}^2}{(\text{QS}^2 + \text{RT}^2)}$
= $\frac{(2\sqrt{2})^2}{[(\sqrt{5})^2 + (\sqrt{5})^2]}$
= $\frac{8}{10}$
= $\frac{4}{5}$
$\therefore$ The value of $\frac{\text{RQ}^2}{(\text{QS}^2 + \text{RT}^2)}$ is $\frac{4}{5}$.
Hence, the correct answer is $\frac{4}{5}$.
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