Question : In $\triangle$ABC, $\angle$A = $90^{\circ}$, BP and CQ are two medians. Then the value of $\frac{BP^2 + CQ^2}{BC^2}$ is:
Option 1: $\frac{4}{5}$
Option 2: $\frac{5}{4}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{3}{5}$
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Correct Answer: $\frac{5}{4}$
Solution :
Since BP and CQ are two medians. So, Q and P are midpoints of AB and AC respectively.
AQ = BQ = $\frac{1}{2}$AB
AP = CP = $\frac{1}{2}$AC
In $\triangle$AQC,
$\angle$A = $90^{\circ}$
By Pythagoras theorem,
⇒ CQ
2
= AC
2
+ QA
2
⇒ 4CQ
2
= 4AC
2
+ 4QA
2
⇒ 4CQ
2
= 4AC
2
+ (2QA)
2
⇒ 4CQ
2
= 4AC
2
+ AB
2
------------(i)
In $\triangle$ BPA,
⇒ BP
2
= BA
2
+ AP
2
⇒ 4BP
2
= 4BA
2
+ 4AP
2
⇒ 4BP
2
= 4BA
2
+ (2AP)
2
⇒ 4BP
2
= 4BA
2
+ AC
2
------------(ii)
Adding (i) and (ii),
⇒ 4CQ
2
+ 4BP
2
= 4AC
2
+ AB
2
+ 4BA
2
+ AC
2
⇒ 4(CQ
2
+ BP
2
) = 5AC
2
+ 5BA
2
= 5(AC
2
+ AB
2
) = 5BC
2
⇒ $\frac{BP^2 + CQ^2}{BC^2}$ = $\frac{5}{4}$
Hence, the correct answer is $\frac{5}{4}$.
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