Question

In a triangle, $ABC$ $\underset{–}{ACB}={90}^{\circ }$ . IF $P$ and $Q$ are points of trisection of $AB$ then $\overline{{CP}^2}+\overline{C{Q}^2}=$?

• A. $\frac{5}{3}(\overline{AB}{)}^2$
• B. $\frac{5}{9}(\overline{AB}{)}^2$
• C. $\frac{5}{4}(\overline{AB}{)}^2$
• D. $\frac{1}{9}(\overline{AB}{)}^2$

Answer 1

The correct option is B $\frac{5}{9}(\overline{AB}{)}^2$

In ${\Delta }^{\prime \prime }AQC$, $CP$ is median

${AC}^2+{CQ}^2=2({CP}^2+x^2)$ [Appllonins theorem]

$\Rightarrow {AC}^2+{CQ}^2={2CP}^2+{2x}^2⟶\left(1\right)$

In ${\Delta }^{\prime \prime }PCB$,

${PC}^2+{CB}^2=({CQ}^2+x^2).2$

$\Rightarrow {PC}^2+{CB}^2=2{CQ}^2+{2x}^2⟶\left(2\right)$

Adding $\left(1\right)$ & $\left(2\right)$, we get

${AC}^2+{CB}^2+{CQ}^2+{PC}^2={2CP}^2+{2CQ}^2+{4x}^2$

$\Rightarrow {\left(AB\right)}^2={CP}^2+{CQ}^2+{4x}^2$

$\Rightarrow {CP}^2+{CQ}^2={5x}^2=\frac{5}{9}{\left(AB\right)}^2$ [B]