Question

In a $\Delta$ ABC ,P,Q,R are the feet of angle bisectors from the vertices to their opposite sides as shown in figure. $\Delta$ PQR is constructed.

If $AB=7$ units, $BC=8$ units, $AC=5$ units. Then the side PQ will be

• A. $\frac{\sqrt{28}}{3}$
• B. $\frac{\sqrt{88}}{3}$
• C. $\frac{\sqrt{78}}{3}$
• D. $\frac{\sqrt{84}}{3}$

Answer 1

Answer: (D)

$\frac{\sqrt{84}}{3}$

We know that $cosC=\left(\frac{a^2+b^2-c^2}{2ab}\right)=\frac{1}{2}$

$AB=7units,BC=8units,AC=5units$

In $\Delta$ PQC
$cosC=\frac{C{Q}^2+P{C}^2-P{Q}^2}{2PC.PQ}$
$\Rightarrow P{Q}^2=P{C}^2+C{Q}^2-PC.PQ$
$\{where.PC=\frac{ab}{b+c}andQC=\frac{ab}{a+c}\}$
so $PQ=\sqrt{\frac{100}{9}+\frac{64}{9}-\frac{80}{9}}=\frac{\sqrt{84}}{3}=\frac{2}{3}\sqrt{21}units$