Question

Tangents at $P,Q,R$ on a circle of radius $r$ from a triangle whose sides are $3r,4r,5r$, then $P{R}^2+R{Q}^2+Q{P}^2=$

• A. $\frac{84}{5}{r}^2$
• B. $\frac{184}{5}{r}^2$
• C. $\frac{176}{5}{r}^2$
• D. $\frac{44r^2}{5}$

Answer 1

The correct option is D $\frac{44r^2}{5}$

$\begin{array}{c}3r,4r,5rarepythagorous\\ triplateABCisrightangletriangle\\ PQ=\sqrt{r^2+r^2}=\sqrt{2}r\\ In\Delta APR\\ \cos A=\frac{A{P}^2+A{R}^2-P{R}^2}{2AP\cdot AR}\\ =\frac{{\left(3r\right)}^2+{\left(3r\right)}^2-{\left(PR\right)}^2}{2\cdot 3r\cdot 3r}\\ =\frac{9r^2+9r^2-P{R}^2}{2\cdot 3r\cdot 3r}\\ 18r^2\cos wA=18r^2-{\left(PR\right)}^{2}P{R}^2=18r^2\left(1-\cos A\right)\\ similarly,in\Delta CQR\\ \cos C=\frac{4r^2+4r^2-Q{R}^2}{2\cdot 2r\cdot 2r}\\ Q{R}^2=8r^2\left(1-\cos C\right)\\ P{Q}^2+Q{R}^2+P{R}^2=2r^2+8r^2\left(1-\cos C\right)+18r^2\left(1-\cos A\right)\\ =2r^2+8r^2\left(1-\frac{3r}{5r}\right)+18r^2\left(1-\frac{4r}{5r}\right)\\ =2r^2+\frac{16r^2}{5}+\frac{{186}^2}{5}\\ =\frac{10r^2+16r^2}{5}+18r^2=\frac{44r^2}{5}\end{array}$