Question

A square $ABCD$ is constructed inside a triangle $PQR$ having sides $10,17and21$ as shown in figure. Find the perimeter of the square $ABCD$.

• A. $28$
• B. $23.2$
• C. $25.4$
• D. $28.8$

Answer 1

The correct option is B $23.2$

$PQ=10,PR=17,QR=21$

$ar\Delta PQR=\sqrt{s(s-a)(s-b)(s-c)}$

$S=\frac{a+b+c}{2}=\frac{10+17+21}{2}=24$

$\therefore$ $ar\Delta PQR=\sqrt{24(24-10)(24-17)(24-21)}$

$=\sqrt{24\times 14\times 7\times 3}=\sqrt{2\times 2\times 2\times 3\times 2\times 7\times 7\times 3}$

$=2\times 2\times 3\times 7=84$

Also, $ar\Delta PQR=\frac{1}{2}\times QR\times h$

$\Rightarrow 84=\frac{1}{21}\times 21\times h\Rightarrow h=\frac{84\times 2}{21}=8$

Area of trap $ABRQ$ = $\frac{1}{2}(AB+QR)\times 32$

Let side of square ABCD of $x.$

Now, $ar.ABRQ=\frac{1}{2}(x+21)\times x$

Also, $ar\Delta PAB=\frac{1}{2}\times PN\times AB=\frac{1}{2}\times (8-x)\times x$

Now, $ar\Delta PAB+arABRQ=ar\Delta PQR$

$\Rightarrow$$\frac{x(8-x)}{2}+\frac{x(x+21)}{2}=84$

$\Rightarrow$ $\frac{8x-x^2+x^2+21x}{2}=84\Rightarrow \frac{29x}{2}=84\Rightarrow x=\frac{168}{29}$

$\therefore$ Perimeter of square ABCD $=$ $4x=\frac{168}{29}\times 4\approx 23.2$