Question

A square $ABCD$ is constructed inside a triangle $PQR$ having sides $10,17$ and $21$ as shown in figure. Find the approximate value of perimeter of the square $ABCD$.

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

Answer 1

The correct option is B $23.2$

First find the Area of triangle PQR:

Applying Heron's formula:

$S=\frac{a+b+c}{2}$

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

Area of triangle PQR = $\sqrt{s(s-a)(s-b)(s-c)}$

=$\sqrt{24(24-10)(24-17)(24-21)}$

=$84$

From this we get height of triangle as 8.

Assume the sides of the square be x.

Then AB = BC = CD = AD = x

Area of small triangle APB = $\frac{x(8-x)}{2}$--- (1)

Area of trapezoid ABRQ = $\frac{x(21+x)}{2}$--- (2)

Adding equation (1) and (2), we get area of triangle PQR,

So, $8x–x^2+21x+x^2=84\times 2$

$29x=168$

$x=5.8$

Hence perimeter of the square ABCD = 4x

= $23.2$