Question

Legs (sides other than the hypotenuse) of a right triangle are of lengths $16cm$ and $8cm$. Find the length of the side of the largest square that can be inscribed in the triangle.

• A. $\frac{16}{3}cm$
• B. $\frac{20}{3}cm$
• C. $\frac{14}{3}cm$
• D. $\frac{22}{3}cm$

Answer 1

The correct option is A $\frac{16}{3}cm$

Let $ABC$ be a right triangle right angles at $B$ with $AB=16cm$ and $BC=8cm$. Then, the largest square $BRSP$ which can be inscribed in this triangle will be as shown in figure.
Let $PB=xcm$. So, $AP=(16-x)cm$.
In $△APS$ and $△ABC$,
$\angle A=\angle A$
$\angle APS=\angle ABC$ (Each ${90}^o$)
So, $△APS\sim △ABC$ ($AA$ similarly)
Therefore, $\frac{AP}{AB}=\frac{PS}{BC}$
or $\frac{16-x}{16}=\frac{x}{8}$
or $128-8x=16x$
or $x=\frac{128}{24}=\frac{16}{3}$
Thus, the side of the required square is of length $\frac{16}{3}cm$