Question

The square of the length of a rod $AB$ is $72{cm}^2$. If we place the rod in the corner of a room, so that the end $A$ is always on the edge between the two walls of the corner and the end $B$ is always on the floor, what is the maximum possible area of the triangle formed by the rod, the edge between the walls and the floor?

• A. $6$ ${cm}^2$
• B. $12$ ${cm}^2$
• C. $18$ ${cm}^2$
• D. $24$ ${cm}^2$

Answer 1

Answer: (C)

$18$ ${cm}^2$

$△ABC$ is max.
when $AB=BC$
${AB}^2+{BC}^2={AC}^2$
$\Rightarrow 2{AB}^2=72$
$\Rightarrow AB=6=BC$
$△ABC=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 6\times 6=18{cm}^2$