Question

The area of an isosceles right angled triangle varies directly as the square of the length of its leg. If the area is $18c{m}^2$ when the length of its leg is $6cm$, find area of the triangle when length of its leg is $5cm$

Answer 1

The statement " $x$ varies directly as $y$ ," means that when $y$ increases, $x$ increases by the same factor. In other words, $y$ and $x$ always have the same ratio that is:

$x=ky$

where $k$ is the constant of variation.

Let the area be $A$ and the length be $l$.

Here, it is given that the area $A$ varies as the square of the length $l^2$ if $A=18$ cm${}^2$ and $l=6$ cm, therefore,

$A=k{l}^2\Rightarrow 18=k{\left(6\right)}^2\Rightarrow 18=k\times 36\Rightarrow k=\frac{18}{36}\Rightarrow k=\frac{1}{2}$

Thus, the general equation is $A=\frac{1}{2}{l}^2$.

Since $l=5$ cm, therefore,

$A=\frac{1}{2}{\left(5\right)}^2=\frac{25}{2}=12.5$

Hence, the area of the triangle when the length of its leg is $5$ cm is $12.5$ cm${}^2$.