Question

If the area of the triangle shown is 108 square centimetres, what is its perimeter in centimetres?

Answer 1

Hint: We first find the formulas for perimeter and semi-perimeter for triangles with sides’ length $x,y,z$. Then we put the values to get the area for the triangle. We simplify the equation and get the quadratic equation to solve. At the end we find the perimeter of the triangle. The formula for finding area in terms of semi perimeter is
$\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
where $s = \dfrac{ a + b + c}{2} $
a,b and c are sides of triangles.

Complete step-by-step answer:
If we take the semi-perimeter of any triangle as $s$ for the sides $x,y,z$, then the area of the triangle is $\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ square units.
We have the sides as $a,a,24$. The semi-perimeter is $s=\dfrac{a+a+24}{2}=a+12$.
Now we place the values in the equation of area to equate with 108.
So, $\sqrt{\left( a+12 \right)\left( a+12-a \right)\left( a+12-a \right)\left( a+12-24 \right)}=108$.
We now simplify the equation to find the quadratic equation.
$ \sqrt{\left( a+12 \right)\left( a+12-a \right)\left( a+12-a \right)\left( a+12-24 \right)}=108$
$ \Rightarrow 12\sqrt{\left( a+12 \right)\left( a-12 \right)}=108 $
$ \Rightarrow 12\sqrt{\left( a+12 \right)\left( a-12 \right)}=108$
$ \Rightarrow \sqrt{{{a}^{2}}-144}=9 $
$\Rightarrow {{a}^{2}}-144=81 $
$ \Rightarrow {{a}^{2}}=225$
$\Rightarrow a=15 $
We will not take the negative value as length can’t be negative.
Therefore, the perimeter is $15+15+24=54$ cm.
So, the correct answer is “54 cm”.

Note: We need to be careful about the area equation where we are using a semi-perimeter instead of perimeter. We can also use that particular formula of $\dfrac{b}{2}\sqrt{{{a}^{2}}-{{\left( \dfrac{b}{2} \right)}^{2}}}$as the area for isosceles triangle where the equal sides’ length is $a$ and the third one is $b$.