Question

The perimeter of an isosceles triangle is 42 cm and its base is $1{\displaystyle \frac{1}{2}}$ times its congruent sides, then the area of the triangle is approximately equals to:

Answer 1

Hint: In this question, we need to determine the area of the isosceles triangle such that the perimeter of an isosceles triangle is 42 cm and its base is $1{\displaystyle \frac{1}{2}}$ times its congruent sides. For this, we will use the property of the isosceles triangle to determine the length of the sides and then, apply Heron's formula to evaluate the area of the triangle.

Complete step-by-step answer:

An isosceles triangle is the one which has two equal sides and the length of the base is different from the length of the equal sides. The equal sides often known as congruent sides.
Let the length of the congruent sides of the isosceles triangle be $x$ .
According to the question, the length of the base of the isosceles triangle is $1{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}$ times of its congruent sides. So, we can write
The length of the base is given as, $b={\displaystyle \frac{3}{2}}x$ .
The sum of all the sides of the triangle is the perimeter of the triangle.
Also, it has been given that the perimeter of the isosceles triangle is 42 cm. So, we can write
 $$\begin{gathered} x+x+{\displaystyle \frac{3}{2}}x=42 \\ \Rightarrow 2x+{\displaystyle \frac{3}{2}}x=42 \\ \Rightarrow {\displaystyle \frac{7}{2}}x=42 \\ \Rightarrow x={\displaystyle \frac{42\times 2}{7}} \\ \Rightarrow x=6\times 2 \\ \Rightarrow x=12\text{ cm} \end{gathered}$$
Hence, the length of the sides of the isosceles triangle are 12 cm, 12 cm and ${\displaystyle \frac{3}{2}}\times 12=18\text{ cm}$ .
Now, using Heron's formula to evaluate the area of the triangle as
 $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where, a, b, and c are the length of the sides of the triangles and $s={\displaystyle \frac{a+b+c}{2}}$ .
So,
 $$\begin{gathered} s={\displaystyle \frac{a+b+c}{2}} \\ ={\displaystyle \frac{12+12+18}{2}} \\ ={\displaystyle \frac{42}{2}} \\ =21 \end{gathered}$$
Substituting the value of ‘s’ in the formula $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ we get,
 $$\begin{gathered} A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-12\right)\left(21-12\right)\left(21-18\right)} \\ =\sqrt{21\times 9\times 9\times 3} \\ =\sqrt{5103} \\ =71.435 \\ =72 \end{gathered}$$
Hence, the area of the isosceles triangle is 71.435 sq. cm. which approximates to 72 sq.cm.
So, the correct answer is “72.sq. cm”.

Note: It is worth noting down here that the Heron's formula can only be used to find the area of the triangle of which the length of each side has been given. However, if the length of the base and the length of the triangles are known then, we can use the formula $A={\displaystyle \frac{1}{2}}\times base\times height$ . Here also, we can determine the length of the height of the triangle and then, apply the above formula.