Question

A traffic signal board, indicating 'SCHOOL AHEAD' , is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is 180cm, what will be the area of the signal board:

Answer 1

Hint:
We are given a traffic signal board, indicating 'SCHOOL AHEAD' is an equilateral triangle, so we will solve the heron's formula by taking all sides equal. After that we are given the perimeter as 180, we will again assume all the 3 sides equal and hence find the side by a using \[p = a + b + c\] . Then we will put the value of a in the formula derived by us.

Complete step by step solution:
According to the question, we are given that side of the equilateral triangular board is a.
Since the triangle is equilateral, all the sides of the triangle are a.
According to herons formula,


It states that the area of a triangle whose sides have lengths a, b and c is
 $A = \sqrt {s(s - a)(s - b)(s - c)} $
where s is the semi-perimeter of the triangle, that is,
 $s = \dfrac{{a + b + c}}{2}$
We know that, perimeter
 \[p = a + b + c\] …(1)

Hence, we can say that
 \[ \Rightarrow s = \dfrac{p}{2}\]

Since all sides are equal in an equilateral triangle, we get
 \[ \Rightarrow s = \dfrac{{a + a + a}}{2}\]
On simplification we get,
 \[ \Rightarrow s = \dfrac{{3a}}{2}\]

Now, using the Heron’s formula, we get the area as
 $ \Rightarrow A = \sqrt {s(s - a)(s - a)(s - a)} $
On simplification we get,
 $ \Rightarrow A = \sqrt {s{{(s - a)}^3}} $ … (2)

Put \[s = \dfrac{{3a}}{2}\] in (2), we get
 $ \Rightarrow A = \sqrt {\dfrac{{3a}}{2}{{(\dfrac{{3a}}{2} - a)}^3}} $

On taking LCM we get,
 \[ \Rightarrow A = \sqrt {\dfrac{{3a}}{2}{{(\dfrac{{3a - 2a}}{2})}^3}} \]

On simplification we get,
 \[ \Rightarrow A = \sqrt {\dfrac{{3a}}{2}{{(\dfrac{a}{2})}^3}} \]

On Multiplication of fractions inside root we get,
 \[ \Rightarrow A = \sqrt {\dfrac{{3{a^4}}}{{{2^4}}}} \]

Since, \[\sqrt {{a^n}b} = {a^{\dfrac{n}{2}}}\sqrt b \] , we get
 \[ \Rightarrow A = {(\dfrac{a}{2})^2}\sqrt 3 \]

On simplification we get,
 \[ \Rightarrow A = \dfrac{{{a^2}\sqrt 3 }}{4}\] …(3)

CASE (When the perimeter is 180 cm) :

From (1), we know that
 \[P = a + b + c\]

Put, p=180, we get
 \[ \Rightarrow 180 = a + b + c\]

Since we are given an equilateral triangle, all sides are equal
 \[ \Rightarrow 180 = a + a + a\]

On simplification we get,
 \[ \Rightarrow 180 = 3a\]

On dividing the equation by 3 we get,
 \[ \Rightarrow a = \dfrac{{180}}{3}\]

Hence, by solving the equation, we get
 \[ \Rightarrow a = 60\]

Put \[a = 60\] in (3) to get the required area, we get
 \[ \Rightarrow A = \dfrac{{{{60}^2} \times \sqrt 3 }}{4}\]

On simplification we get,
 \[ \Rightarrow A = \dfrac{{3600 \times \sqrt 3 }}{4}\]

On division we get,
 \[ \Rightarrow A = 900 \times \sqrt 3 \]

We know that, hence \[\sqrt 3 = 1.732\]
 \[ \Rightarrow A = 900 \times 1.732\]

On mulriplication we get,
 \[ \Rightarrow A = 1558.8\,c{m^2}\]

Hence, the required area is \[A = 1558.8\,c{m^2}\]

Note:
This is a direct formula based application of heron's formula, the trick here was knowing about the equilateral triangles, here are some important properties of equilateral triangles which may be used in these types of questions:
(1)All sides are equal in an equilateral triangle
(2) All angles are equal to \[{60^\circ }\] in an equilateral triangle
(3)The median, perpendicular, and angle bisector are the same for a vertex in an equilateral triangle and they divide the triangle into equal areas.