Question

A signboard is in the shape of an equilateral triangle with side 'a'. Find the area of the signboard using Heron's formula. If the perimeter of the board is 210 cm, what will be its area ?​

Answer 1

Given :-

A signboard is in the shape of an equilateral triangle with side 'a'. Find the area of the signboard using Heron's formula. If the perimeter of the board is 210 cm

To Find :-

Area using heron's formula

Solution :-

In an equilateral triangle

Perimeter = a + a + a

210 = a + a + a

210 = 3a

210/3 = a

70 = a

Therefore

Side are 70 cm, 70 cm and 70 cm

Now,

Semiperimeter = Perimeter/2

Semiperimeter = 210/2

Semiperimeter = 105 m

Area = √s(s - a)(s - b)(s - c)

Area = √105(105 - 70)(105 - 70)(105 - 70)

Area = √105 × 35 × 35 × 35

Area = √4501875 cm

Area = 2121.76 cm²

Hence

Area is 2121.76 cm²

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

A signboard is in the shape of an equilateral triangle with side 'a'.

Let triangle is represented as ABC.

So that,

AB = a units

BC = a units

CA = a units

Now,

We know that,

$\underline{\boxed{\sf Semi \ perimeter, \: s \: = \: \dfrac{ \: Perimeter \: }{2}}}$

So,

$\rm :\longmapsto\:s = \dfrac{a + a + a}{2}$

$\rm :\longmapsto\:s = \dfrac{3a}{2}$

Let assume that,

AB, a = a units

BC, b = a units

CA, c = a units

Now, we know that,

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

So, on substituting the values of s, a, b and c, we get

$\rm :\longmapsto\:Area = \sqrt{\bigg(\dfrac{3a}{2} \bigg) \bigg(\dfrac{3a}{2} - a\bigg) \bigg(\dfrac{3a}{2} - a\bigg) \bigg(\dfrac{3a}{2} - 2\bigg) }$

$\rm :\longmapsto\:Area = \sqrt{\bigg(\dfrac{3a}{2} \bigg) \bigg(\dfrac{3a - 2a}{2} \bigg) \bigg(\dfrac{3a - 2a}{2}\bigg) \bigg(\dfrac{3a - 2a}{2}\bigg) }$

$\rm :\longmapsto\:Area = \sqrt{\bigg(\dfrac{3a}{2} \bigg) \bigg(\dfrac{a}{2} \bigg) \bigg(\dfrac{a}{2}\bigg) \bigg(\dfrac{a}{2}\bigg) }$

$\rm :\longmapsto\:Area = \bigg(\dfrac{a}{2} \bigg) \bigg(\dfrac{a}{2} \bigg) \sqrt{3}$

$\bf\implies \:Area = \dfrac{ \sqrt{3} }{4} {a}^{2} \: sq. \: units$

Now,

Further it is given that if the perimeter of board is 210 cm

So, Let assume that side of board be 'a' cm

So, 3a = 210

It implies, a = 70 cm

Now, we have proved above that area of equilateral triangle of side a is

$\bf\implies \:Area = \dfrac{ \sqrt{3} }{4} {a}^{2} \:$

So, on substituting the value of a, we get

$\rm :\longmapsto\:Area = \dfrac{ \sqrt{3} }{4} \times {(70)}^{2}$

$\rm :\longmapsto\:Area = \dfrac{ \sqrt{3} }{4} \times 70 \times 70$

$\rm :\longmapsto\:Area = \sqrt{3} \times 35 \times 35$

$\rm :\longmapsto\:Area = 1225 \sqrt{3} \: {cm}^{2}$

$\rm :\longmapsto\:Area = 1225 \times 1.732$

$\bf :\longmapsto\:Area = 2121.70 \: {cm}^{2}$