Heron's Formula

We know that,

Each side of the equilateral triangle is equal.

Given,

Length of each side of an equilateral triangle = a cm.

Perimeter of traffic signal board (equilateral triangle) = sum of all the sides = a + a + a = 3a cm.

By formula,

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{3a}{2}$ cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units, where a, b and c are sides of triangle.

Substituting values we get :

$A = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big)} \\[1em] = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big)} \\[1em] = \sqrt{\dfrac{3a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2}} \\[1em] = \sqrt{\dfrac{3a^4}{16}} \\[1em] = \dfrac{\sqrt{3}}{4}a^2.$

Given,

Perimeter = 180 cm

∴ 3a = 180

⇒ a = $\dfrac{180}{3}$ = 60 cm.

Substituting value of a, we get :

$\text{Area of triangle (A)} = \dfrac{\sqrt{3}}{4}a^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 60^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 3600 \\[1em] = 900\sqrt{3}\text{ cm}^2.$

Hence, the area of the signal board is $900\sqrt{3}$ cm².

Here a, b and c are the sides of the triangle.

Let a = 122 m, b = 22 m and c = 120 m

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2}$

s = $\dfrac{a + b + c}{2} = \dfrac{122 + 22 + 120}{2} = \dfrac{264}{2}$ = 132 m.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$\text{Area of one wall} = \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \\[1em] = \sqrt{132 \times 10 \times 110 \times 12} \\[1em] = \sqrt{1742400} \\[1em] = 1320 \text{ m}^2.$

We know that,

The rent of advertising per year = ₹ 5000 per m²

So,

The rent of one complete triangular wall for 1 month

= $\dfrac{\text{Rent per sq. unit} \times \text{ Area}}{12}$

= $\dfrac{(5000 \times 1320)}{12} = 11 \times 5000$ = ₹ 5,50,000.

∴ The rent of one wall for 3 months = ₹ 5,50,000 x 3 = ₹ 16,50,000.

Hence, the rent paid by the company = ₹ 16,50,000.

Let a, b and c be the sides of the triangle.

Let a = 11 m, b = 6 m and c = 15 m.

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{a + b + c}{2}$

= $\dfrac{11 + 6 + 15}{2} = \dfrac{32}{2}$ = 16 m.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{16(16 - 11)(16 - 6)(16 - 15)} \\[1em] = \sqrt{16 \times 5 \times 10 \times 1} \\[1em] = \sqrt{800} \\[1em] = 20\sqrt{2} \text{ m}^2.$

Hence, area of the wall painted in colour = $20\sqrt{2}$ m².

Let a, b and c be the sides of the triangle.

Let a = 18 cm, b = 10 cm.

Given,

Perimeter = 42 cm

∴ a + b + c = 42

⇒ 18 + 10 + c = 42

⇒ 28 + c = 42

⇒ c = 42 - 28 = 14 cm.

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{42}{2}$ = 21 cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\[1em] = \sqrt{21 \times 3 \times 11 \times 7} \\[1em] = \sqrt{4851} \\[1em] = 21\sqrt{11} \text{ cm}^2$

Hence, area of triangle = $21\sqrt{11}$ cm².

Given,

Sides of a triangle are in the ratio of 12 : 17 : 25.

Let sides of the triangle are :

a = 12x, b = 17x and c = 25x.

Given,

Perimeter = 540 cm

∴ 12x + 17x + 25x = 540 cm

⇒ 54x = 540 cm

⇒ x = $\dfrac{540}{54}$

⇒ x = 10 cm

⇒ a = 12 × 10 = 120 cm,

⇒ b = 17 × 10 = 170 cm,

⇒ c = 25 × 10 = 250 cm.

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{540}{2}$ = 270 cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{270(270 - 120)(270 - 170)(270 - 250)} \\[1em] = \sqrt{270 \times 150 \times 100 \times 20} \\[1em] = \sqrt{81000000} \\[1em] = 9000 \text{ cm}^2.$

Hence, area of triangle = 9000 cm².

Length of equal sides (a and b) = 12 cm

Let third side of triangle (c) = x cm

Given,

Perimeter of triangle = 30

∴ 12 + 12 + x = 30

⇒ 24 + x = 30

⇒ x = 30 - 24

⇒ x = 6 cm.

∴ c = 6 cm.

By formula,

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{30}{2}$ = 15 cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\[1em] = \sqrt{15 \times 3 \times 3 \times 9} \\[1em] = \sqrt{1215} \\[1em] = 9\sqrt{15} \text{ cm}^2.$

Hence, area of triangle = $9\sqrt{15}$ cm².