Question

Rohan constructed a paper windmill using 4
triangles. For AGOH the sides are in the ratio 25 :
17:12 and for ACOD, the sides are 13 cm, 84 cm
and 85 cm. For AAOB, he took all sides of equal
length.
​

Answer 1

$\mathbb\pink{AREA~~OF~~∆GOH~=~9000mm²}$

$\large\bold{Given:}$

∆GOH's side's ration = 25 : 17 : 12

Perimeter = 540mm

$\large\bold{To~Find:}$

Area of the Triangle ( GOH )

$\large\bold{Formulas:}$

$\red{\longrightarrow}$Perimeter of a Triangle

⇒ side1 + side2 + side3 = Perimeter

$\red{\longrightarrow}$Heron's Formula

⇒ ${Area \: of \: a \: trianle \: = \: \sqrt{(s)(s - a)(s - b)(s - c)} }$

$\large\bold{Calculation:}$

Let's assume the triangle's sides as $\pink{25x,~17x,~12x}$

Then, the perimeter = 25x + 17x + 12x = $\pink{54x}$

According to question :

54x = 540

x = 540 ÷ 54 = 10

$\boxed{\rm{∴~x~=~10}}$

25x = 25 × 10 = $\pink{250mm}$

17x = 17 × 10 = $\pink{170mm}$

12x = 12 × 10 = $\pink{12mm}$

$\mathrm{We're~having~the~required~values~to~find~Area}$

$\large\underline\mathrm\purple{Heron's~Formula }$

$Area \: of \: a \: trianle \: = \: \sqrt{(s)(s - a)(s - b)(s - c)}$

s = semiperimeter

a = side 1

b = side 2

c = side 3

$\large\underline\mathrm\purple{Let's~find~the~Area}$

s = 540 ÷ 2 = 270

a = 250

b = 170

c = 120

$= \sqrt{(270)(270 - 250)(270 - 170)(270 - 120) \: } \\ \\ = \sqrt{270 \times 20 \times 100 \times 150 \: } \\ \\ = \sqrt{10 \times 3 \times 3 \times 3 \times 10 \times 2 \times 10 \times 2 \times 5 \times 10 \times 3 \times 5} \\ \\ = \: 10 \times 10 \times 2 \times 3 \times 3 \times 5 \\ \\ = \: 9000$

$\large\bold{Solution:}$

$\boxed{\mathrm{\red{Area~of~∆GOH~=~9000mm²}}}$

Answer 2

$\mathbb\pink{AREA~~OF~~∆GOH~=~9000mm²}$

$\large\bold{Given:}$

∆GOH's side's ration = 25 : 17 : 12

Perimeter = 540mm

$\large\bold{To~Find:}$

Area of the Triangle ( GOH )

$\large\bold{Formulas:}$

$\red{\longrightarrow}$Perimeter of a Triangle

⇒ side1 + side2 + side3 = Perimeter

$\red{\longrightarrow}$Heron's Formula

⇒ ${Area \: of \: a \: trianle \: = \: \sqrt{(s)(s - a)(s - b)(s - c)} }$

$\large\bold{Calculation:}$

Let's assume the triangle's sides as $\pink{25x,~17x,~12x}$

Then, the perimeter = 25x + 17x + 12x = $\pink{54x}$

According to question :

54x = 540

x = 540 ÷ 54 = 10

$\boxed{\rm{∴~x~=~10}}$

25x = 25 × 10 = $\pink{250mm}$

17x = 17 × 10 = $\pink{170mm}$

12x = 12 × 10 = $\pink{12mm}$

$\mathrm{We're~having~the~required~values~to~find~Area}$

$\large\underline\mathrm\purple{Heron's~Formula }$

$Area \: of \: a \: trianle \: = \: \sqrt{(s)(s - a)(s - b)(s - c)}$

s = semiperimeter

a = side 1

b = side 2

c = side 3

$\large\underline\mathrm\purple{Let's~find~the~Area}$

s = 540 ÷ 2 = 270

a = 250

b = 170

c = 120

$= \sqrt{(270)(270 - 250)(270 - 170)(270 - 120) \: } \\ \\ = \sqrt{270 \times 20 \times 100 \times 150 \: } \\ \\ = \sqrt{10 \times 3 \times 3 \times 3 \times 10 \times 2 \times 10 \times 2 \times 5 \times 10 \times 3 \times 5} \\ \\ = \: 10 \times 10 \times 2 \times 3 \times 3 \times 5 \\ \\ = \: 9000$

$\large\bold{Solution:}$

$\boxed{\mathrm{\red{Area~of~∆GOH~=~9000mm²}}}$