Question

The side of a tringle are in the ratio 4: 5:6 and its perimeter is 45 cm .Find the area of the triangle.​

Answer 1

the side of a triangle are in ratio 4:5:6

And its perimeter is 45 CM .

so,

To Find The Area ,

let the ratio Be in X .

4x,5x,6x

Now ,

according to the given question,

$4x + 5x + 6x = 45$

$15x = 45$

$x = \frac{45}{15} = 3$

so, X=3

$4x = 4 \times 3 = 12$

$5x = 5 \times 3 = 15$

$6x = 6 \times 3 = 18$

$\small\huge{{ \triangle}}$

sides according to the given figure .

Answer 2

$\underline\mathfrak\pink{Given:-}$

• The side of a tringle are in the ratio 4 : 5 : 6
• its perimeter is 45 cm.

$\large\underline\mathfrak{To \: find:-}$

• Find the area of the triangle. ....?

$\large\underline\mathfrak\pink{Solutions:-}$

Let the sides of triangle be x.

So the side will be 4x, 5x, and 6x.

• a = 4x
• b = 5x
• c = 6x

✰ Perimeter = a + b + c

$\: \: \: \: \: \leadsto \: \: \: {45} \: \: = \: \: {4x} \: + \: {5x} \: + \: {6x}$

$\: \: \: \: \: \leadsto \: \: \: {45} \: \: = \: \: {15x}$

$\: \: \: \: \: \leadsto \: \: \: {x} \: \: = \: \: \frac{45}{15}$

$\: \: \: \: \: \leadsto \: \: \: {x} \: \: = \: \: {3}$

Therefore, Each side are :-

⟹ 4x = 4 × 3 = 12

⟹ 5x = 5 × 3 = 15

⟹ 6x = 6 × 3 = 18

Now,

Area of ∆ABC

The sides of a triangle are:-

• a = 12 cm
• b = 15 cm
• c = 18 cm

$\: \: \: \: \: \therefore \: \: Area \: \: of \: \: triangle \: \: by \: \: heroes's \: \: formula:-$

$\: \: \: \: \: \sqrt{s \: (s \: - \: a) \: (s \: - \: b) \: (s \: - \: c)}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{a \: + \: b \: + \: c}{2}$

⟹ a ⇢ 12 cm

⟹ b ⇢ 15 cm

⟹ c ⇢ 18 cm

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{{12} \: + \: {15} \: + \: {18}}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{45}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: {22.5}$

$\: \: \: \: \: \therefore \: \: \: Area \: \: of \: \: triangular \: \: field:-$

$\: \: \: \: \: \leadsto \: \: \sqrt{{22.5} \: ({22.5} \: - \: {12}) \: ({22.5} \: - \: {15}) \: ({22.5} \: - \: {18})}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{22.5} \: \times \: {(10.5)} \: \times \: {(7.5)} \: \times \: {(4.5)}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{7973.4}$

$\: \: \: \: \: \leadsto \: \: {89.29} \: {cm}^{2}$

$\: \: \: \: \: \: Hence, \\ \: \: \therefore \: \: Area \: \: of \: \: triangle \: \: \leadsto \: \: {89.29} \: {cm}^{2}$

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