Question

The perimeter of the triangular field is 135cm while the sides of the perimeter are in ratio 25:17:12, find the area of the triangular field.(Find area using herons formula)​

Answer 1

Solution :

Given :

• The sides of the perimeter are in ratio 25:17:12
• The perimeter of the triangular field = 135cm

Explanation :

Let the ratio be x

As we know that formula of the perimeter of triangle :

• Side + Side + Side

A/q

➝ 25x + 17x + 12x = 135

➝ 54x = 135

➝ x = 135/54

➝ x = 2.5 cm

So,

• 1st side of ∆ = 25x = 25 × 2.5cm = 62.5 cm
• 2nd side of ∆ = 17x = 17 × 2.5cm = 42.5cm
• 3rd side of ∆ = 12x = 12 × 2.5cm = 30cm

Now,

Using Herons Formula :

Semi perimeter = Side + Side + Side/2

Semi perimeter = 62.5 + 42.5 + 30/2

Semi perimeter = 135/2

Semi perimeter = 67.5 cm

&

Area of triangle = √s(s-a)(s-b)(s-c)

Area of ∆ = √67.5(67.5 - 62.5)(67.5 - 42.5)(67.5 - 30)

Area of ∆ = √67.5(5)(25)(37.5)

Area of ∆ = √326406.25

Area of ∆ = 562.5cm²

Thus,

The are of triangular field will be 562.5cm² .

Answer 2

$\red{\bigstar}$ G I V E N

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• The perimeter of the triangular field is 135 cm

• The ratio of sides of triangular field is 25:17:12

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$\blue{\bigstar}$ T OㅤF I N D

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• Area of the triangular field ?

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$\purple{\bigstar}$ S O L U T I O N

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• Let sides of the triangular field be 25x, 17x, 12x

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$\quad\odot\:\underline{\boxed{\sf{Perimeter_{(triangular\:field)} = Sum\:of\:all\:sides\:of\:triangular\:field}}}$

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Putting all known values ::

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$\sf \quad \dashrightarrow\quad 135 = 25x + 17x + 12x$

$\\ \sf \quad \dashrightarrow\quad 54x = 135$

$\\ \sf \quad \dashrightarrow\quad x = {\cancel{\dfrac{135}{54}}}$

$\\ \sf \quad \dashrightarrow\quad \pink{x = 2.5\:cm}$

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$\green{\bigstar}$ H E N C E

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Sides of triangular field ::

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• 1st side = 25x = 25 × 2.5 = 62.5 cm

• 2nd side = 17x = 17 × 2.5 = 42.5 cm

• 3rd side = 12x = 12 × 2.5 = 30 cm

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We know that ::

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$\quad\odot\:\underline{\boxed{\sf{Semi\:Perimeter_{(triangular\:field)}\:(s) = \dfrac{Perimeter_{(triangular\:field)}}{2}}}}$

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Putting all known values ::

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$\sf \quad \dashrightarrow\quad Semi\:Perimeter_{(triangular\:field)}\:(s) = \dfrac{135}{2}$

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After cancelling 135 with 2, we get ::

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$\sf \quad \dashrightarrow\quad\pink{Semi\:Perimeter_{(triangular\:field)}\:(s) = 67.5\:cm}$

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Now, finding area of triangular field ::

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Using heron's formula ::

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$\quad\odot\:\underline{\boxed{\sf{Area_{(triangular\:field)} = \sqrt{s\:(s - a)\:(s - b)\:(s - c)}}}}$

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• Where a, b, and c are sides of triangular field.

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Putting all known values ::

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$\sf \quad \dashrightarrow\quad Area_{(triangular\:field)} = \sqrt{67.5\:(67.5 - 62.5)\:(67.5 - 42.5)\:(67.5 - 30)}$

$\\ \sf \quad \dashrightarrow\quad Area_{(triangular\:field)} = \sqrt{67.5\:\times\:5\:\times\:25\:\times\:37.5}$

$\\ \sf \quad \dashrightarrow\quad Area_{(triangular\:field)} = \sqrt{316406.25}$

$\\ \sf \quad \dashrightarrow\quad \pink{Area_{(triangular\:field)} = 562.5\:cm^2}$

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$\therefore\:{\underline{\sf{Hence,\:area\:of\:triangular\:field\:is\:\bf{562.5\:cm^2}}}}$

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$\orange{\bigstar}$ N O T E

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