Question

Question: The perimeter of a triangular field is 135 cm and its sides are in the ratio 25:17:12. Find its area .
*Formula to be used : Heron's formula*



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Answer 1

$\sf\underline{Question-}$

The perimeter of a triangular field is 135 cm and its sides are in the ratio 25:17:12. Find its area.

$\sf\underline{Given-}$

• The perimeter of triangle field is 135cm.
• Sides of perimeter are in ratio 25:17:12 .

$\sf\underline{Asked-}$

• Area of the triangular field.

$\sf\underline{Solution-}$

Let the ratio be : $\sf{x}$

|Formula of triangle :- Side + Side + Side|

$\sf\underline{According \: to \: the \: Q.}$

$\sf{25x : 17x : 12x}$

$\sf{54x = 135}$

$\sf{x=}$ $\dfrac{135}{54}$

$\sf\blue{ x = 2.5cm}$

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$\sf{Therefore,}$

1st side of the triangle : 62.5

2nd side of the triangle : 42.5cm

3rd side of the triangle : 30cm

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$\sf{Now,}$ By using heron's formula , we'll found out the semi-perimeter.

Semi perimeter-

$\sf\dfrac{Side+Side+Side}{2}$

$\mapsto$ $\dfrac{62.5 + 42.5 + 30}{2}$

$\mapsto$ $\dfrac{135}{2}$

$\sf\underline{Semi \: Perimeter \: = 67.5cm}$

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Area∆ = $\sqrt{s(s - a)} (s - b)(s - c)$

Area∆ = $\sqrt{67.5(5)(25)(37.5)}$

Area∆ =$\sqrt{326406.25}$

$\mapsto$ $\sf\underline\blue{Area ∆ = 526.5cm^2}$

➛ Accordingly , the area of the triangular field is $\sf\underline{526.5cm^2}$ .

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