if sides of a triangle are in ratio 3:5:7 . Find the area of triangle whose perimeter is 142.5​

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1. Let the sides of the triangle be 3x, 5x and 7x and we’ve Perimeter of triangle as 142.5 so:-

   • Ratio of sides = 3:5:7

   [tex] \colon\implies{\pmb{\sf{ Perimeter_{( \Delta )} = a+b+c }}} \\ \\ \colon\implies{\sf{ 142.5 = 3x+5x+7x }} \\ \\ \\ \colon\implies{\sf{ \cancel{142.5} = \cancel{15}x }} \\ \\ \\ \colon\implies{\sf{ x = 9.5}} \\ [/tex]

   Therefore,

   • three Sides are 28.5 , 47.5 and 66.5

   Now, We also Know that we’ve to find the Area of the triangle using Heron‘s Formula as:-

   [tex] \maltese \ {\blue{\pmb{\underline{\boxed{\sf{ Area _ {( \Delta )} = \sqrt{ s(s-a)(s-b)(s-c) } }}}}}} \\ \\ \colon\implies{\sf{Where, s = \dfrac{a+b+c}{2} }} \\ \\ \colon\implies{\sf{ s = \dfrac{28.5+47.5+66.5}{2} }} \\ \\ \colon\implies{\sf{ s = \dfrac{142.5}{2} }} \\ \\ \colon\implies{\sf{ s = 71.25 }} \\ \\ \\ \colon\implies{\sf{ Now, Area _ {( \Delta )} = \sqrt{ 71.25(71.25-28.5)(71.25-47.5)(71.25-66.5) } }} \\ \\ \\ \colon\implies{\sf{ Area _ {( \Delta )} = \sqrt{ 71.25(42.75)(23.75)(4.75) } }} \\ \\ \\ \colon\implies{\sf{ Area _ {( \Delta )} = \sqrt{ 71.25 \times 42.75 \times 23.75 \times 4.75 } }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = \sqrt{ 343619.82 } }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = 586.19 \ unit^2 }} \\ [/tex]

   Hence,

   The Area of the ∆ will be 586.19 unit².

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