The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th
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1 Answer

Answer :

(c) 67.5 mGiven 2s = a + b + c ⇒ 240 = 78 + 50 + Third side ⇒ Third side = 240 m – 128 m = 112 m.∴ Area of Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)= \(\sqrt{120(120-78)(120-50)(120-112)}\)= \(\sqrt{120 imes42 imes70 imes8}\) = 1680Also, Area of Δ = \(rac{1}{2}\)x b x h ∴ \(rac{1}{2}\) x 50 x h = 1680 ⇒ h = \(rac{1680}{25}\) = 67.5 m.
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