in a∆ABC AB=15 BC=13 AC=14 find the area and hence its altitude on AC
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Answered by
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ΔABC
Given:
AB= 15 cm
BC= 13 cm
AC= 14 cm
∴Area of Δ = Heron's Formula
∴Area of Δ = √s(s-a)(s-b)(s-c) [ whole under root }
∴Semi perimeter = perimeter / 2
∴Perimeter = 15 + 13 + 14 = 42cm
∴S = 21cm
∴s-a = 21 - 15 = 6cm
∴s-b = 21 - 13 = 8cm
∴s-c = 21 - 14 = 7cm
∴√21(6)(8)(7) = √7056
∴Area = 84cm²
Altitude= ??
∴Area of Δ = 1/2 *bh
( lets take one of the side as a base of a Δ so e take 14 )
∴ 84cm² = 1/2 * 14 * h
∴84cm² = 7h
∴h = 84/7
∴h = 12cm
Given:
AB= 15 cm
BC= 13 cm
AC= 14 cm
∴Area of Δ = Heron's Formula
∴Area of Δ = √s(s-a)(s-b)(s-c) [ whole under root }
∴Semi perimeter = perimeter / 2
∴Perimeter = 15 + 13 + 14 = 42cm
∴S = 21cm
∴s-a = 21 - 15 = 6cm
∴s-b = 21 - 13 = 8cm
∴s-c = 21 - 14 = 7cm
∴√21(6)(8)(7) = √7056
∴Area = 84cm²
Altitude= ??
∴Area of Δ = 1/2 *bh
( lets take one of the side as a base of a Δ so e take 14 )
∴ 84cm² = 1/2 * 14 * h
∴84cm² = 7h
∴h = 84/7
∴h = 12cm
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Answered by
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Step-by-step explanation:
see guys or भैस
the area for this triangle is 84
and the altitude is 7cm from AC
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