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Ingus Ullemany wo sides.(b)The perimeter of a triangular piece of land is 1020 m and its area is 3570 m². If one of its lengthsis 20 m, find the lengths of remaining two sides. |
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Answer» The perimeter of a triangular piece of land is 1020m and it’s area is 3570 sq mtr. One of its length is 20M. What is length of remaining 2 sides? Perimeter = 1020, Area = 3570, one side = 20 Let a,B and c(20) be three sides of triangle and let X be perimeter/2(1020/2=510). Using Heron's FORMULA , Area = square root [x(x-a)×(x-b)×(x-c)] A^2 = [x(x-a)×(x-b)×(x-c)] 3570^2 = [510(510-a)×(510-b)×(510-20)] 3570^2 = [510×490(510-a)×(510-b)] 3570^2 = [249900×(510-a)×(510-b)] As a+b+c = 1020, a+b+20 = 1020, a = 1000-b, b = 1000-a 3570^2 = [249900×(510-a)×(510-(1000-a))] 3570^2 = [249900×(510-a)×(510-1000+a)] 3570^2 = [249900×(510-a)(-490+a)] 3570^2 = 249900×(-249900+510a+490a -a^2) 3570^2 = 249900×(-249900+1000a-a^2) 12744900/249900 = -249900+1000a-a^2 51 = -249900 +1000a-a^2 a^2-1000a+249900+51=0 a^2-1000a+249951=0 a^2 -1000a+249951+49=49 a^2-1000a+250000=49 (a-500)×(a-500)=7×7 a-500=7 a=507 b=1000-507=493 Hence, the lengths of remaining sides are 507 and 493 metres. |
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