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The perimeter of a triangular piece of land is 1020 m and its area is 3570 m². If one of its lengths
is 20 m, find the lengths of remaining two sides.
Answers
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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