Question

The longest side of a triangle has length (2−1) cm. The other sides have lengths (−1) cm and (+1) cm. Given that the largest angle is 120°,
find:
a. the value of .
b. the area of the triangle

Answer 1

the longest side of a triangle has length (2x-1). the other sides have lengths (x-1)cm and (x+1)cm. given that the largest angle is 120, work out the value of x

and the area of the triangle.

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The longest side is opposite the largest angle.

Use the Law of Cosines:

(2x-1)^2 = (x-1)^2 + (x+1)^2 - 2(x-1)(x+1)cos(120)

4x^2 - 4x + 1 = 2x^2 + 2 + x^2 - 1

4x^2 - 4x + 1 = 3x^2 + 1

x^2 - 4x = 0

x = 0

x = 4

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Sides = 7, 3 and 5

Use Heron's Law:

s = perimeter/2 = 7.5

Area = sqrt(s(s-a)*(s-b)*(s-c))

Area = sqrt(7.5*5.5*2.5*0.5)

Area = sqrt(51.5625)

Area =~ 7.18 sq cm

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