In-a-triangle-with-one-angle-of-120-the-lengths-of-the-sides-form-an-AP-If-the-length-of-the-greatest-side-is-7-cm-the-area-of-the-triangle-is-

Question Number 43641 by gunawan last updated on 13/Sep/18

In a triangle with one angle of 120 °,

the lengths of the sides form an AP .

If the length of the greatest side is

7 cm ., the area of the triangle is

Answered by $@ty@m last updated on 13/Sep/18

L e t c > b > a

c^{2} = a^{2} + b^{2} - 2 a b \cos C

7^{2} = a^{2} + b^{2} - 2 a b \cos 120

49 = a^{2} + b^{2} + a b (∵ \cos 120 = - \frac{1}{2})

\dots \dots (1)

∵ a, b, c a r e i n A P

∴ 2 b = a + c

\Rightarrow 2 b = a + 7

\Rightarrow a = 2 b - 7 \dots (2)

∴ f r o m (1)

49 = {(2 b - 7)}^{2} + b^{2} + (2 b - 7) b

49 = 4 b^{2} - 28 b + 49 + b^{2} + 2 b^{2} - 7 b

7 b^{2} - 35 b = 0

7 b (b - 5) = 0

∴ b = 5

\Rightarrow a = 2 \times 5 - 7 = 3

∴ A r e a o f △ = \sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{\frac{15}{2} (\frac{15}{2} - 3) (\frac{15}{2} - 5) (\frac{15}{2} - 7)}

= \frac{1}{4} \times \sqrt{15 \times 9 \times 5 \times 1}

= \frac{1}{4} \times 5 \times 3 \sqrt{3}

= \frac{15}{4} \sqrt{3} s q . u n i t

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