Question

area of triangle ABC,a=√2,b=√3,c=√6+√2/2​

Answer 1

Given :- a = √2, b = √3 , c = (√6 + √2)/2 .

To Find :-

• Area of ∆ABC = ?

Solution :-

we know that,

→ cos θ = (a² + b² - c²)/2ab

so,

→ cos B = (a² + c² - b²)/2ac

putting given values we get,

→ cos B = [(√2)² + (√6 + √2/2)² - (√3)²] / 2 * √2 * (√6 + √2)/2

→ cos B = [2 + {(8 + 2√12)/4} - 3] / (√12 + 2)

→ cos B = (2 + 2 + √3 - 3 ) / (2 + 2√3)

→ cos B = (1+ √3)/2(1 + √3)

→ cos B = (1/2)

→ cos B = cos 60°

→ B = 60°

then,

→ Area ∆ABC = (1/2) * a * c * sin B

→ Area ∆ABC = (1/2) * √2 * {(√6 + √2)/2} * sin 60°

→ Area ∆ABC = √2 * (√6 + √2)/4 * (√3/2)

→ Area ∆ABC = √6(√6 + √2)/8

→ Area ∆ABC = (6 + √12)/8

→ Area ∆ABC = (6 + 2√3)/8

→ Area ∆ABC = (3 + √3)/4 (Ans.)

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