Question

If in a triangle ABC, AB = 5 units, AB = 5 units, ∠B = `cos^-1 (3/5)` and radius of circumcircle of ΔABC is 5 units, then the area (in sq.units) of ΔABC is  ______.

Options

• `6 + 8sqrt(3)`

• `8 + 2sqrt(2)`

• `4 + 2sqrt(3)`

• `10 + 6sqrt(2)`

Answer 1

If in a triangle ABC, AB = 5 units, AB = 5 units, ∠B = `cos^-1 (3/5)` and radius of circumcircle of ΔABC is 5 units, then the area (in sq.units) of ΔABC is  `underlinebb(6 + 8sqrt(3))`.

Explanation:

Given that, ∠B = `cos^-1 (3/5)`, R = 5 units

∴ cosB = `3/5` ⇒ sinB = `4/5`

Now, `b/(sinB)` = 2R ⇒ b = 2RsinB

= `2(5)4/5`

∴ b = 8

Now, By cosine rule

cosB = `(a^2 + c^2 - b^2)/(2ac) = 3/5`

⇒ `(a^2 + 25 - 64)/(10a) = 3/5`

⇒ a² – 39 = 6a

⇒ a² – 6a – 39 = 0

⇒ a = `(6 +- 8sqrt(3))/2`

⇒ a = `(6 +- 8sqrt(3))/2 = 3 +- 4sqrt(3)`

⇒ a = `3 + 4sqrt(3)`, a = `3 - 4sqrt(3)`

a = `3 - 4sqrt(3) < 0`

∴ a = `3 - 4sqrt(3)` can not be possible (since length of side can not be negative

∴ a = `3 + sqrt(3)`

Now, area of ΔABC = `(abc)/(4R) = ((3 + 4sqrt(3))(8)(5))/(4(5))`

∴ Δ = `(6 + 8sqrt(3))`sq.units