Question

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.

Answer 1

Let ABC be an equilateral triangle and O be the interior point and AQ, BR and CP are the perpendiculars drawn from point O such that OQ = 10 cm, OR = 6 cm, OP = 14 cm.

Let sides of an equilateral triangle be a cm.

Area of ΔOAB = `1/2 xx AB xx OP`

= `(1/2 xx a xx 14)cm^2`

= 7a cm²   ...(i)

Area of ΔOBC = `1/2 xx BC xx OQ`

= `(1/2 xx a xx 10)cm^2`

= 5a cm²   ...(ii)

Area of ΔOAC = `1/2 xx AC xx OR`

`(1/2 xx a xx 6)cm^2`

= 3a cm²  ...(iii)

∴ Area of ΔABC = Area of (ΔOAB + ΔOBC + ΔOAC)

= (7a + 5a + 3a) cm²

= 15a cm²  ...(iv) [From (i), (ii) and (iii)]

Area of an equilateral ΔABC = `sqrt(3)/4 a^2`  ...(v)

From (iv) and (v),

`sqrt(3)/4 a^2 = 15a`

`\implies a = (60sqrt(3))/3 = 20sqrt(3)`

Substituting `a = 20sqrt(3)` in (v), we get

Area of ΔABC = `sqrt(3)/4 (20sqrt(3))^2`

= `300sqrt(3)`

Thus, the area of an equilateral triangle is `300sqrt(3)  cm^2`.