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 = ${\displaystyle \frac{1}{2}\times AB\times OP}$

= ${\displaystyle (\frac{1}{2}\times a\times 14)c{m}^2}$

= 7a cm²   ...(i)

Area of ΔOBC = ${\displaystyle \frac{1}{2}\times BC\times OQ}$

= ${\displaystyle (\frac{1}{2}\times a\times 10)c{m}^2}$

= 5a cm²   ...(ii)

Area of ΔOAC = ${\displaystyle \frac{1}{2}\times AC\times OR}$

${\displaystyle (\frac{1}{2}\times a\times 6)c{m}^2}$

= 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 = ${\displaystyle \frac{\sqrt{3}}{4}{a}^2}$  ...(v)

From (iv) and (v),

${\displaystyle \frac{\sqrt{3}}{4}{a}^2=15a}$

${\displaystyle \Rightarrow a=\frac{60\sqrt{3}}{3}=20\sqrt{3}}$

Substituting ${\displaystyle a=20\sqrt{3}}$ in (v), we get

Area of ΔABC = ${\displaystyle \frac{\sqrt{3}}{4}{\left(20\sqrt{3}\right)}^2}$

= ${\displaystyle 300\sqrt{3}}$

Thus, the area of an equilateral triangle is ${\displaystyle 300\sqrt{3} c{m}^2}$.