G-1 Insane Equilateral Triangle

Geometry Level pending

Equilateral triangle $MNO$ is inscribed in equilateral triangle $ABC$ with $MN \perp BC$ . If the area of $\bigtriangleup MNO$ is $12 \text{ cm}^{2}$ , what is the area of $\bigtriangleup ABC$ in $\text{cm}^{2}$ ?

The answer is 36.

1 solution

Brian Dela Torre
Jan 28, 2016

The area of $\bigtriangleup$ MNO is 12 cm which has side measuring 4 $\sqrt[4]{3}$ .

Since $MN \perp BC$ and $\angle$ C = 60° then ,

$\bigtriangleup$ NMC, $\bigtriangleup$ MOB and $\bigtriangleup$ OAN is in form of 30°60°90° triangle and they have equal areas.

Hence the area of $\bigtriangleup$ NMC is 4 $\sqrt[4]{3}$ $\times$ $\frac{4\sqrt[4]{3^{3}}}{3}$ = 8

We will multiply the resulting area to 3 because there are three equal triangles which $\bigtriangleup$ NMC is one of them. The product obtained will be added to the middle triangle $\bigtriangleup$ DEF. The answer

(8 $\times$ 3) + 12 = $\boxed{36 cm^{2}}$

G-1 Insane Equilateral Triangle

The answer is 36.

1 solution

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