Question

from point in the interior of an equilateral triangle , perpendicular are drawn on each of its three sides.the length of the perpendiculars are 14cm,10cm and 6cm.find the area of the triangle

Answer 1

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Answer 2

Area of the triangle is $\bold{300 \sqrt{3} \mathrm{cm}^{2}}$

Solution:

The sides of an equilateral triangle ABC is denoted by S that is sides $A C=B C=C A=S$

The triangle Diagram is attached below.

Now as we can see the three perpendiculars as $O X \perp \mathrm{AB}, O Z \perp \mathrm{AC}, 0 \mathrm{Y} \perp \mathrm{BC}$

The perpendiculars are denoted as OX=14cm; OZ=6cm; OY=10cm

Therefore, the area of the triangle ABC should be equal to the sum of areas of triangles AOC, COB, BOA

Let us equate the areas, we get:

area of ABC=area of AOC+area of COB+area of BOA  

$\frac{\sqrt{3}}{4} S^{2}=\frac{1}{2}(O Z+A C)+\frac{1}{2}(O X+A B)+\frac{1}{2}(O Y+B C)$

$\frac{\sqrt{3}}{4} S^{2}=\frac{1}{2}((O Z+A C)+(O X+A B)+(O Y+B C))$

$=\frac{1}{2}(A C)(O X+O Z+O Y)=\frac{1}{2} S(14+6+10)$

$\frac{\sqrt{3}}{4} S^{2}=\frac{1}{2} S(30) ; \quad S=\frac{60}{\sqrt{3}}$

Therefore, the length of the sides of the triangle ABC is$\frac{60}{\sqrt{3}} cm=34.64cm.$

Area of the triangle is $=\frac{\sqrt{3}}{4} \times \frac{60}{\sqrt{3}} \times \frac{60}{\sqrt{3}}$

$=300 \sqrt{3} \mathrm{cm}^{2}$.