Question

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

Answer 1

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Here is your answer

Let the side of equilateral triangle is x cm

Now In Triangle AOC,

Area of Triangle = $\frac {1}{2} \times base \times altitude$


$= \frac{1}{2} \times ac \times od \\ \\ = \frac{1}{2} \times x \times 10 \\ \\ = 5x \: {cm}^{2}$

In Triangle BOC,

Area of Triangle = $\frac {1}{2} \times bc \times of$


$= \frac{1}{2} \times x \times 6 \\ \\ = 3x \: {cm}^{2}$
In Triangle AOB,

Area of Triangle = $\frac{1}{2} \times ab \times oe$


$= \frac{1}{2} \times x \times 14 \\ \\ = 7x \: {cm}^{2}$
Area of Triangle = Area of Triangle AOC + Area of Triangle BOC + Area of Triangle AOB


$= (5x + 3x + 7x) {cm}^{2} \\ \\ =15x \: {cm}^{2}$
As we know that,

Area of a equilateral triangle = $\frac{\sqrt{3}}{4} (side){}^{2}$

$15x = \frac{ \sqrt{3} }{4} x {}^{2} \\ \\ x = \frac{15 \times 4}{ \sqrt{3} } \\ \\ x = \frac{60}{1.732} \\ \\ x = 34.64 \: cm {}^{2}$
Therefore, Area of equilateral triangle is $34.64 {cm}^{2}$

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Answer 2

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

Answer :

Please Refer the attachment for the diagram.

Step by step explanation :

We know that,

Sides of an equilateral triangle are equal.

Let all the sides be named 'x'.

Now,

Let's find the areas of the triangles.

Area of ∆ HAB = 1/2 × AB × HF

= 1/2 × 'x' × 14

= 7x cm^2

Area of ∆ HBC = 1/2 × BC × HD

= 1/2 × 'x' × 10

= 5x cm^2

Area of ∆ HAC = 1/2 × AC × HE

= 1/2 × 'x' × 6

= 3x cm^2

________________________

Now,

Area of ∆ HAB + Area of ∆ HBC + Area of ∆ HAC = Area of equilateral ∆ ABC

= 7x + 5x + 3x

= 15x cm^2

Now,

Using Heron's Formula,

Semi perimeter = ( x + x + x ) /2

s = (3x) / 2 cm

Now,

Area of equilateral triangle ABC

$\tt{ = \sqrt{s(s - a)(s - b)(s - c)}}$

$\tt{= \sqrt{ \frac{3x}{2}( \frac{3x}{2} - x)(\frac{3x}{2} - x)(\frac{3x}{2} - x) } }$

$\tt {= \sqrt{ \frac{3x}{2} \times \frac{x}{2} \times \frac{x}{2} \times \frac{x}{2}}}$

$\tt{= \frac{ \sqrt{3} }{4} x {}^{2} }$

Now,

$\tt{ \frac{ \sqrt{3} }{4} x {}^{2} =15x }$

$\tt{x {}^{2} = \frac{15x \times 4}{3}}$

$\tt{x = \frac{60 \sqrt{3} }{3}}$

$\tt{x = 20 \sqrt{3} cm}$

Now,

Area of equilateral triangle ABC

$\tt{= \frac{ \sqrt{3} }{4} \times (20 \sqrt{3}) {}^{2} }$

$\tt{= \frac{ \sqrt{3} }{4} \times 400 \times 3}$

$\tt{= \sqrt{3} \times 100 \times 3}$

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

√3 = 1.73

$\tt{ = 300 \times 1.73}$

$\tt {= 519 \: cm {}^{2} }$

_________________________

Area of the triangle is 519 sq cm