Question

From a point in the interior of an equilateral triangle the perpendicular distances of the sides √3 cm, 2√3 cm and 5√3. What is the perimeter (in cm) of the triangle ?​

Answer 1

Answer:

$\huge\underline\bold {Answer:}$

Ler each side of the triangle be a cm.

Area of triangle ABC = Area of triangle AOB + Area of triangle AOC + Area of triangle BOC

$= > \frac{ \sqrt{3} }{4} a {}^{2} = \frac{1}{2} \times a \times \sqrt{3} + \frac{1}{2} \times a \times 2 \sqrt{3} + \frac{1}{2} \times a \times 5 \sqrt{3}$

$= > \frac{ \sqrt{3} }{4} a {}^{2} = \frac{1}{2} a( \sqrt{3} + 2 \sqrt{3} + 5 \sqrt{3} )$

$= > \frac{ \sqrt{3} }{4}a {}^{2} = \frac{1}{2} \times a \times 8 \sqrt{3} \\ = > \frac{ \sqrt{3} }{4} a = \frac{1}{2} \times 8 \sqrt{3}$

$= > a = \frac{8 \sqrt{3} }{2} \times \frac{4}{ \sqrt{3} } \\ = > a = 16$

Therefore, perimeter of the triangle

= 16 × 3 = 48 cm.

Answer 2

Answer:

Let each side of equilateral △ABC be

′

a

′

cm

In △ABC,OD=

3

cm

OE=2

3

cm

OF=5

3

cm

Now, ar△ABC=ar△BOC+ar△COA+ar△AOB

⇒

4

3

a

2

=

2

1

[BC×OD+AC×DE+AB×OF]=

2

1

[a×

3

+a×2

3

+a×5

3

]

⇒

4

3

a

2

=

2

1

×8a

3

⇒a=16cm

∴Perimeter of △ABC=3a=3×16=48cm