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:

=93•53 cm

Step-by-step explanation:

Let ABC be an equilateral triangle in which AB=BC=AC=x cm and the perpendicular distances of three sides from the point O are OP=6 cm,OQ=9 cm and OR = 12 cm.

Line segments OA,OB and OC are joined.

A/q

ar(∆ABC)=ar(∆AOB)+ar(∆BOC)+ar(∆AOC)

=>(√3/4)×(side)²=1/2×AB×OP+1/2×BC×OQ

+1/2×AC×OR

=>(√3/4)×x²=1/2 × x × 6+1/2 × x × 9+ 1/2 × x × 12

=>(√3/4)×x²=x/2(6+9+12)

=>(√3/4)× x=27/2

=>x= (27/2)×(4/√3)

=>x=9√3×2

=>x= 18√3 cm

, Hence,the perimeter of the equilateral∆ABC= 3×side= 3× x

= 3×18√3

=54√3

=54×1•732

=93•528

=93•53 cm,