Question

three circles of radius 2 cm is inscribed in a equilateral triangle. then find the area of triangle.

Answer 1

SOLUTION:

we know that, when a line touces externally then angle between center and intersection point is 90°.

In Right ΔAOB;

∠OAB = 30° [Think yourself, : ) ]

using trigonometry ratio;

Let ∠OAB = Ф

tan Ф = OB/AB

=> tan 30° = $\frac{2}{AB}$

=> $\frac{1}{√3}$ = $\frac{2}{AB}$

=> AB = 2√3 cm

Similarly, CD = 2√3 cm

AD = AB + BC + CD

AD = 2√3 + 4 + 2√3 = 4(1 +√3) cm

∵Area of equilateral triangle = √3/4 side²

∴ Area of ΔADX = √3/4 × $[4(1 + √3)^{2}$ unit²

= √3/4 × 16 (48 + 1 + 8√3)cm²

= 4√3 (49 + 8√3) cm²

= 196√3 + 96 cm²

=  4( 24 + 49√3) cm²

Hence, Area of equilateral Triangle is 4( 24 + 49√3) cm²

____________________

© Amrit_____