question_answer

Three circles of radii 4 cm, 6 cm and 8 cm touch each other pairwise externally. The area of the triangle formed by the line segments joining the centres of the three circles, is

A) \[6\sqrt{6}\,c{{m}^{2}}\]                    

B) \[24\sqrt{6}\,c{{m}^{2}}\]

C) \[144\sqrt{13}\,c{{m}^{2}}\]               

D) (d)\[12\sqrt{105}\,c{{m}^{2}}\]

Correct Answer: B

Solution :

From figure, AB = 12 cm, BC = 14 cm and CA = 10 cm

Area of \[\Delta ABC\]

\[=\sqrt{s\,(s-a)(s-b)(s-c)}\]

\[=\sqrt{18\,(18-14)(18-12)(18-10)}\]

\[\left[ \because s=\frac{a+b+c}{2} \right]\]

\[=\sqrt{18\times 4\times 6\times 8}=24\sqrt{6}\,c{{m}^{2}}\]