question_answer

If a right circular cone of vertical height 24 cm has a volume of \[\frac{5}{2}(10+5\sqrt{3})sq.cm\], then the area of its curved surface is:       

A)  \[4m\times 2m\]                     

B)  \[625c{{m}^{2}}\]              

C)  \[125c{{m}^{2}}\]           

D)  \[150c{{m}^{2}}\]       

Correct Answer: C

Solution :

Let radius of base and slant height be r and \[\text{ }A-\left( s \right),B-\left( r \right),C-\left( p \right),D-\left( q \right)\] respectively. Vertical height, h = 24 cm Volume \[(A)=\sqrt{s(s-a)(s-b)(s-c)},\] \[s=\frac{a+b+c}{2}.\]   \[\frac{\sqrt{3}}{2}{{(side)}^{2}}.\] \[=\sqrt{3}\]   \[\Delta ABC=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times AD\] \[\Delta ABC\] \[\Delta ABC=\frac{1}{2}\times BC\times AB\] Slant height, \[ABCD=AB\times BC=l\times b\] Area of curved surface \[=l+b+l+b\] \[=2l+2b\]