Question

In the given figure, AB = BE = 7 cm such that ABE is a quadrant of a circle, BC = 9 cm, BD = 15 cm,
CF ⊥ AD such that CF = 8 cm. If the given figure is rotated about the side AD, then the surface area of the resulting solid is

• A. $(178\pi +15\sqrt{82})c{m}^2$
• B. $(178+15\sqrt{82})\pi c{m}^2$
• C. $(178+15\sqrt{82})c{m}^2$
• D. $1408\pi c{m}^2$

Answer 1

The correct option is B $(178+15\sqrt{82})\pi c{m}^2$


When the given figure is rotated about the side AD, the solid formed contains a hemisphere, frustum and a cone.

Let $l_{1}and{l}_2$ be the slant height of the frustum and the cone respectively.

Now, in $\Delta CDF,$

$l_2=DF=\sqrt{(CD{)}^2+(CF{)}^2}$

$=\sqrt{6^2+8^2}$

$=\sqrt{36+64}$

= 10 cm

Also, for frustrum,

$l_1=\sqrt{h^2+(r_1-r_2{)}^2}$

$\Rightarrow l_1=\sqrt{9^2+(8-7{)}^2}$

$\Rightarrow l_1=\sqrt{81+1}$

$\Rightarrow l_1=\sqrt{82}cm$

∴ Surface area of the solid formed

= CSA of hemisphere + CSA of frustum + CSA of cone

$=2\pi (7{)}^2+\pi (7+8)l_1+\pi (8)l_2$

$=2\pi (7{)}^2+\pi (15)\sqrt{82}+\pi (8\left)\right(10)$

$=98\pi +15\sqrt{82}\pi +80\pi$

$=(178+15\sqrt{82})\pi c{m}^2$

Hence, the correct answer is option (b).