Match the two columns.
$\begin{matrix} C o l u m n - I & C o l u m n - I I (A) T o t a l s u r f a c e a r e a o f c y l i n d e r i s & (p) I t s t o t a l s u r f a c e a r e a - b a s e a r e a (B) C u r v e d s u r f a c e a r e a o f h e m i s p h e r e & (q) H a l f o f t h e c i r c u m f e r e n c e o f B a s e \times h e i g h t^{2} + b a s e^{2} (C) T o t a l s u r f a c e a r e a o f a c o n e i s & (2 \times b a s e a r e a) + (c i r c u m f e r e n c e o f c i r c l e b a s e \times h e i g h t) (D) C u r v e d s u r f a c e a r e a o f a c o n e i s & (s) \frac{a r e a o f b a s e \times (s l a n t h e i g h t + b a s e r a d i u s)}{r a d i u s o f c i r c u l a r b a s e} \end{matrix}$
Choose the correct option:

(A) \to q; (B) \to r; (C) \to s; (D) \to p

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(A) \to r; (B) \to p; (C) \to s; (D) \to q

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(A) \to p; (B) \to r; (C) \to q; (D) \to s

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(A) \to s; (B) \to q; (C) \to p; (D) \to r

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Solution

The correct option is B $(A) \to r; (B) \to p; (C) \to s; (D) \to q$
(A) TSA of cylinder $= 2 π r^{2} + 2 π r h$
$= 2 π r^{2} + 2 π r h$
$2 \times b a s e a r e a + c i r c u m f e r e n c e o f b a s e \times h e i g h t$
(B) CSA of hemisphere $= 2 π r^{2}$
$= 3 π r^{2} - π r^{2}$
= TSA of hemisphere – base area
(C) TSA of cone $= π r (r + l) = \frac{π r^{2} (r + l)}{r}$
$= \frac{a r e a o f b a s e \times (r a d i u s + h e i g h t)}{r a d i u s}$
(D) CSA of cone $= π r l$
$= \frac{1}{2} \times 2 π r \times \sqrt{h^{2} + r^{2}} [∵ l^{2} = h^{2} + r^{2}]$
$= \frac{1}{2} o f c i r c u m f e r e n c e \times \sqrt{h e i g h t^{2} + b a s e^{2}}$
$∴$ option (b) is correct.

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