Question

If both radius and height of a right circular cone are doubled then ratio of its original CSA(curved surface area) to new CSA is what?

PLEASE GIVE FULL SOLUTIONS WITH ALL FORMULAS AND REASONS

Answer 1

Answer:

• Required Ratio is 1 : 8

Explanation:

Let,

Initial radius of cone be r

Initial height of cone be h

then,

Initial CSA of cone would be

→ Initial CSA of cone = 1/3 π r² h  ____eqn(1)

Now,

When radius and height are doubled then,

New radius of cone = 2 r

New height of cone = 2 h

then,

New CSA of cone would be,

→ New CSA of cone = 1/3 π (2r)² (2h)

→ New CSA of cone = 1/3 π × 4r² × 2 h

→ New CSA of cone = 8/3 π r² h   ____eqn(2)

Now,

Ratio of original CSA and new CSA would be

→ initial CSA / new CSA = ( 1/3 π r² h ) / ( 8/3 π r² h )

→ Initial CSA / new CSA = 1 / 8

Therefore,

• Ratio of its original CSA to new CSA is 1 : 8.

Answer 2

Answer:

Let the Original Radius of cone be r

Let the Original Height of cone be h

⠀

A.T.Q.

⠀

New Radius of cone be 2r (i.e. Doubled)

New Height of cone be 2h (i.e. Doubled)

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Original\:CSA_{cone}:New\:CSA_{cone}\\\\\\:\implies\sf \dfrac{1}{3}\pi r^{2}h : \dfrac{1}{3}\pi (2r)^{2}(2h)\\\\{\scriptsize\qquad\bf{\dag}\:\:\frak{Canceling \:\: \bigg(\frac{1}{3}\pi \bigg) \:\:from \:\:both \:\:sides}}\\\\:\implies\sf r^{2}h : (2r)^{2}(2h)\\\\\\:\implies\sf r^{2}h :4r^{2} \times 2h\\\\\\:\implies\sf r^{2}h : 8r^{2}h\\\\{\scriptsize\qquad\bf{\dag}\:\:\frak{Canceling \:\:(r^{2}h)\:\:from \:\:both \:\:sides}}\\\\:\implies\underline{\boxed{\sf 1:8}}$

⠀

$\therefore\:\underline{\textsf{Hence, required ratio of CSA is \textbf{1:8}}}.$

$\rule{180}{1.5}$

$\boxed{\begin{minipage}{6 cm}\bigstar \:\underline{\textbf{Formulae Related to Cone :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:CSA = \pi rl\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:TSA = Area\:of\:Base + CSA\\{\quad\:\:\:\qquad=\pi r^2+\pi rl}\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\dfrac{1}{3}\pi r^2h\end{minipage}}$