Question

The lateral surface area is twice as big as the base area of a cone.if the height of the cone is 9,what is the total surface area?
1.9π
2.27π
3.81π
4.90π​

Answer 1

$\huge\sf\pink{Answer}$

☞ Total Surface Area of Cone is (3) 81π

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$\huge\sf\blue{Given}$

✭ Lateral Surface Area is twice as big as the base area of a cone.

✭ Height of cone is

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$\huge\sf\gray{To \:Find}$

◈ Its Total Surface Area?

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$\huge\sf\purple{Steps}$

$\large\sf\star Diagram\star$

$\setlength{\unitlength}{0.99cm}\begin{picture}(6, 4)\linethickness{0.26mm}\qbezier(5.8,2.0)(5.8,2.3728)(4.9799,2.6364)\qbezier(4.9799,2.6364)(4.1598,2.9)(3.0,2.9)\qbezier(3.0,2.9)(1.8402,2.9)(1.0201,2.6364)\qbezier(1.0201,2.6364)(0.2,2.3728)(0.2,2.0)\qbezier(0.2,2.0)(0.2,1.6272)(1.0201,1.3636)\qbezier(1.0201,1.3636)(1.8402,1.1)(3.0,1.1)\qbezier(3.0,1.1)(4.1598,1.1)(4.9799,1.3636)\qbezier(4.9799,1.3636)(5.8,1.6272)(5.8,2.0)\put(0.2,2){\line(1,0){2.8}}\put(3.2,4){\sf{9}}\put(3,2){\line(0,2){4.5}}\put(1.5,1.7){\sf{r}}\qbezier(.2,2.05)(.7,3)(3,6.5)\qbezier(5.8,2.05)(5.3,3)(3,6.5)\put(1,4){\sf l}\put(3,2.02){\circle*{0.15}}\put(2.7,2){\dashbox{0.01}(.3,.3)}\end{picture}$

$\leadsto\textsf{Lateral Surface Area = 2(Base Area)}\\\\ \leadsto\sf \pi rl=2 \times (\pi r^2)\\\\ \leadsto\sf \pi rl=2\pi r^2\\\\ \leadsto\sf l = \dfrac{2\pi r^2}{\pi r}\\\\ \leadsto\sf l =2r \qquad -eq(1)$

Using Pythagoras Theorem

$\twoheadrightarrow\sf l^2=h^2+r^2\\\\{\scriptsize\qquad\bf{\dag}\:\:\textsf{Using value of \:l\: from eq. (1) \& h = 9}}\\\\ \twoheadrightarrow\sf (2r)^2=(9)^2+(r)^2\\\\ \twoheadrightarrow\sf 4r^2 = 81 + r^2\\\\ \twoheadrightarrow\sf 4r^2 - r^2 = 81\\\\ \twoheadrightarrow\sf 3r^2 = 81\\\\ \twoheadrightarrow\sf r^2 = \dfrac{81}{3}\qquad -eq(2)$

Total Surface Area of the Cone

$\dashrightarrow\sf\:\:TSA=Lateral\: Surface\:Area+Base\:Area\\\\ \dashrightarrow\sf\:\:TSA=2Base\:Area+Base\:Area\\\\ \dashrightarrow\sf\:\:TSA=3Base\:Area\\\\ \dashrightarrow\sf\:\:TSA=3 \pi r^2\\\\{\scriptsize\qquad\bf{\dag}\:\:\textsf{Using value of r ^2 from eq. (2)}}\\\\\dashrightarrow\sf\:\:TSA = 3\times \pi \times \dfrac{81}{3}\\\\ \sf\orange{\dashrightarrow{\sf TSA = 81\pi}}$

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Answer 2

Given

• LSA of cone = 2 (base area of cone)
• height of cone , h = 9 units

To find

• TSA of cone

Formula used

● Formula for Lateral surface area of cone

$\red{\bigstar}\boxed{\sf{LSA\:of\:cone=\pi r l }}$

● Formula for Base area of cone

$\red{\bigstar}\boxed{\sf{Base\:area\:of\:cone=\pi r^2}}$

● Formula for TSA of cone

$\red{\bigstar}\boxed{\sf{TSA\:of\:cone=\pi r ( r + l )}}$

● Relation between radius(r) , slant height(l) and perpendicular height (h) of right circular cone using pythagoras theorem

$\red{\bigstar}\boxed{\sf{l^2=h^2+r^2}}$

( where r is radius , h is height , l is slant height of cone )

Figure

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• In the figure
• h = 9 is perpendicular height of cone
• l is slant height of cone
• r is the radius of cone

Solution

Finding the relation between radius and slant height of cone

As given

$\implies\sf{LSA\:of\:cone=2(base\:area\:of\:cone)}\\\\\implies\sf{\pi r l = 2 \pi r^2}\\\\\implies\boxed{\red{\sf{l=2 \; r\:\:\:\:\:\:...eqn(1)}}}$

Calculating the radius and slant height of cone

By Pythagoras theorem in the figure

$\implies\sf{l^2=r^2+h^2}\\\\\sf{using\:eqn\:(1)}\\\\\implies\sf{(2r)^2=r^2+(9)^2}\\\\\implies\sf{4r^2=r^2+81}\\\\\implies\sf{3r^2=81}\\\\\implies\sf{r^2=27}\\\\\implies\boxed{\red{\sf{r=3\sqrt{3}\;units }}}$

so,

$\implies\boxed{\red{\sf{l=2\;r=2(3\sqrt{3})=6\sqrt{3}\;units}}}$

Calculating the TSA of cone

$\implies\sf{TSA\:of\:cone=\pi r (r + l )}\\\\\implies\sf{TSA\:of\:cone=\pi \times 3\sqrt{3}\times \left( \;3\sqrt{3 }\;+\;6\sqrt{3} \;\right)}\\\\\implies\sf{TSA\:of\:cone=\pi \times 3\sqrt{3} \times \left(\; 9 \sqrt{3}\;\right)}\\\\\implies\boxed{\boxed{\large{\red{\sf{TSA\:of\:cone=81\;\pi \;\:\:\;sq.\;units}}}}}$

Hence, the correct option is

(3) 81 π .