Question

Please any Brainly Star or Moderator Do this.

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

Answer:

⋆ DIAGRAM :

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

⠀

☢ Given : Lateral surface area is twice as big as the Base Area of a cone.

⠀

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

$\rule{150}{1}$

☯ $\underline{\textsf{By Using Pythagoras Theorem :}}$

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

$\rule{200}{2}$

☯ $\underline{\textsf{Total Surface Area of the Cone :}}$

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

⠀

$\therefore\:\underline{\textsf{Total Surface Area of Cone is 3) \textbf{81 \pi }}}.$

$\rule{150}{2.2}$

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