Question

A toy has two parts. One part is cone shaped and has radius of 5 cm. and it is placed on the top of a hemisphere having similar radius. Total height of toy is 17 cm. Find the total surface area of the toy? ​

Answer 1

Given :

• Radius of cone shaped portion is 5 cm.
• The conical portion is placed over a hemisphere having radius similar as the cone.
• Total height of the toy is 17 cm.

To Find :

• Total Surface Area of the toy.

Solution :

The radius of the hemisphere is 5 cm.

Total Height of the toy = 17 cm.

Height of the cone will be difference between total height of the toy and radius of the cone.

$\sf{Height_{cone}\:=\:Total\:Height_{toy}\:-\:Radius_{cone}}$

$\longrightarrow$ $\sf{Height_{cone}\:=\:17\:-5}$

$\longrightarrow$ $\sf{Height_{cone}\:=\:12\:cm}$

Radius of cone = 5 cm.

Since, we have to find the curved surface area, we require the slant height of the cone.

Formula :

$\large{\boxed{\bold{l^2\:=\:r^2\:+\:h^2}}}$

Where,

• l = slant height
• r = radius
• h = perpendicular height

Block in the available data,

$\longrightarrow$ $\sf{l^2=5^2+12^2}$

$\longrightarrow$ $\sf{l^2=25+144}$

$\longrightarrow$ $\sf{l^2=169}$

$\longrightarrow$ $\sf{l=\sqrt{169}}$

$\longrightarrow$ $\sf{l=13\:cm}$

•°• Slant height of the cone = 13 cm

Now using the formula of CSA of cone.

Formula :

$\large{\boxed{\bold{CSA\:=\:\pi\:r\:l}}}$

Block in the available data,

$\longrightarrow$ $\sf{CSA\:=\:\pi\:\times\:5\:\times\:13}$

$\longrightarrow$ $\sf{CSA\:=\:\pi\:65}$

$\longrightarrow$ $\sf{CSA\:=\:65\:\pi\:\:\:(1)}$

Now, calculating the curved surface area of the hemisphere.

Radius = 5 cm

Formula :

$\large{\boxed{\bold{CSA\:=\:2\:\pi\:r^2}}}$

Block in the available data,

$\longrightarrow$ $\sf{CSA\:=\:2\:\pi\:\times\:5^2}$

$\longrightarrow$ $\sf{CSA\:=\:2\:\pi\:\times\:25}$

$\longrightarrow$ $\sf{CSA\:=\:50\:\pi\:\:\:(2)}$

Now, the total surface area of the toy will be sum of equation 1 and 2.

$\sf{TSA_{toy}\:=\:CSA_{cone}\:+\:CSA_{hemisphere}}$

$\longrightarrow$ $\sf{TSA_{toy}\:=\:65\:\pi\:+\:50\:\pi}$

$\longrightarrow$ $\sf{TSA_{toy}\:=\:\pi\:(65\:+\:50)}$

$\longrightarrow$ $\sf{TSA_{toy}\:=\:\pi\:(115)}$

$\longrightarrow$ $\sf{TSA_{toy}\:=\:\dfrac{22}{7}\:\times\:115}$

$\longrightarrow$ $\sf{TSA_{toy}\:=\:22\:\times\:16.42}$

$\longrightarrow$ $\sf{TSA_{toy}\:=\:361.24}$

$\large{\boxed{\bold{\purple{Total\:Surface\:Area\:of\:Toy\:=\:361.24\:cm^2}}}}$

Answer 2

AnswEr:

$\boxed{\setlength{\unitlength}{.8mm}\begin{picture}(55,25)\thicklines\put(35,18){\vector(2,-1){20}}\put(20,18){\vector(-2,-1){20}}\put(17,21){\sf\large\underline{\bigstar \:Toy}}\put(0,1){\sf{Hemisphere}}\put(25,1){+}\put(43,1){\sf{Cone}}\end{picture}}$

$\bf{Toy}\begin{cases}\textsf{Radius = 5cm}\\\textsf{Height = 17cm}\end{cases}$

$\bf{Cone}\begin{cases}\textsf{Radius = 5cm}\end{cases}$

$\bf{Hemisphere}\begin{cases}\textsf{Radius = 5cm}\\\textsf{Height = 17 - 5 = 12cm}\end{cases}$

$\rule{170}1$

$\underline{\bigstar\:\sf{Total \: surface \: area \: of \: toy:}}$

$\normalsize\ : \implies\sf\ T.S.A_{toy} = C.S.A_{cone} + C.S.A_{Hemisphere} \\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \pi rl + 2 \pi r^{2} \\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \pi r \sqrt{r^2 + h^2} + 2 \pi r^{2} \\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \pi \left(r \sqrt{r^2 + h^2} + 2 \times\ r^{2} \right) \\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \pi \left( r\sqrt{(5)^2 + (12)^2} + 2 \times\ 5 \times\ 5 \right)\\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \pi \left(5 \sqrt{25 + 144} + 2 \times\ 25 \right)\\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \pi \left( 5 \times\ 13 + 50 \right)\\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \frac{22}{7} \left( 65 + 50 \right)\\\\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \frac{22}{7} \left( 115 \right)\\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \frac{22 \times\ 115}{7}\\\\\\\normalsize\ : \implies\sf\ T.S.A_{toy} = \frac{\cancel{2530}}{\cancel{7}} = 361.42\\\\\\\normalsize\ : \implies{\underline{\boxed{\bf \green{T.S.A_{toy} = 361.42 \: cm^{2} }}}}$