Question

A toy is in the form of a cone.radius 3.5 cm. sourmounted by a hemisphere of same radius. total height of toy is 15. 5 cm .Find the TSA of toy.​

Answer 1

Answer :

$\boxed{\bf{\large{\ {{214.5 \: cm{ }^{2}}}}}}$

Step-by-step explanation :

$\large{\bf{\text{\ {{GIVEN \: :-}}}}}$

Radius of cone = 3.5 cm

Radius of hemisphere = 3.5 cm

Height of toy = 15.5 cm

$\large{\bf{\text{\ {{TO \: FIND \: :-}}}}}$

Total surface area of toy

$\large{\bf{\text{\ {{FORMULA \: USED \: :-}}}}}$

$curved \: surface \: area \: of \: cone \: = \pi\: r l$

$curved \: surface \: area \: of \: hemisphere \: = 2\pi \: {r}^{2}$

$\large{\bf{\text{\ {{SOLUTION \: :-}}}}}$

The toy is a combination of a hemisphere and a cone.

Given :-

Radius of cone = 3.5 cm

Radius of hemisphere = 3.5 cm

Height of toy = 15.5 cm

According to the figure :-

So, AD = 15.5 cm

OC = OD = OB = 3.5 cm

OA = AD-OD = 15.5 - 3.5 = 12 cm

Now, TSA of toy,

CSA of cone + CSA of hemisphere

$\pi \: r \: l \: + \: 2\pi \: {r}^{2} = \: \pi \: r \: \sqrt{ {h}^{2} + {r}^{2} } + 2\pi \: {r}^{2} \: \: \: \\ => \: \frac{22}{7} \times 3.5 \times \sqrt{(12) {}^{2} + (3.5) {}^{2} } + 2 \times \frac{22}{7} \times ( {3.5)}^{2} \\ => \: 11 \sqrt{144 + 12.25} + 22 \times 3.5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ => \: 11 \sqrt{156.25} + 11 \times 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ => \: 11(12.5) + 77 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ => \: 137.5 + 77 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ => \: 214.5 \: cm {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

So, the total surface area of the toy is $214.5 \: {cm}^{2}$.

Answer 2

$\LARGE\textbf{\underline{\underline{Given :-}}}$

$\large\texttt{↦ Radius of the cone = 3.5 cm .}\\\\\large\texttt{↦ Radius of the hemisphere = 3.5 cm .}\\\\\large\texttt{↦ Height of the toy = 15.5 cm .}$

$\LARGE\textbf{\underline{\underline{To find :-}}}$

$\large\texttt{↦ T.S.A. of the toy = ?}$

$\LARGE\textbf{\underline{\underline{Formula :-}}}$

$\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{green}{ C.S.A_{ ( \: Cone \: ) } = \pi r l }}}\\\\\\\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{green}{ C.S.A _ { ( \: Hemisphere \: ) } = 2 \pi r ^ {2} }}}$

$\LARGE\textbf{\underline{\underline{Solution :-}}}$

$\large\textsf{Given}\begin{cases}\textsf\textcolor{purple}{↦ Radius of the cone = 3.5 cm }\\\\\textsf\textcolor{purple}{↦ Radius of the hemisphere = 3.5 cm }\\\\\textsf\textcolor{purple}{Height of the toy = 15.5 cm }\end{cases}$

$\large: \:\Longrightarrow\textsf{BO = OC = OD = 3.5 cm......... ( All are the radius of the hemisphere)}\\\\\\\large: \:\Longrightarrow\textsf{AD = AO + OD}\\\\\\\large: \:\Longrightarrow\textsf{15.5 = AO + 3.5}\\\\\\\large: \:\Longrightarrow\textsf{- AO = 3.5 - 15.5}\\\\\\\large: \:\Longrightarrow\textsf{- AO = - 12 }\\\\\\\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{AO = 12 cm .}}}$

• We have to find the slant height ( l ) of the cone , so that we can find the C.S.A ( Curved Surface Area ) of the cone .

• To find the slant height we have to use the Pythagoras theorem .

• As the height of the cone and the radius of the cone make a angle of 90° with eachother so by using the Pythagoras theorem we can find the Slant height of the cone.

$\large\bigstar\textsf\textcolor{orange}{\: \: \: By Using Pythagoras theorem :-}\\\\\\\large: \:\Longrightarrow\textsf{AC² = OC² + AO²}\\\\\\\large: \:\Longrightarrow\textsf{AC² = 3.5² + 12²}\\\\\\\large: \:\Longrightarrow\textsf{AC² = 12.25 + 144}\\\\\\\large: \:\Longrightarrow\textsf{AC² = 156.25}\\\\\\\large: \:\Longrightarrow\textsf{ AC = \sqrt{156.25} }\\\\\\\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{AC = 12.5 cm .}}}$

$\large\bigstar\textsf\textcolor{orange}{\: \: \: C.S.A_{( \: Cone \: ) } = \cfrac{22}{7} × 3.5 × 12.5 }\\\\\\\large: \:\Longrightarrow\textsf{ = \cfrac{22}{7} × 43.75 }\\\\\\\large: \:\Longrightarrow\textsf{ = \cancel\cfrac{22}{7} × \cancel{43.75} }\\\\\\\large: \:\Longrightarrow\textsf{= 22 × 6.25}\\\\\\\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{ C.S.A _ { ( \: Cone \: ) }= 137.5 \: cm^{2} }}}$

$\large\bigstar\textsf\textcolor{orange}{\: \: \: C.S.A _{ ( \: Hemisphere \: ) }= 2 × \cfrac{22}{7} × 3.5 ^ {2} }\\\\\\\large: \:\Longrightarrow\textsf{ = 2 × \cfrac{22}{7}× 12.25 }\\\\\\\large: \:\Longrightarrow\textsf{ = \cfrac{22}{7} × 24.5 }\\\\\\\large: \:\Longrightarrow\textsf{ \cancel\cfrac{22}{7} ×\cancel{24.5} }\\\\\\\large: \:\Longrightarrow\textsf{= 22 × 3.4}\\\\\\\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{ C.S.A_{ ( \: Hemisphere \: )}= 74.8 \: cm^{2} }}}$

$\large\bigstar\textsf\textcolor{orange}{\: \: \: T.S.A. _ { ( \: Toy \: ) } = C.S.A_{ ( \: Cone + Hemisphere \: ) } }\\\\\\\large: \:\Longrightarrow\textsf{= 137.5 + 74.8 }\\\\\\\large: \:\Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{ T.S.A_ { ( \: Toy \: ) } = 212.3 \: cm^{2} }}}$