Question

Rasheed got a playing top(lattu) as his birthday present which surprisingly had no colour on it. he want to colour it with caryons .the top is shaped like a cone surmounted by a hemisphere the entire top is 5 cm in height and the diameter of top is 3.5cm find the area he has to colour?
(Take = π=22/7)



Good answers only​

Answer 1

$\large{\underbrace{\underline{\mathrm{Understanding\:the\:concept}}}}$

Rasheed got a playing top on his birthday present which have no colour in it. Now rasheed wants to colour it. The shape of the top is like a cone surmounted by its hemisphere. The entire top is 5cm in height and the color by its diameter is 3.5cm, Now we have to find the area of the top that he wants to color.

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$\huge\bold{Lets\:solve}$

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$\large \sf { TSA\:of\:the\:toy\:=CSA\:of\:the\:hemisphere+CSA\:of\:cone}$

Now, the curved surface area of the hemisphere

$\large \sf \color{pink} \frac{1}{2} (4\pi \: r^{2} ) = 2\pi \: r^{2}$

$\large \sf\color{cyan} (2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} )cm^{2}$

Also, the height of cone = Height of top – Height (radius) hemispherical part.

$\large\sf\color{pink} = (5 - \frac{3.5}{2} )cm = 3.25 \: cm$

So the slant height of cone$(l)$

$\large \sf \color{cyan}= \sqrt{r} ^{2} + h ^{2} = \sqrt{( \frac{3.5} {2} } )^{2} + (3.25)^{2} cm = 3.7cm$

Therefore, CSA of the cone = π rl

$\large \sf \color{pink}( \frac{22}{7} \times \frac{3.5}{2} \times 3.7)cm^{2}$

This gives the surface of area top as

$\large \sf \color{cyan} = (2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} )cm^{2} + ( \frac{22}{7} \times \frac{3.5}{2} \times 3.7)cm ^{2}$

$\large \sf \color{pink} = \frac{22}{7} \times \frac{3.5}{2} (3.5 + 3.7)cm^{2} = \frac{11}{2} \times (3.5 + 3.7cm ^{2} ) = 39.6 \: cm ^{2}$

　　　　　　 $\large{\underline{\boxed{\mathfrak{= 39.6 cm^2 (approx.)}}}}$

You may note that "total surface of area top" is not the sum of the total surface area of the cone and the hemisphere.

Answer 2

Answer:

The area Rasheed has to colour is 39.05 cm².

Step-by-step explanation:

${\underline{\underline{\bf{Given:}}}}$

• Rasheed want to colour a playing top.
• The structure of a playing top is like a cone surmounted by a hemisphere.
• The total height of the entire top is 5 cm.
• Diameter of the top is 3.5 cm.

${\underline{\underline{\bf{To\: find:}}}}$

• The total area he has to colour.

${\underline{\underline{\bf{Formulae\:to\:be\:used}}}}$

$\underline{\boxed{\sf{{CSA}_{(Cone)}=\pi rl\:sq.units}}}$

$\underline{\boxed{\sf{{Slant\: height}_{(Cone)}=\sqrt{r^2+h^2}\:units}}}$

$\underline{\boxed{\sf{{CSA}_{(Hemisphere)}=2\pi r^2\:sq.units}}}$

$\:\:\:\:\:\:\:\:\:\sf{\bullet\:r=radius}$

$\:\:\:\:\:\:\:\:\:\sf{\bullet\:l=slant\:height}$

$\:\:\:\:\:\:\:\:\:\sf{\bullet\:h=height}$

${\underline{\underline{\bf{Solution:}}}}$

‎ ‎ ‎ ‎ ‎ ‎We are said that the structure of the top is like a cone surmounted by a hemisphere. So, the area to be coloured equals the sum of CSA of cone and CSA of hemisphere.

‎ ‎ ‎ ‎ ‎ ‎Why not TSA? TSA is isn't adviced to use here because, the conical shape is already surmounted by the hemispherical shape. So, the position of base of the hemisphere and cone is inside the the surface area of the top and we can't colour it. Hence, we use the formulas of CSA.

$\:$

${\boxed{\sf{CSA\:of\:cone}}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎$\leadsto{\sf{\purple{\pi rl\:sq.units}}}$

Here, we are said to take the value of π is 22/7. And the measures of radius and slant height are unknown. So, first let's find them!

Radius:

‎ ‎ ‎ ‎ ‎$\mapsto{\bf{\dfrac{1}{2}\times Diameter}}$

$\\\implies{\sf{\dfrac{1}{2}\times 3.5}}$

$\\\implies\underline{\sf{1.75\:cm}}$

Slant height:

‎ ‎ ‎ ‎ ‎$\mapsto{\bf{\sqrt{r^2+h^2}}}$

The measure of height of the cone equals:

$\rightarrow{\sf{Total\:height-Radius}}$

$\rightarrow{\sf{5-1.75}}$

$\rightarrow{\sf{3.25\:cm}}$

Now, by inserting the obtained measures in the formula:

$\\\:\:\:\:\:\::\implies{\sf{\sqrt{(1.75)^2+(3.25)^2}}}$

$\\\:\:\:\:\:\::\implies{\sf{\sqrt{3.0625+10.5625}}}$

$\\\:\:\:\:\:\::\implies{\sf{\sqrt{13.625}}}$

$\\\:\:\:\:\:\::\implies\underline{\sf{3.6\:cm}}$

$\:$

We've obtained the measures of radius and slant height. So, we can proceed finding the answer by substituting the obtained measures in the formula of CSA of cone:

$\\ \:\:\:\:\:\::\implies{\sf{\pi rl\:sq.units}}$

$\\ \:\:\:\:\:\:\:\:\::\implies{\sf{\dfrac{22}{7}\times 1.75\times 3.69}}$

$\\ \:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{5.5\times 3.6}}$

$\\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies\underline{\bf{19.8\:cm^2}}$

$\:$

${\boxed{\sf{CSA\:of\: hemisphere}}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎$\leadsto{\sf{\purple{2\pi r^2\:sq.units}}}$

By inserting the measures,

$\\ \:\:\:\:\:\::\implies{\sf{2\times \dfrac{22}{7}\times (1.75)^2}}$

$\\ \:\:\:\:\:\:\:\:\::\implies{\sf{2\times \dfrac{22}{7}\times 3.0625}}$

$\\ \:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{2\times 9.625}}$

$\\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies\underline{\bf{19.25\:cm^2}}$

$\:$

${\boxed{\sf{Area\:to\:be\:coloured}}}$

$\\\leadsto{\bf{\purple{CSA\:of\:cone+CSA\:of\: hemisphere}}}$

$\\ \:\:\:\:\:\::\implies{\sf{19.8+19.25}}$

$\\ \:\:\:\:\:\:\:\:\:\::\implies{\underline{\underline{\frak{\red{39.05\:cm^2}}}}}$

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