Question

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Chapter Surface area and volumes
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Answer 1

Radius of cone = r/2 units

Slant height of cone = 2l units

We know that :-

$\underline{ \boxed{ \large \tt T.S.A = \pi \: r \: l +\pi {r}^{2} }} \\ \\ \tt \: where : r \: = radius \\ \tt \: \: \: \: l = slant \: height$

T.S.A = Total Surface Area

Now , we will find TSA of cone whose radius is r/2 and slant height is 2l.

$\large \tt \implies \: T.S.A = (\pi \times \dfrac{r}{2} \times 2 l) +\pi {(\dfrac{r}{2})}^{2}$

$\large \tt \implies \: T.S.A = (\pi \times \dfrac{r}{1} \times l) +\pi {(\dfrac{ {r}^{2} }{4})}^{}$

$\large \tt \implies \: T.S.A = \pi r l + \dfrac{\pi \: {r}^{2}}{4}$

taking out πr common :-

$\large \tt \implies \: T.S.A = \pi \: r ( l + \dfrac{ \: r}{4})$

Answer :

=> Option second πr(l+r/4) is correct

$\large \tt \underline{\red{Additional-Information:}}$

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = (4/3)πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Step-by-step explanation:

Radius of cone = r/2 units

Slant height of cone = 2l units

We know that :-

\begin{gathered} \underline{ \boxed{ \large \tt T.S.A = \pi \: r \: l +\pi {r}^{2} }} \\ \\ \tt \: where : r \: = radius \\ \tt \: \: \: \: l = slant \: height\end{gathered}

T.S.A=πrl+πr

2

where:r=radius

l=slantheight

T.S.A = Total Surface Area

Now , we will find TSA of cone whose radius is r/2 and slant height is 2l.

\large \tt \implies \: T.S.A = (\pi \times \dfrac{r}{2} \times 2 l) +\pi {(\dfrac{r}{2})}^{2}⟹T.S.A=(π×

2

r

×2l)+π(

2

r

)

2

\large \tt \implies \: T.S.A = (\pi \times \dfrac{r}{1} \times l) +\pi {(\dfrac{ {r}^{2} }{4})}^{}⟹T.S.A=(π×

1

r

×l)+π(

4

r

2

)

\large \tt \implies \: T.S.A = \pi r l + \dfrac{\pi \: {r}^{2}}{4}⟹T.S.A=πrl+

4

πr

2

taking out πr common :-

\large \tt \implies \: T.S.A = \pi \: r ( l + \dfrac{ \: r}{4})⟹T.S.A=πr(l+

4

r

)

Answer :

=> Option second πr(l+r/4) is correct

\large \tt \underline{\red{Additional-Information:}}

Additional−Information:

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = (4/3)πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²