Question

Find

(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high.

(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)​

Answer 1

Answer:

i) curved surface area = 59.4cm2

ii) 95.04 m2 steel was used

Step-by-step explanation:

i)Diameter of a cylindrical(closed) petrol tank=4.2 m

Radius= 4.2/2  =2.1 m

Height=4.5 m

Curved Surface area=2πrh=2× 22/7 ×2.1×4.5=59.4sq.m

ii)Total surface area=2πr(r+h)

=2× 22/7  ×2.1×(2.1+4.5)

=2×22/7 ×2.1×6.6

=87.12 sq.m

Let the amount of steel used in making the tank=x

Steel wasted= x/12

Given:Amount of steel required−Amount of steel wasted=T.S.A of the tank.

⇒x− x/12

=87.12

⇒  12x−x/12  =87.12

11x/12   =87.12

x= 12/11×87.12=95.04sq.m

∴95.04sq.m steel was used in actual while making such a tank.

Answer 2

Given :

† Height = 4.5m

†Diameter = 4.2m

To find :

† C.S.A of Cylinder

† Steel was actually used if 1/12 of the Steel actually used was wasted in making the tank.

Solution :

Radius= 2.1 m

Using Formula :

$\bf\boxed { \green{C.S.A.=2\pi{rh}}} \red \bigstar</p><p>$

Putting Values :

$: \implies\sf{{2}\times\dfrac{22}{7}\times\dfrac{21}{10}\times\dfrac{45}{10}}$

$: \implies\sf \: {\cancel\dfrac{297}{5}}</p><p>$

$: \implies\sf\bold{{59.4m}^{2}</p><p>}$

Using Formula :

$\bf\boxed{ \red{T.S.A=2\pi{rh(r+h)}}} \: \blue \bigstar$

Putting Values :

$: \implies\sf{2\times\dfrac{22}{7}\times\dfrac{21}{10}\times{(2.1+4.5)}} : \:$

$:\implies\sf{\dfrac{66}{5}\times\dfrac{(66)}{10}}$

$:\implies\sf{6.6\times{6.6}}$

$:\implies\sf\bold{{87.12m}^{2}}$

Now :

$:\implies\sf\bold{Let\:actually\:steel\:used=x}$

$:\implies\sf{Wasted\:Steel=\dfrac{1x}{12}}$

A.T.Q :

$:\implies\sf{x-\dfrac{1x}{12}=87.12}$

$:\implies\sf{\dfrac{12x-x}{12}=87.12}$

$:\implies\sf{\dfrac{11x}{12}=87.12}$

$:\implies\sf{x=87.12\times\dfrac{12}{11}}$

$:\implies\sf\bold{95.04{m}^{2}}$