Question

Please explain example 26 .spam answers will be deleted on the spot.

Answer 1

Answer:

2.24 m

Step-by-step explanation:

Given, Diameter of tank = 17.5 cm.

∴ radius(r) = 8.75 m.

Width of embankment = 4 m.

∴ Radius of tank with embankment(R) = 8.75 + 4 = 12.75 m.

Height of embankment = 2 m.

∴ Volume of soil spread on the embankment:

= πh(R² - r²)

= 22/7 * 2(R + r)(R - r)

= (22/7) * 2 * (12.75 + 8.75)(12.75 - 8.75)

= (22/7) * 2 * (21.5)(4)

=  3784/7

∴ Volume of the tank = πr²h

⇒ 3784/7 = (22/7) * (8.75)²  h

⇒ 172 = (8.75)² * h

⇒ h = 2.246

⇒ h = ~2.24 m.

Therefore,depth of the tank = 2.24 m.

Hope it helps!        

Answer 2

HIIIII BUDDY!!!!!



Let the radius be r


Radius of the tank=
$= 1 \div 2 \times 17.5$




$= 35 \div 4$


Let d meters be the depth of the well which is dug out



Volume of the soil dug
$= \pi \: r {}^{2} h$



$= \pi \times (35 \div 4) {}^{2} \times d \: \: m {}^{3}$


INTERNAL RADIUS OF THE EMBANKMENT
$= 35 \div 4$



width of the embankment=4 m



Therefore, the external radius of the embankment=
$= (35 \div 4 + 4)m$



$= 51 \div 4 \: \: m$



Height of the embankment= 2m


Therefore, the volume of the soil used in forming the embankment around the circular tank
$= \pi(r1 {}^{2} - r2 {}^{2} )$


$= \pi(r1 + r2)(r1 - r2)$


$= \pi(51 \div 4 + 35 \div 4) \times 2 \: m {}^{3}$


$= \pi \times 86 \div 4 \times 4 \times 2 \: m{}^{3}$


$= \pi \times 86 \times 2 \: m {}^{3}$





According to the question ,


$= \pi \times 35 \div 4 \times 35 \div 4 \times d$




$= \pi \times 86 \times 2$




$d = 172 \times 16 \div 35 \times 35$





$d = 2.247 \: m$





Therefore, the depth of the tank dug out=2.25m(approximately)





THANKS



I THINK IT HELPS.....



@Garu1678