Question

A well of diameter 4m is dug 14m deep. The earth taken out is spread evenly all around the well to form a 40cm high embankment. find the width of the embankment.

Answer 1

Well

d = 4m

r = 2m

H = 14m

Embankment

R = r + w

(w is the width of the embankment)

Therefore, w = R - r

h = 40cm

= 2/5 m

Volume of well = pi(r^2)H

= 22/7 × 2 × 2 × 14

= 176 cm^3

Volume of embankment = volume of well

Pi(R^2)h - pi(r^2)h = pi(r^2)H

Pi h(R^2 - r^2) = pi(r^2)H

h(R+r)(R-r) = (r^2)H

Substitute the values of R and w

H(r+w+r)(w) = r^2H

2/5 (2r+w)(w) = 2×2×14

(2×2 + w)(w) = 4×14× 5/2

(4+w)w = 140

w^2 + 4w -140 = 0

w^2+14w - 10w - 140 = 0

(w+14)(w-10) = 0

w+14=0 w-10=0

w = - 14 w = 10

Therefore width of embankment is 10 m

Answer 2

Answer:

$\red { Width \: of \: the \: embankment (w)}\green {= 10\:m }$

Step-by-step explanation:

Dimensions of a well:

Diameter (d) = 4m,

$Radius (r) = \frac{d}{2} = \frac{4}{2} = \blue {2 \:m}$

$Depth \: of \: the \: well (H) = 14 \:m$

Dimensions of the embankment:

Height(h) = 40 cm

= 0.4 m ,

Let the width of the embankment = w m,

Radius of the embankment = R m,

R = r + w ,

R - r = w,

If the earth taken out is spread evenly all around the well then

Volume of the mud around the well = Volume the mud dug

$\pi (R+r)(R-r) h = \pi r^{2}H$

$\implies \pi ( r+w+r)wh = \pi r^{2}H$

$\implies (2r+w)wh = r^{2}H$

$\implies (2\times 2+w)\times w\times 0.4 = 2^{2}\times 14$

$\implies (4+w)w = \frac{4\times 14}{0.4}$

$\implies w^{2}+4w= 140$

$\implies w^{2}+4w-140=0$

$\implies w^{2}+14w-10w-140=0$

$\implies w(w+14)-10(w+14)=0$

$\implies (w+14)(w-10)=0$

$\implies w+14 = 0\:Or \: w-10 = 0$

$\implies w = -14 \:or \: w = 10$

/* Width of the embankment should not be negative

Therefore.,

$\red { Width \: of \: the \: embankment (w)}\green {= 10\:m }$

•••♪