The inner diameter of a circular well is $3.5$ m. It is $10$ m deep. Find its inner curved surface area

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Solution

Given, diameter of well is $3.5$ m and depth is $10$ m
Then radius of well $=$ $\frac{3.5}{2} = 1.75$ m
And height of well $= 10$ m
Then curved surface area of well $=$ $2 π r h$
$= 2 \times \frac{22}{7} \times 1.75 \times 10$
$=$ $2 \times 22 \times 0.25 \times 10$
$= 110$ sq. m

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Q. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.

The inner diameter of a circular well is $3.5 m .$ It is $10 m$ deep. Find :

$(i)$ inner curved surface area.

$(i i)$ the cost of plastering this curved surface at the rate of $R s . 40$ per $m^{2}$ .

The inner diameter of a circular well is 3.5 m . It is 10 m Deep find it's curved surface area. The cost of plastering this curved surface at the rate of Rs.40. per m square.

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The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find its inner curved surface area

Given, diameter of well is 3.5 m and depth is 10 mThen radius of well = 3.52=1.75 mAnd height of well =10 mThen curved surface area of well = 2πrh=2×227×1.75×10= 2×22×0.25×10=110 sq. m

The inner diameter of a circular well is $3.5$ m. It is $10$ m deep. Find its inner curved surface area

Given, diameter of well is $3.5$ m and depth is $10$ m
Then radius of well $=$ $\frac{3.5}{2} = 1.75$ m
And height of well $= 10$ m
Then curved surface area of well $=$ $2 π r h$
$= 2 \times \frac{22}{7} \times 1.75 \times 10$
$=$ $2 \times 22 \times 0.25 \times 10$
$= 110$ sq. m