KJSEA 2025 Maths Q10 — Surface Area of a Cylinder (Water Contact)

KJSEA 2025 Grade 9 Measurement

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The Question

“A cylindrical bucket has a diameter of 40 cm and is filled with water up to a height of 42 cm. The bucket is open at the top. Find the total surface area of the bucket that is in contact with the water, using the calculator value of pi.”

1

Decide which surfaces touch the water

Picture where the water sits inside the bucket. It rests on the flat circular base at the bottom and presses against the curved wall all the way around, up to the water height. The bucket is open at the top, so the water has no lid to touch. This means we count two surfaces only: the circular base and the curved wall — not a top.

Reading the situation correctly is the whole question; missing or adding a surface changes the answer.

2

Find the radius from the diameter

Every cylinder formula uses the radius, not the diameter. The radius is half the diameter, so halve the 40 cm across the top to get the radius.

r=402=20 cmr = \frac{40}{2} = 20 \text{ cm}
3

Work out the area of the circular base

The base is a flat circle, so its area is pi times the radius squared. Square the radius first, then multiply by pi using the calculator value.

Abase=πr2=π×202=1256.64 cm2A_{\text{base}} = \pi r^{2} = \pi \times 20^{2} = 1256.64 \text{ cm}^{2}
4

Work out the curved wall area

Unroll the curved wall in your mind and it becomes a rectangle: its height is the water height and its length is the circumference of the circle. That gives the curved surface area as 2 times pi times radius times height.

Acurved=2πrh=2×π×20×42=5277.87 cm2A_{\text{curved}} = 2 \pi r h = 2 \times \pi \times 20 \times 42 = 5277.87 \text{ cm}^{2}
5

Add the two areas together

The water touches both surfaces at once, so the total contact area is the base plus the curved wall added together.

A=Abase+Acurved=1256.64+5277.87=6534.51 cm2A = A_{\text{base}} + A_{\text{curved}} = 1256.64 + 5277.87 = 6534.51 \text{ cm}^{2}

Final Result

The surface area of the bucket in contact with the water is 6,534.51 cm² — option C.

Why this method works

The method works because you break a curved 3D shape into pieces whose areas you already know. A flat circular base is just a circle, so its area is pi r squared. The curved wall may look hard, but if you imagine slitting it and rolling it flat it becomes a rectangle whose height equals the water level and whose length equals the distance around the circle (the circumference, 2 pi r). Multiplying those sides gives 2 pi r h. Because the bucket is open at the top, no lid touches the water, so you deliberately leave out the second circle you would include for a fully closed cylinder. Adding only the surfaces the water actually meets gives the correct contact area.

✓

Base 1256.64 plus curved wall 5277.87 equals 6534.51 cm², matching the sum, and only two surfaces were counted because the top is open.