I'm unsure if I miss any pieces in the calculation for its surface area...


The diagram shows the uniform cross-section of a solid paper weight ABCDE, which is in the shape of a trapezium with a semi-circular hole ABC cut out of it. It is given that AC is parallel to DE, AC = $7$ cm, CD = $15$ cm, DE = $21$ cm, AE = $13$ cm, and the height of the trapezium is $12$ cm. The given height is $30$ cm.

The area for the cross section should be:

Area of trapezium - Area of semi-circle = $$0.5*(7+21)*(12) - 0.5*\pi*(3.5)^2 = 148.7 \ \text{cm}^2$$

Total Surface Area:

Area of side 15 + area of side 21 + area of side 13 + area of top and area of bottom + area of semi-cylinder =

$$(15*30) + (21*30) + (13*30) + (2*148.7) + \pi*(3.5)^2 + \pi*3.5*30 = 2135.2 \text{cm}^2$$

The answer for the surface area is $2430 \ \text{cm}^2$, but the closest I've got is $2135.2 \ \text{cm}^2$ so what did I miss?

enter image description here

  • $\begingroup$ Please use MathJax to write your questions. Could you please also write the problem statement in text instead of in an image? $\endgroup$
    – Toby Mak
    May 25, 2020 at 5:37
  • $\begingroup$ Where does "$30$" come from in your calculation? Nothing that you've given mentions how long the paper weight is. Also, you never state the actual question you are trying to solve anywhere. Apparently it is to find the total surface area of the paperweight, but you don't state it. If you want us to help you, the first thing to do is to fully state the problem you want help with. $\endgroup$ May 25, 2020 at 13:36
  • $\begingroup$ @PaulSinclair I've added the full question. I can't solve for the surface area part.. $\endgroup$
    – user234568
    May 25, 2020 at 15:20
  • $\begingroup$ @user234568 I think your answer is correct. $\endgroup$
    – Lee
    May 25, 2020 at 15:42
  • $\begingroup$ You rounded the cross-sectional area incorrectly: $148.75$ rounds up to $148.8$. $\endgroup$
    – Toby Mak
    Jan 31, 2021 at 4:26

1 Answer 1


Think of the solid as having front and back, left and right, and top and bottom faces. The surface area is made up of the two cross-sectional areas (front and back), two flat rectangles (left and right), a half-cylindrical face (top), and a flat bottom face (bottom).

Thus the surface area is:

$$2(148.8) + 15 \cdot 30 + 13 \cdot 30 + \frac{1}{2} (2 \pi \cdot 3.5) \cdot 30+21\cdot30=2100 \ \text{cm}^2$$

to $3$ significant figures.

The $\pi \cdot (3.5)^2$ shouldn't be there as that region is just empty space, and everything else is correct.


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