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