Surface Area of Prisms and Pyramids worksheet for Grade 7, featuring nine geometric figures with dimensions for calculating surface area.

Here are the step-by-step solutions for finding the surface area of each figure.

General Rule:
To find the Surface Area (SA), we add the area of all the faces together.
* For Prisms: $SA = (2 \times \text{Area of Base}) + (\text{Perimeter of Base} \times \text{Height})$
* For Pyramids: $SA = \text{Area of Base} + \text{Area of Triangular Faces}$

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1) Triangular Prism
* Base: Triangle with base $2\text{ ft}$ and height $2\text{ ft}$. The other two sides are not labeled, but looking at problem 6 and 8 which are identical, and standard problems of this type, it is likely an isosceles triangle or right triangle. Let's look closer. Actually, usually, if only one height is given inside, we calculate the base area. For the lateral area, we need the perimeter.
* Wait, let's look at the diagram again. It shows a triangular prism. The base is a triangle with base $2$ and height $2$. The length of the prism is $4$.
* We need the lengths of the other two sides of the triangle to get the perimeter. In many textbook problems like this, if it looks like an isosceles triangle with height bisecting the base, the sides would be $\sqrt{1^2 + 2^2} = \sqrt{5} \approx 2.24$. However, often these diagrams imply specific side lengths. Let's look at Problem 6 and 8. They are the same shape. Problem 8 has no units, just numbers. Problem 1 has units.
* Let's re-examine Problem 1. The triangle has a base of $2$ and a height of $2$. It doesn't explicitly give the slant height or side lengths. However, in similar problems on this worksheet (like #3 and #4), the slant height is given. Here it is missing. Let's assume it's a right triangle? No, the height line is in the middle. Let's assume it is an isosceles triangle where the height splits the base into two $1\text{ ft}$ segments.
* Side length $= \sqrt{1^2 + 2^2} = \sqrt{5} \approx 2.236\text{ ft}$.
* Perimeter $= 2 + 2.236 + 2.236 = 6.472\text{ ft}$.
* Area of two bases $= 2 \times (0.5 \times 2 \times 2) = 4\text{ sq ft}$.
* Lateral Area $= \text{Perimeter} \times \text{Length} = 6.472 \times 4 = 25.888\text{ sq ft}$.
* Total SA $= 4 + 25.89 = 29.89\text{ sq ft}$.
* *Alternative interpretation:* Is it possible the side lengths are integers? If the height was different... No, let's stick to the calculation. $\sqrt{5}$ is irrational. Let's check if there's a simpler interpretation. Maybe the "2 ft" labels on the bottom refer to the two equal sides? No, they are under the segments.
* Let's look at Problem 6. Same dimensions. Problem 8. Same dimensions.
* Let's check Problem 2. Base $3$, Height not given? Ah, in Problem 2, the tick marks suggest it's equilateral? Or maybe the height is derived? No, usually height is given. Wait, in Problem 2, there is a right angle symbol at the bottom vertex of the triangle? No, that's for the prism height.
* Let's look really closely at Problem 1. The "2 ft" and "2 ft" are labeling the two segments of the base created by the altitude. This confirms the total base is $4\text{ ft}$? No, the dimension lines are separate. It looks like the total base is $2+2=4$? Or is the base $2$ and the height $2$? The label "2 ft" is under the left half and "2 ft" under the right half? That would make the base $4$. Let's assume the base is $4$ and height is $2$.
* If base $= 4$, half-base $= 2$. Height $= 2$. Side $= \sqrt{2^2+2^2} = \sqrt{8} \approx 2.828$.
* Perimeter $= 4 + 2.828 + 2.828 = 9.656$.
* Lateral Area $= 9.656 \times 4 = 38.624$.
* Base Area $= 0.5 \times 4 \times 2 = 4$. Two bases $= 8$.
* Total $= 46.62$.
* Let's try another interpretation. Maybe the base is $2$ total, and the height is something else? No, the arrows clearly mark the segments.
* Let's look at Problem 3. Base $8$, height $6$. Half base $4$. Side $\sqrt{4^2+6^2} = \