Mrs. Newells Math: Surface Area and Volume of Pyramids Unit - Free Printable

Let's solve each question on the quiz step by step and explain the reasoning.

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Question 1: Which solid corresponds to the net shown?

Net Description:
The net shows a central polygon with five triangular faces attached to its sides. The base is a pentagon, and there are five triangles connected to each side of the pentagon.

This matches a pentagonal pyramid (a pyramid with a pentagon base and five triangular faces).

✔ Answer: B. Pentagonal Pyramid

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Question 2: Which solid corresponds to the net shown?

Net Description:
The net shows a triangle as the base, with three other triangles attached to each side. This forms a triangular pyramid (tetrahedron), which has four triangular faces.

But the options are:
- A. Square Pyramid
- B. Pentagonal Pyramid
- C. Hexagonal Pyramid
- D. None of the above

Since it's a triangular pyramid, none of the listed options match exactly — square, pentagonal, or hexagonal pyramids have different bases.

✔ Answer: D. None of the above

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Question 3: Find the slant height of the square pyramid.

Given:
- Base edge = 10 units
- Height of pyramid = 12 units (from apex to center of base)
- We need slant height (distance from apex to midpoint of a base edge)

Step-by-step:

1. The slant height forms a right triangle with:
- Vertical leg = height of pyramid = 12 units
- Horizontal leg = half of base edge = 10 ÷ 2 = 5 units

2. Use Pythagorean Theorem:
$$
\text{Slant height} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13
$$

✔ Answer: B. 13 units

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Question 4: Find the height of the triangular pyramid.

Given:
- Slant height (l) = 15 cm
- Base edge = 18√3 cm (this is the length of the side of the equilateral triangle base)
- We need the height of the pyramid (vertical distance from apex to base center)

Assuming this is a regular triangular pyramid (regular tetrahedron), but more likely a regular pyramid with equilateral triangle base.

We can use geometry:

1. The centroid of an equilateral triangle divides the median in a 2:1 ratio.
2. First, find the height of the base triangle:
$$
h_{\text{base}} = \frac{\sqrt{3}}{2} \times 18\sqrt{3} = \frac{\sqrt{3}}{2} \times 18\sqrt{3} = \frac{3}{2} \times 18 = 27 \text{ cm}
$$
So, the median is 27 cm, and the distance from centroid to vertex is:
$$
\frac{2}{3} \times 27 = 18 \text{ cm}
$$

2. Now, consider the right triangle formed by:
- Slant height = 15 cm (hypotenuse)
- Distance from centroid to base edge midpoint = 18 cm (but wait — actually, we want from apex to centroid)

Wait! Let's re-evaluate.

Actually, slant height is from apex to midpoint of a side, not to a vertex.

So:
- The distance from centroid to the midpoint of a side is the inradius of the equilateral triangle.

For an equilateral triangle of side $ s $, the inradius $ r = \frac{s}{2\sqrt{3}} $

$$
r = \frac{18\sqrt{3}}{2\sqrt{3}} = 9 \text{ cm}
$$

Now, we have a right triangle:
- Hypotenuse = slant height = 15 cm
- One leg = inradius = 9 cm
- Other leg = height of pyramid = ?

Use Pythagorean Theorem:
$$
h^2 + 9^2 = 15^2 \\
h^2 + 81 = 225 \\
h^2 = 144 \\
h = 12 \text{ cm}
$$

✔ Answer: D. 12 cm

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Question 5: Find the volume of the square pyramid.

Given:
- Base edge = 12 in
- Slant height (l) = 10 in

We need volume:
$$
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
$$

But we’re given slant height, not vertical height.

So first, find vertical height using Pythagorean Theorem.

In a square pyramid:
- Half the base = 12 ÷ 2 = 6 in
- Slant height = 10 in
- So:
$$
h^2 + 6^2 = 10^2 \\
h^2 + 36 = 100 \\
h^2 = 64 \\
h = 8 \text{ in}
$$

Now compute volume:
- Base area = $ 12 \times 12 = 144 \text{ in}^2 $
- Volume:
$$
V = \frac{1}{3} \times 144 \times 8 = 48 \times 8 = 384 \text{ in}^3
$$

✔ Answer: B. 384 in³

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Question 6: Find the surface area of the triangular pyramid.

Given:
- Base edge = 6√3 in
- Height of pyramid (vertical) = 4 in
- But we need surface area = lateral area + base area

Assume it's a regular triangular pyramid (all faces are equilateral triangles). But let’s check.

Alternatively, perhaps the height of the pyramid is 4 in, and we need to find slant height to compute lateral area.

But wait — the diagram shows a right triangle with:
- Vertical height = 4 in
- Base = 3√3 in (half of 6√3?)
- Then slant height?

Wait — look at the labels:
- Height of pyramid = 4 in
- Base edge = 6√3 in
- The dotted line from apex to base center is 4 in
- From center to midpoint of side is 3√3 in? Wait, let’s compute.

For equilateral triangle of side $ s = 6\sqrt{3} $, the inradius (distance from centroid to side) is:
$$
r = \frac{s}{2\sqrt{3}} = \frac{6\sqrt{3}}{2\sqrt{3}} = 3 \text{ in}
$$

So, in the right triangle:
- Vertical leg = 4 in
- Horizontal leg = 3 in
- So slant height = ?
$$
l = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ in}
$$

Now, each lateral face is a triangle with:
- Base = $ 6\sqrt{3} $
- Height = slant height = 5 in

Area of one lateral face:
$$
A = \frac{1}{2} \times 6\sqrt{3} \times 5 = 15\sqrt{3}
$$

Three lateral faces:
$$
3 \times 15\sqrt{3} = 45\sqrt{3}
$$

Now, base area (equilateral triangle):
$$
A = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 36 \times 3 = \frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3}
$$

Total surface area:
$$
45\sqrt{3} + 27\sqrt{3} = 72\sqrt{3} \text{ in}^2
$$

✔ Answer: C. 72√3 in²

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✔ Final Answers:

| Question | Answer |
|--------|--------|
| 1 | B |
| 2 | D |
| 3 | B |
| 4 | D |
| 5 | B |
| 6 | C |

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Let me know if you'd like a visual explanation or any clarification!