Solved Sheet1 Surface Area of Triangular Pyramids Find the | Chegg.com - Free Printable

Let's solve each of the three regular triangular pyramids (also known as regular tetrahedrons if all faces are equilateral triangles, but here they may be regular triangular pyramids, meaning the base is an equilateral triangle and the lateral faces are congruent isosceles triangles).

We'll use the formula for the surface area of a triangular pyramid:

> Surface Area = Area of Base + Lateral Surface Area

Since it’s a regular triangular pyramid:
- The base is an equilateral triangle
- The three lateral faces are congruent isosceles triangles

So,

> Surface Area = Area of Base + 3 × Area of One Lateral Face

---

We’ll go through each problem step by step.

---

🔷 Problem 1

Given:
- Base side = 12 ft
- Height from apex to base = 10 ft (this is the slant height of the lateral face)
- But we also see a dashed line labeled 10 ft from the apex to the midpoint of a base edge — this is the slant height of the triangular face.
- Also, there’s a height of the base triangle shown as 10.4 ft? Wait — let's check carefully.

Wait — actually, in the diagram:
- The base is an equilateral triangle with side 12 ft.
- From the apex to the center of the base (dashed line) is not the slant height.
- But the dashed line from apex to midpoint of base edge is 10 ft, and that is the slant height of the lateral face.
- And the height of the base triangle is not given directly, but we can calculate it.

But wait — the height of the equilateral triangle base is:
$$
h_{\text{base}} = \frac{\sqrt{3}}{2} \times \text{side} = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3} \approx 10.392 \text{ ft}
$$
But in the diagram, there's a dashed line from the apex down to the base, perpendicular to the base — that's the height of the pyramid, which is 10 ft?

Wait — no! Let’s clarify the diagram:

Looking at the first figure:
- The vertical dashed line from apex to base is labeled 10 ft — this is the height of the pyramid (from apex to centroid of base).
- But the slant height (from apex to midpoint of a base edge) is not given directly.

Wait — actually, there's a right triangle drawn inside one lateral face:
- It shows a triangle with base 6 ft (half of 12 ft), and hypotenuse 10 ft? No — wait.

Actually, look again: the dashed line from apex to the midpoint of a base edge is labeled 10 ft, and the base of that triangle is 6 ft (half of 12 ft). So this is the slant height of the lateral face.

So:
- Slant height $ l = 10 $ ft
- Base side $ s = 12 $ ft

Now, we need:
1. Area of base (equilateral triangle):
$$
A_{\text{base}} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} \approx 62.35 \text{ ft}^2
$$

2. Area of one lateral face (isosceles triangle with base 12 ft and height 10 ft):
$$
A_{\text{lateral}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 10 = 60 \text{ ft}^2
$$

3. Total surface area:
$$
SA = A_{\text{base}} + 3 \times A_{\text{lateral}} = 36\sqrt{3} + 3 \times 60 = 36\sqrt{3} + 180
$$

Using $ \sqrt{3} \approx 1.732 $:
$$
36 \times 1.732 = 62.352
$$
$$
SA \approx 62.352 + 180 = 242.352 \text{ ft}^2
$$

But let’s keep it exact unless asked otherwise.

✔ Answer for #1: $ \boxed{180 + 36\sqrt{3}} $ ft² ≈ 242.35 ft²

---

🔷 Problem 2

Given:
- Base side = 14 m
- Height of pyramid (from apex to base) = 13 m
- But we need slant height to find lateral face area.

Wait — in the diagram, there's a right triangle formed:
- Vertical leg = 13 m (height of pyramid)
- Horizontal leg = distance from centroid to midpoint of base edge.

For an equilateral triangle with side $ s = 14 $ m:
- The height of the base triangle is:
$$
h = \frac{\sqrt{3}}{2} \times 14 = 7\sqrt{3} \approx 12.124 \text{ m}
$$
- The centroid divides this height in ratio 2:1, so distance from centroid to base vertex is $ \frac{2}{3} \times 7\sqrt{3} = \frac{14\sqrt{3}}{3} $
- But we need distance from centroid to midpoint of side → that is $ \frac{1}{3} \times 7\sqrt{3} = \frac{7\sqrt{3}}{3} $

But wait — in the diagram, there's a dashed line from apex to midpoint of base edge, and it's labeled 12 m.

Ah! Yes — the slant height is 12 m.

So:
- Slant height $ l = 12 $ m
- Base side $ s = 14 $ m

Now compute:

1. Base area:
$$
A_{\text{base}} = \frac{\sqrt{3}}{4} \times 14^2 = \frac{\sqrt{3}}{4} \times 196 = 49\sqrt{3} \approx 84.87 \text{ m}^2
$$

2. Lateral face area (triangle with base 14 m, height 12 m):
$$
A_{\text{lateral}} = \frac{1}{2} \times 14 \times 12 = 84 \text{ m}^2
$$

3. Total surface area:
$$
SA = 49\sqrt{3} + 3 \times 84 = 49\sqrt{3} + 252
$$

Approximate:
$$
49 \times 1.732 = 84.868
$$
$$
SA \approx 84.868 + 252 = 336.868 \text{ m}^2
$$

✔ Answer for #2: $ \boxed{252 + 49\sqrt{3}} $ m² ≈ 336.87 m²

---

🔷 Problem 3

Given:
- Base side = 8 yd
- Slant height = 7 yd (from apex to midpoint of base edge)
- There's a dashed line from apex to base, labeled 12 yd — but that’s the height of the pyramid, not slant height.

Wait — actually, the dashed line from apex to midpoint of base edge is labeled 7 yd — that’s the slant height.

Also, the base is equilateral triangle with side 8 yd.

So:
- Base side $ s = 8 $ yd
- Slant height $ l = 7 $ yd

Compute:

1. Base area:
$$
A_{\text{base}} = \frac{\sqrt{3}}{4} \times 8^2 = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3} \approx 27.71 \text{ yd}^2
$$

2. Lateral face area:
$$
A_{\text{lateral}} = \frac{1}{2} \times 8 \times 7 = 28 \text{ yd}^2
$$

3. Total surface area:
$$
SA = 16\sqrt{3} + 3 \times 28 = 16\sqrt{3} + 84
$$

Approximate:
$$
16 \times 1.732 = 27.712
$$
$$
SA \approx 27.712 + 84 = 111.712 \text{ yd}^2
$$

✔ Answer for #3: $ \boxed{84 + 16\sqrt{3}} $ yd² ≈ 111.71 yd²

---

✔ Final Answers:

1) Surface Area = $ \boxed{180 + 36\sqrt{3}} $ ft² ≈ 242.35 ft²
2) Surface Area = $ \boxed{252 + 49\sqrt{3}} $ m² ≈ 336.87 m²
3) Surface Area = $ \boxed{84 + 16\sqrt{3}} $ yd² ≈ 111.71 yd²

Let me know if you'd like these rounded to nearest whole number or decimal place!