https://file.aiquickdraw.com/mathbot/file/1726797289075dwnqfrs1.jpg

Let's analyze and solve each part of the question based on the image.

### Given Information:
- **Pyramid X** has a square base of \(10 \, \text{cm} \times 10 \, \text{cm}\).
- Each triangular face has a height of \(13 \, \text{cm}\).
- The height of the pyramid (from base to vertex) is \(h \, \text{cm}\).

#### Part (a): Show that \(h = 12 \, \text{cm}\).
We are given that the triangular faces have a height of 13 cm. Let's calculate the height \(h\) of the pyramid.

1. Consider a right triangle formed by the height \(h\) of the pyramid, half the base of the pyramid \(\frac{10}{2} = 5 \, \text{cm}\), and the slant height (height of the triangular face) \(13 \, \text{cm}\).
   
2. Using the Pythagorean theorem:
\[
h^2 + 5^2 = 13^2
\]
\[
h^2 + 25 = 169
\]
\[
h^2 = 169 - 25 = 144
\]
\[
h = \sqrt{144} = 12 \, \text{cm}
\]
Thus, \(h = 12 \, \text{cm}\) is correct.

---

#### Part (b): Find the height of pyramid Y.
Pyramid Y is geometrically similar to pyramid X, and its volume is twice the volume of X. To find the height of Y, we first need to relate their volumes and heights.

1. **Volume of a pyramid** is given by:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
For pyramid X, the base area \(A\) is:
\[
A = 10 \times 10 = 100 \, \text{cm}^2
\]
Thus, the volume of X is:
\[
V_X = \frac{1}{3} \times 100 \times 12 = 400 \, \text{cm}^3
\]
Since pyramid Y has twice the volume of X:
\[
V_Y = 2 \times V_X = 2 \times 400 = 800 \, \text{cm}^3
\]

2. Since pyramid Y is geometrically similar to pyramid X, the volumes of similar shapes are proportional to the cube of the ratio of their corresponding lengths. Let the height of pyramid Y be \(h_Y\). Then:
\[
\frac{V_Y}{V_X} = \left(\frac{h_Y}{h_X}\right)^3
\]
Substituting the known values:
\[
\frac{800}{400} = \left(\frac{h_Y}{12}\right)^3
\]
\[
2 = \left(\frac{h_Y}{12}\right)^3
\]
Taking the cube root of both sides:
\[
\frac{h_Y}{12} = \sqrt[3]{2}
\]
\[
h_Y = 12 \times \sqrt[3]{2} \approx 12 \times 1.26 = 15.1 \, \text{cm}
\]
Thus, the height of pyramid Y is approximately \(15.1 \, \text{cm}\).

---

#### Part (c): Find the radius of the sphere.
The volume of a sphere is equal to the volume of pyramid X. The volume of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
We already know the volume of pyramid X is \(400 \, \text{cm}^3\), so:
\[
\frac{4}{3} \pi r^3 = 400
\]
Solving for \(r^3\):
\[
r^3 = \frac{400 \times 3}{4 \pi} = \frac{1200}{4 \pi} = \frac{300}{\pi}
\]
\[
r^3 \approx \frac{300}{3.1416} \approx 95.49
\]
Taking the cube root of both sides:
\[
r \approx \sqrt[3]{95.49} \approx 4.6 \, \text{cm}
\]
Thus, the radius of the sphere is approximately \(4.6 \, \text{cm}\).

---

### Summary of Answers:
- (a) \(h = 12 \, \text{cm}\)
- (b) The height of pyramid Y is approximately \(15.1 \, \text{cm}\).
- (c) The radius of the sphere is approximately \(4.6 \, \text{cm}\).

Let me know if you need more details on any step!

### Relative Questions:
1. How does the height of a pyramid relate to the volume and the dimensions of its base?
2. Why do volumes of similar pyramids relate to the cube of the scale factor?
3. What is the significance of the Pythagorean theorem in 3D geometry?
4. How is the volume of a sphere derived from its radius?
5. What are other real-life applications of geometric similarity in pyramids and spheres?

**Tip:** Always verify your answers with units to avoid confusion, especially in volume or area-based calculations.

An upright pyramid X has a square base measuring 10 cm by 10 cm. Its triangular faces are congruent and each has a height of 13 cm. The height of the pyramid is h cm. (a) Show that h = 12 cm. (b) A geometrically similar pyramid Y has twice the volume of X. Find the height of Y. (c) A sphere has the same volume as pyramid X. Find the radius of the sphere.

Solve a geometry problem involving the height of a pyramid, the volume of geometrically similar pyramids, and calculating the radius of a sphere. Includes application of the Pythagorean theorem and formulas for pyramid and sphere volumes. Suitable for Grades 9-11.

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