Right Square Pyramid 648D3C

1. **Problem Statement:** We have a right square pyramid with a surface area of $50.75$ square inches and a base length of $3.5$ inches. 2. **Understanding the Pyramid:** A right square pyramid has a square base and four triangular faces meeting at a single apex above the center of the base. 3. **Surface Area Formula:** The surface area $S$ of a right square pyramid is given by: $$S = B + L$$ where $B$ is the area of the square base and $L$ is the lateral surface area. 4. **Calculate Base Area $B$:** Since the base is a square with side length $3.5$ inches, $$B = (3.5)^2 = 12.25 \text{ square inches}$$ 5. **Lateral Surface Area $L$:** The lateral surface area is the sum of the areas of the four triangular faces. Each triangle has base $3.5$ inches and slant height $l$ (unknown). The area of one triangle is: $$\frac{1}{2} \times 3.5 \times l = 1.75l$$ So, $$L = 4 \times 1.75l = 7l$$ 6. **Use Surface Area to Find Slant Height $l$:** Given total surface area $S = 50.75$, $$50.75 = 12.25 + 7l$$ Subtract base area: $$50.75 - 12.25 = 7l$$ $$38.5 = 7l$$ Divide both sides by 7: $$\cancel{7}l = \frac{38.5}{\cancel{7}}$$ $$l = 5.5 \text{ inches}$$ 7. **Diagram Sketch:** - Draw a square base with side length labeled $3.5$ inches. - Draw four triangular faces meeting at the apex. - Label the slant height $l = 5.5$ inches on one triangular face. This completes the sketch and labeling of the right square pyramid with the given dimensions.