Question

here is a square pyramid and its net. the lateral faces are congruent triangles. the base (shaded) is a square. (all lengths are in millimeters.) (a) find the area of the base of the pyramid. 25 mm² (b) find the area of one lateral face of the pyramid. 14.75 mm² (c) use the net to find the lateral surface area of the pyramid. the base is not included. \boxed{} mm² (d) use the net to find the total surface area of the pyramid. \boxed{} mm²

Explanation:

Step1: Solve part (a) - Base Area

The base is a square with side length \( s = 5 \) mm (from the net, the square's side is 5). The area of a square is \( A = s^2 \).
\( A = 5^2 = 25 \) \( \text{mm}^2 \).

Step2: Solve part (b) - Lateral Face Area

A lateral face is a triangle. The base of the triangle is the side of the square (\( b = 5 \) mm), and the height of the triangle is \( h = 5.9 \) mm (from the net). The area of a triangle is \( A = \frac{1}{2} \times b \times h \).
\( A = \frac{1}{2} \times 5 \times 5.9 = 14.75 \) \( \text{mm}^2 \).

Step3: Solve part (c) - Lateral Surface Area

There are 4 congruent lateral faces. So lateral surface area is \( 4 \times \) area of one lateral face.
From part (b), one lateral face area is \( 14.75 \) \( \text{mm}^2 \). So \( 4 \times 14.75 = 59 \) \( \text{mm}^2 \).

Step4: Solve part (d) - Total Surface Area

Total surface area is base area + lateral surface area. From part (a), base area is \( 25 \) \( \text{mm}^2 \), from part (c), lateral surface area is \( 59 \) \( \text{mm}^2 \). So \( 25 + 59 = 84 \) \( \text{mm}^2 \).

Answer:

(a) \( 25 \) \( \text{mm}^2 \)
(b) \( 14.75 \) \( \text{mm}^2 \)
(c) \( 59 \) \( \text{mm}^2 \)
(d) \( 84 \) \( \text{mm}^2 \)