Question

1. label the all the lengths of the net and calculate the surface area. 2. determine the surface area of the polyhedron below. 3. determine the surface area of the polyhedron below. 4. determine the surface area of the triangular prism below. 5. determine the surface area of the pyramid below.

Explanation:

1. For the rectangular - prism in question 2:

• The net of a rectangular prism has 6 faces. Let the length \(l = 12\) cm, width \(w = 6\) cm, and height \(h = 3\) cm.
• The surface - area formula of a rectangular prism is \(SA=2(lw + lh+wh)\).
• Step 1: Calculate \(lw\):
• \(lw=12\times6 = 72\) \(cm^{2}\).
• Step 2: Calculate \(lh\):
• \(lh = 12\times3=36\) \(cm^{2}\).
• Step 3: Calculate \(wh\):
• \(wh=6\times3 = 18\) \(cm^{2}\).
• Step 4: Calculate the surface area:
• \(SA = 2(72 + 36+18)=2\times126 = 252\) \(cm^{2}\).

1. For the triangular prism in question 3:

• The triangular prism has 2 triangular faces and 3 rectangular faces.
• For the triangular faces: The base of the triangle \(b = 5\) cm and height \(h_{1}=4\) cm. The area of a triangle is \(A_{\triangle}=\frac{1}{2}bh\). So, \(A_{\triangle}=\frac{1}{2}\times5\times4 = 10\) \(cm^{2}\), and the total area of the 2 triangular faces is \(2\times10 = 20\) \(cm^{2}\).
• For the rectangular faces:
• The first rectangular face has dimensions \(10\) cm (length) and \(5\) cm (width), so its area \(A_{1}=10\times5 = 50\) \(cm^{2}\).
• The second rectangular face has dimensions \(10\) cm and \(5\) cm, so its area \(A_{2}=10\times5 = 50\) \(cm^{2}\).
• The third rectangular face has dimensions \(10\) cm and \(\sqrt{4^{2}+(\frac{5}{2})^{2}}=\sqrt{16 + 6.25}=\sqrt{22.25}\) cm. But we can also calculate the surface - area using the perimeter of the base times the height of the prism. The perimeter of the triangular base \(P=5 + 5+ \sqrt{4^{2}+(\frac{5}{2})^{2}}+\sqrt{4^{2}+(\frac{5}{2})^{2}}\). A simpler way is to note that the three rectangular - face areas are \(A_{1}=10\times5\), \(A_{2}=10\times5\), \(A_{3}=10\times5\) (since the prism is symmetric). The total area of the rectangular faces is \(10\times(5 + 5+5)=150\) \(cm^{2}\).
• The surface area of the triangular prism is \(SA=20 + 150=170\) \(cm^{2}\).

1. For the triangular - based pyramid in question 4:

• The base is a right - triangle with legs \(a = 9\) cm and \(b = 12\) cm, so the area of the base \(A_{base}=\frac{1}{2}\times9\times12 = 54\) \(cm^{2}\).
• The slant heights: For the face with base \(9\) cm, using the Pythagorean theorem, the slant height \(l_{1}=\sqrt{12^{2}+(\frac{9}{2})^{2}}=\sqrt{144+\frac{81}{4}}=\sqrt{\frac{576 + 81}{4}}=\sqrt{\frac{657}{4}}\). For the face with base \(12\) cm, the slant height \(l_{2}=\sqrt{9^{2}+(\frac{12}{2})^{2}}=\sqrt{81 + 36}=\sqrt{117}\).
• Another way: The three triangular faces:
• The first triangular face with base \(9\) cm and height \(15\) cm has area \(A_{1}=\frac{1}{2}\times9\times15=\frac{135}{2}=67.5\) \(cm^{2}\).
• The second triangular face with base \(12\) cm and height \(15\) cm has area \(A_{2}=\frac{1}{2}\times12\times15 = 90\) \(cm^{2}\).
• The third triangular face (the base) has area \(A_{3}=\frac{1}{2}\times9\times12 = 54\) \(cm^{2}\).
• The surface area \(SA=67.5+90 + 54=211.5\) \(cm^{2}\).

1. For the square - based pyramid in question 5:

• The base is a square with side length \(s = 6\) cm, so the area of the base \(A_{base}=6\times6 = 36\) \(cm^{2}\).
• Each triangular face has base \(b = 6\) cm and height \(h = 10\) cm. The area of each triangular face is \(A_{\triangle}=\frac{1}{2}\times6\times10 = 30\) \(cm^{2}\).
• Since there are 4 triangular faces, the total area of the triangular faces is \(4\times30 = 120\) \(cm^{2}\).
• The surface area of the square - based pyramid is \(SA=36+120 = 156\) \(cm^{2}\).

Answer:

1. For the rectangular prism in question 2: \(252\) \(cm^{2}\)
2. For the triangular prism in question 3: \(170\) \(cm^{2}\)
3. For the triangular - based pyramid in question 4: \(211.5\) \(cm^{2}\)
4. For the square - based pyramid in question 5: \(156\) \(cm^{2}\)