A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #2 # and the pyramid's height is #8 #. If one of the base's corners has an angle of #(3pi)/8#, what is the pyramid's surface area?
T S A = 86.835
Area of parallelogram base
Area of
Area of
Lateral surface area =
Total surface area =Area of parallelogram base + Lateral surface area
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To find the surface area of the pyramid, we need to calculate the area of the base and the area of the four triangular faces.

Area of the base (parallelogram): Area = base × height Area = 7 × 2 = 14 square units

To find the area of each triangular face: We first need to find the length of the slant height (l) using the Pythagorean theorem: l^2 = h^2 + (1/2 * base)^2 l^2 = 8^2 + (1/2 * 7)^2 l^2 = 64 + 12.25 l^2 = 76.25 l ≈ √76.25 ≈ 8.72 units
Now, we can find the area of one triangular face using the formula: Area of one triangular face = (1/2) × base × height Area of one triangular face = (1/2) × 7 × 8 ≈ 28 square units
Since there are four triangular faces, the total area of the triangular faces is: Total area of triangular faces = 4 × 28 = 112 square units

Total surface area of the pyramid: Total surface area = area of base + area of triangular faces Total surface area = 14 + 112 = 126 square units
Therefore, the surface area of the pyramid is 126 square units.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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