A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #6 # and #7 # and the pyramid's height is #2 #. If one of the base's corners has an angle of #(5pi)/6#, what is the pyramid's surface area?
T S A = 70.4258
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, you need to calculate the areas of its different components and then sum them up. The surface area of a pyramid includes the area of the base and the lateral surface area.

Calculate the area of the base, which is a parallelogram. You can use the formula for the area of a parallelogram: ( \text{Area}_{\text{base}} = \text{base length} \times \text{height} ). In this case, since the base is a parallelogram, you'll need to use one of the diagonals as the height. The length of the diagonal can be found using the Pythagorean theorem.

Calculate the lateral surface area of the pyramid. Since the lateral faces of the pyramid are triangles, you can find their area using the formula for the area of a triangle: ( \text{Area}_{\text{lateral}} = \frac{1}{2} \times \text{base length} \times \text{slant height} ). The slant height can be found using trigonometric relationships.

Sum up the areas of the base and the lateral faces to get the total surface area of the pyramid.
Use these steps to find the surface area of the given pyramid.
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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): Given that the base is a parallelogram with side lengths of 6 and 7, we can find its area using the formula for the area of a parallelogram: ( \text{Area}{\text{base}} = \text{base} \times \text{height} ). Therefore, ( \text{Area}{\text{base}} = 6 \times 2 = 12 ).

Area of Each Triangular Face: The height of the pyramid is given as 2. We can find the length of the slant height of each triangular face using the given dimensions and the height of the pyramid. For this, we can use the Pythagorean theorem because the slant height, height, and side length form a right triangle. The slant height, ( l ), is given by ( l = \sqrt{h^2 + (\frac{1}{2} \text{base})^2} ). Plugging in the values, we get ( l = \sqrt{2^2 + (\frac{1}{2} \times 6)^2} = \sqrt{4 + 9} = \sqrt{13} ). Now, the area of each triangular face is ( \frac{1}{2} \times \text{base} \times \text{height} ), where the base is the side length of the parallelogram base, and the height is the slant height. So, ( \text{Area}_{\text{triangle}} = \frac{1}{2} \times 6 \times \sqrt{13} = 3\sqrt{13} ).

Total Surface Area: There are four triangular faces, each with an area of ( 3\sqrt{13} ), and one base with an area of 12. Therefore, the total surface area is the sum of the areas of the base and the four triangular faces. ( \text{Total Surface Area} = 4 \times (3\sqrt{13}) + 12 = 12\sqrt{13} + 12 ).
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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.
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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