A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #1 # and the pyramid's height is #4 #. If one of the base's corners has an angle of #pi/3#, what is the pyramid's surface area?
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To find the surface area of the pyramid, we need to calculate the areas of the base and the four triangular faces.

Area of the base (parallelogram): Area = base * height Area = 8 * 1 Area = 8 square units

Now, let's find the lengths of the sides of the triangular faces. Since the base is a parallelogram, opposite sides are equal. So, we have two types of triangles:
 Triangles with base length 8 and height 4.
 Triangles with base length 1 and height 4.

For the triangles with base length 8:
 Area = (1/2) * base * height
 Area = (1/2) * 8 * 4
 Area = 16 square units

For the triangles with base length 1:
 Area = (1/2) * base * height
 Area = (1/2) * 1 * 4
 Area = 2 square units

Since there are four triangular faces, the total surface area of the pyramid is: Total Surface Area = Area of base + 4 * Area of triangular faces = 8 + 4(16 + 2) = 8 + 4(18) = 8 + 72 = 80 square units
Therefore, the surface area of the pyramid is 80 square units.
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To find the surface area of the pyramid, we first need to calculate the area of the base and the lateral (slant) surface area.
The area of a parallelogram base with sides of lengths (a) and (b) is given by (A_{\text{base}} = a \times b).
In this case, (a = 8) and (b = 1), so (A_{\text{base}} = 8 \times 1 = 8).
The lateral surface area of a pyramid is given by (A_{\text{lateral}} = \frac{1}{2} \times \text{perimeter of base} \times \text{slant height}).
The slant height of the pyramid can be found using the height and the side length of the base. In a right triangle formed by the height, the slant height, and half the base's diagonal, the slant height is the hypotenuse, and the height and half the base's diagonal are the other two sides. Thus, the slant height can be found using the Pythagorean theorem:
[\text{Slant height} = \sqrt{\text{height}^2 + \left(\frac{1}{2} \times \sqrt{a^2 + b^2}\right)^2}]
Substitute the given values to find the slant height:
[\text{Slant height} = \sqrt{4^2 + \left(\frac{1}{2} \times \sqrt{8^2 + 1^2}\right)^2} = \sqrt{16 + \left(\frac{1}{2} \times \sqrt{65}\right)^2} = \sqrt{16 + \left(\frac{1}{2} \times 8.06\right)^2} = \sqrt{16 + 16.06^2} \approx \sqrt{16 + 258.24} \approx \sqrt{274.24} \approx 16.56]
The perimeter of the base is (2a + 2b), which in this case is (2(8) + 2(1) = 16 + 2 = 18).
Now, we can calculate the lateral surface area:
[A_{\text{lateral}} = \frac{1}{2} \times 18 \times 16.56 \approx 9 \times 16.56 \approx 149.04]
The total surface area of the pyramid is the sum of the base area and the lateral surface area:
[A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}} = 8 + 149.04 = 157.04]
So, the surface area of the pyramid is approximately (157.04) 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.
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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