A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #5 # and the pyramid's height is #6 #. If one of the base's corners has an angle of #pi/3 #, what is the pyramid's surface area?

Answer 1

T S A = 74.9435.

#CH = 4 * sin (pi/3) = 3.4641#
Area of parallelogram base #= a * b1 = 5*3.4641 = color(red)(17.3205)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (5/2)^2)= 6.5#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.5= #color(red)(13)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(4/2)^2 )= 6.3246#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*6.3246 = color(red)( 15.8115)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 13)+ (2* 15.8115) = color(red)(57.623)#

Total surface area =Area of parallelogram base + Lateral surface area # = 17.3205 + 57.623 = 74.9435#

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Answer 2

The surface area of the pyramid can be calculated by finding the area of the base and adding the area of the four triangular faces.

The area of the base, which is a parallelogram, can be calculated using the formula for the area of a parallelogram: base times height.

The height of the parallelogram is given by the height of the pyramid, which is 6 units.

The area of the base is (4 \times 6 = 24) square units.

Next, to find the area of each triangular face, we first need to find the length of the slant height. This can be done using the Pythagorean theorem.

The slant height, (l), can be found using (l = \sqrt{h^2 + (\frac{1}{2}b)^2}), where (h) is the height of the pyramid and (b) is the length of one side of the base.

Substituting the given values, we get (l = \sqrt{6^2 + (\frac{1}{2} \times 4)^2} = \sqrt{36 + 4} = \sqrt{40}).

The area of each triangular face is given by ( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times \sqrt{40}) for the two triangular faces with base 4 and ( \frac{1}{2} \times 5 \times \sqrt{40}) for the two triangular faces with base 5.

Thus, the total surface area of the pyramid is (24 + 2 \times \frac{1}{2} \times 4 \times \sqrt{40} + 2 \times \frac{1}{2} \times 5 \times \sqrt{40} = 24 + 4\sqrt{40} + 5\sqrt{40} = 24 + 9\sqrt{40}) square units.

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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