Asked by no name

Find surface area of a square pyramid with “a” of .5cm and “l” of .8cm. Round to tenth and proper units.

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Answered by no name
is that correct
Answered by GPT 3.5
The surface area of a square pyramid can be calculated using the formula:

Surface area = Base area + (0.5 × Perimeter of base × Slant height)

To calculate the surface area, we need to find the base area and the slant height.

The base area of a square pyramid is given by the formula:

Base area = a^2

where a is the length of one side of the base.

Given that a = 0.5 cm, we can calculate the base area:

Base area = (0.5 cm)^2 = 0.25 cm^2

Now, let's calculate the slant height. The slant height can be found using the Pythagorean theorem:

l^2 = (0.5/2)^2 + h^2

where l is the slant height and h is the height of the pyramid.

Given that l = 0.8 cm, we can solve for h:

0.8^2 = (0.5/2)^2 + h^2
0.64 = 0.25/4 + h^2
0.64 = 0.0625 + h^2
h^2 = 0.5775
h ≈ √0.5775
h ≈ 0.759 cm

Now we can calculate the surface area using the formula:

Surface area = Base area + (0.5 × Perimeter of base × Slant height)

The perimeter of the base can be calculated as:

Perimeter of base = 4 × a

Perimeter of base = 4 × 0.5 cm = 2 cm

Surface area = 0.25 cm^2 + (0.5 × 2 cm × 0.759 cm)

Surface area = 0.25 cm^2 + 0.759 cm^2

Surface area ≈ 1.009 cm^2

Rounded to the nearest tenth, the surface area of the square pyramid is approximately 1.0 cm^2.
Answered by GPT 3.5
I apologize for the incorrect calculation. Let's recalculate the surface area correctly.

Base area = a^2 = (0.5 cm)^2 = 0.25 cm^2

Perimeter of base = 4a = 4(0.5 cm) = 2 cm

To find the slant height, we can use the Pythagorean theorem:

l^2 = (0.5/2)^2 + h^2
= (0.25 cm)^2 + (0.8 cm)^2
= 0.0625 cm^2 + 0.64 cm^2
= 0.7025 cm^2

l ≈ √0.7025 cm ≈ 0.837 cm

Now, we can calculate the surface area:

Surface area = Base area + (0.5 × Perimeter of base × Slant height)
= 0.25 cm^2 + (0.5 × 2 cm × 0.837 cm)
= 0.25 cm^2 + 0.837 cm^2
= 1.087 cm^2

Rounded to the nearest tenth, the surface area of the square pyramid is approximately 1.1 cm^2.

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