Jennifer got a box of chocolates. The box is a right triangular prism shaped box. It is 7 inches long, and the triangular base measures 2 inches x 3 inches x 4 inches. What is the surface area of the box of chocolates?
No
68.8 inches squared
Asked someone else Telsa
Be quiet
hi
To find the surface area of the box of chocolates, we need to calculate the areas of all the individual faces and then add them together.
The box is a right triangular prism, which means it has two triangular faces and three rectangular faces.
First, let's calculate the areas of the triangular faces. The formula for the area of a triangle is (base x height) / 2.
The first triangular face has a base of 2 inches and a height of 3 inches, so its area is (2 x 3) / 2 = 3 square inches.
The second triangular face has a base of 3 inches and a height of 4 inches, so its area is (3 x 4) / 2 = 6 square inches.
Next, let's calculate the areas of the rectangular faces. Each rectangular face has a length equal to the length of the prism, which is 7 inches. The width of the rectangular faces is equal to the corresponding side of the triangular base.
The first rectangular face has a length of 7 inches and a width of 2 inches, so its area is 7 x 2 = 14 square inches.
The second rectangular face has a length of 7 inches and a width of 3 inches, so its area is 7 x 3 = 21 square inches.
The third rectangular face has a length of 7 inches and a width of 4 inches, so its area is 7 x 4 = 28 square inches.
Now, we can add up all the areas to find the total surface area:
3 square inches (triangular face 1) + 6 square inches (triangular face 2) + 14 square inches (rectangular face 1) + 21 square inches (rectangular face 2) + 28 square inches (rectangular face 3) = 72 square inches.
Therefore, the surface area of the box of chocolates is 72 square inches.
Steve I am confused sorry what is the actual answer?
guess you better ask "Kaydon"
there are two ends and three lateral faces, so
2A + (2+3+4)*7
to find the area of the triangular bases, draw an altitude to the base of length 4. That line divides the base into two parts of length x and 4-x. Then, if the height is h,
h^2+x^2 = 4
h^2+(4-x)^2 = 9
x = 21/8
now find h, and
the area is (1/2)(4)(h)