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 in x 3 in x 4 in. What is the surface area of the box of chocolates?
what is the answer
To find the surface area of the box, we need to calculate the areas of each of its faces.
The box is in the shape of a right triangular prism, so it has 3 rectangular faces and 2 triangular faces.
Let's find the areas of these faces step-by-step:
1. Area of the triangular base:
The base of the box is a triangle with sides measuring 2 inches, 3 inches, and 4 inches.
To find the area of a triangle, we can use Heron's formula:
Semi-perimeter (s) = (2 + 3 + 4)/2 = 9/2 = 4.5 inches
Area (A) = √(s(s-a)(s-b)(s-c)) where a, b, and c are the lengths of the sides.
A = √(4.5(4.5-2)(4.5-3)(4.5-4))
A = √(4.5 * 2.5 * 1.5 * 0.5)
A = √(5.625)
A ≈ 2.37 square inches
2. Area of the rectangular faces:
Each rectangular face has dimensions of 7 inches (length) by one of the side lengths of the triangular base. So, we need to calculate the areas of the three rectangular faces separately.
Face 1: Area = 7 * 2 = 14 square inches
Face 2: Area = 7 * 3 = 21 square inches
Face 3: Area = 7 * 4 = 28 square inches
3. Area of the triangular faces:
Each triangular face has the dimensions of the two remaining side lengths of the triangular base and the height of the triangular prism. In this case, the height is 7 inches.
Face 4: Area = (1/2) * 2 * 7 = 7 square inches
Face 5: Area = (1/2) * 3 * 7 = 10.5 square inches
Now, let's find the total surface area by summing up the areas of all the faces:
Total Surface Area = Area of the triangular base + Area of the rectangular faces + Area of the triangular faces
Total Surface Area = 2.37 + 14 + 21 + 28 + 7 + 10.5
Total Surface Area = <<85.87=85.87>>85.87 square inches
Therefore, the surface area of the box of chocolates is approximately 85.87 square inches.
To find the surface area of a right triangular prism, we need to calculate the area of each face and then add them up.
The triangular base of the box has sides measuring 2 inches, 3 inches, and 4 inches. To find the area of the triangular base, we can use the formula for the area of a triangle, which is 1/2 times base times height.
Area of the triangular base = 1/2 * base * height = 1/2 * 2 in * 3 in = 3 in^2.
The triangular base has two faces, so the total area of the triangular base is 2 * 3 in^2 = 6 in^2.
The two rectangular faces of the box have dimensions that match the length and width of the triangular base. They have dimensions of 2 in x 3 in and 2 in x 4 in.
Area of the first rectangular face = length * width = 2 in * 3 in = 6 in^2.
Area of the second rectangular face = length * width = 2 in * 4 in = 8 in^2.
Finally, we have the top and bottom faces of the box, which are both right triangles with dimensions of 2 in x 4 in.
Area of the top face = 1/2 * base * height = 1/2 * 2 in * 4 in = 4 in^2.
Area of the bottom face = 1/2 * base * height = 1/2 * 2 in * 4 in = 4 in^2.
Now, we can add up the areas of all the faces to find the total surface area of the box.
Total surface area = 6 in^2 (triangular base) + 6 in^2 (rectangular faces) + 4 in^2 (top face) + 4 in^2 (bottom face)
Total surface area = 20 in^2.
Therefore, the surface area of the box of chocolates is 20 square inches.
The easy part is to find the areas of the 3 rectangles, they are:
2x7, 3x7, and 4x7.
The hard part is to find the area of the triangular base.
Since it is not right-angled, 2^2 + 3^2 ≠ 4^2, we need more complex methods.
method one: Heron's formula
area = √(s(s-a)(s-b)(s-c)), where s is 1/2 the perimeter, and a, b, and c are the three sides.
area = √(4.5 * 2.5 * 1.5 * .5) = appr 2.905
method two: Use the cosine law to find one of the angles Ø, then area = (1/2)(a)(b)sinØ, where Ø is the contained angle between sides a and b.
I will pick the angle opposite side 2
2^2 = 3^2 + 4^2 - 2(3)(4)cosØ
cosØ = 21/24 = ....
Ø = 28.955°
area = (1/2)(4)(3)sin28.955 = appr 2.905 , same as before
method 3
make a sketch with 4 as the base, draw an altitude to meet the base, call it h. Let the base be cut into parts x, and 4-x
You now have 2 right-angled triangles:
x^2 + h^2 = 3^2 and (4-x)^2 + h^2 = 2^2
solve for h and use (1/2)(4)(h), you must get 2.905
Add up your three rectangles plus your two triangles areas.