Question

6 6.2.14 practice 6 find the surface area of the polyhedron that can be assembled from this net. type your answer in the box. square inches

Explanation:

Step1: Identify the shapes in the net

The net consists of two congruent triangles, three rectangles (two of size \(10 \times 4\) and one of size \(12 \times 4\)), and wait, actually let's re - examine. Wait, the triangular part: the triangle has a base? Wait, no, looking at the net, there are two congruent triangles (the ones with sides 10, 10 and base related to the trapezoid? Wait, no, let's break down the areas.

First, the two triangular faces: each triangle has a base? Wait, the height of the triangle is 8 in, and the base? Wait, the middle rectangle is 12 in (height) and 4 in (width). Wait, maybe the triangles: the area of a triangle is \(\frac{1}{2}\times base\times height\). Wait, the two triangles: let's see, the slant sides are 10 in, the height (altitude) of the triangle is 8 in, so we can find the base of the triangle using Pythagoras? Wait, no, maybe the base of the triangle is 12 in? Wait, no, the middle rectangle is 12 in (length) and 4 in (width). Wait, perhaps the net is for a polyhedron with two triangular faces and three rectangular faces? Wait, no, looking at the net: there are two triangles (the top and the other side), and three rectangles? Wait, no, let's count the shapes:

- Two congruent triangles: each with base 12 in? Wait, no, the height of the triangle is 8 in, and the hypotenuse? Wait, no, the triangle has a height of 8 in, and the base? Wait, maybe the base of the triangle is 12 in? Wait, no, let's calculate the area of each part:

1. Area of the two triangular faces:
Each triangle has a base of 12 in? Wait, no, the middle rectangle is 12 in (length) and 4 in (width). Wait, the triangle: the height is 8 in, and the base? Wait, maybe the base of the triangle is 12 in? Wait, no, let's use the formula for the area of a triangle: \(A=\frac{1}{2}\times base\times height\). Wait, if the height of the triangle is 8 in, and the slant side is 10 in, then the base of the triangle (half - base? Wait, no, maybe the triangle is isoceles with two sides 10 in, height 8 in, so the base of the triangle is \(2\times\sqrt{10^{2}-8^{2}}=2\times6 = 12\) in. Ah, yes! So each triangle has a base of 12 in and height of 8 in. So area of one triangle is \(\frac{1}{2}\times12\times8 = 48\) square inches. So two triangles: \(2\times48 = 96\) square inches.

2. Area of the rectangular faces:
- The middle rectangle (the one between the two triangles) has dimensions \(12\times4\), so area \(12\times4 = 48\) square inches.
- Then there are two rectangles with dimensions \(10\times4\): area of each is \(10\times4 = 40\) square inches, so two of them: \(2\times40 = 80\) square inches.
- Then there is another rectangle? Wait, no, looking at the net, there are three rectangles? Wait, no, the bottom part: two rectangles of \(10\times4\) and one of \(12\times4\)? Wait, no, the net has:

Wait, the net:

- Two triangles (top and the other side)
- Three rectangles? Wait, no, let's count the number of rectangles:

Wait, the vertical rectangles: one is \(12\times4\), and two are \(10\times4\), and then the bottom two rectangles? Wait, no, the bottom part has two rectangles of \(10\times4\) (since the length is 10 in and width 4 in) and one rectangle? Wait, no, the total rectangles:

Wait, maybe I made a mistake. Let's re - examine the net:

The net consists of:

- Two congruent triangles: each with base 12 in and height 8 in.
- Three rectangles? Wait, no, looking at the net, there are:

1. Two triangles: area of each triangle \(A_{triangle}=\frac{1}{2}\times12\times8 = 48\), so two triangles: \(2\times48 = 96\)
2. Three rectangles? Wait, no, the middle rectangle (the one connecting the two triangles) is \(12\times4\), area \(12\times4 = 48\)
3. Then there are two rectangles of \(10\times4\) (the ones on the sides of the middle rectangle) and two rectangles of \(10\times4\) at the bottom? Wait, no, the bottom part has two rectangles of \(10\times4\) (since the length is 10 in and width 4 in) and one rectangle? Wait, no, the net:

Wait, the total surface area is the sum of the areas of all the faces.

Wait, let's list all the faces:

- Two triangular faces: area of each is \(\frac{1}{2}\times12\times8 = 48\), so \(2\times48 = 96\)
- Three rectangular faces? Wait, no, looking at the net, there are:

Wait, the vertical rectangles:

- One rectangle with dimensions \(12\times4\): area \(12\times4 = 48\)
- Two rectangles with dimensions \(10\times4\): area \(2\times(10\times4)=80\)
- Two rectangles with dimensions \(10\times4\) at the bottom? Wait, no, the bottom part has two rectangles of \(10\times4\) (since the length is 10 in and width 4 in) and one rectangle? Wait, no, the net has:

Wait, maybe the net is composed of:

- Two triangles (area \(2\times\frac{1}{2}\times12\times8 = 96\))
- Three rectangles: one \(12\times4\) (area 48), two \(10\times4\) (area \(2\times40 = 80\)), and two \(10\times4\)? Wait, no, I think I missed. Wait, the bottom part has two rectangles of \(10\times4\) (since the length is 10 in and width 4 in) and one rectangle? Wait, no, the total number of rectangular faces:

Wait, let's count the number of rectangles:

- The middle rectangle (connecting the two triangles): \(12\times4\)
- Two rectangles on the sides of the middle rectangle (the ones with length 10 in and width 4 in): \(2\times(10\times4)\)
- Two rectangles at the bottom (also with length 10 in and width 4 in): \(2\times(10\times4)\)? Wait, no, that can't be. Wait, maybe the net is for a polyhedron with two triangular bases and three rectangular lateral faces? No, that would be a triangular prism, but a triangular prism has two triangular bases and three rectangular lateral faces. Wait, in a triangular prism, the lateral faces are rectangles with dimensions equal to the sides of the triangle and the length of the prism.

Wait, let's confirm:

In a triangular prism, the surface area is \(2\times\) (area of triangular base) \(+\) (perimeter of triangular base) \(\times\) length of the prism.

Ah! That's a better approach.

So, first, find the area of the triangular base:

The triangular base is an isoceles triangle with height \(h = 8\) in, and the equal sides (the slant sides) \(l=10\) in.

Using Pythagoras, the base of the triangle (let's call it \(b\)): the half - base \(x\) satisfies \(x^{2}+h^{2}=l^{2}\), so \(x=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\) in. So the full base \(b = 2x=12\) in.

Area of triangular base \(A_{triangle}=\frac{1}{2}\times b\times h=\frac{1}{2}\times12\times8 = 48\) square inches.

Perimeter of triangular base \(P=b + 2l=12+2\times10 = 32\) in.

The length of the prism (the distance between the two triangular bases) is the width of the rectangular faces, which is 4 in.

So the lateral surface area (area of the three rectangular faces) is \(P\times\) length \(=32\times4 = 128\) square inches.

Total surface area \(=2\times A_{triangle}+ \) lateral surface area \(=2\times48+128=96 + 128=224\)? Wait, no, that can't be, because when we look at the net, the rectangles:

Wait, no, in the net, the rectangles:

- The three lateral faces of the triangular prism: one with dimensions \(12\times4\) (the base of the triangle and the length of the prism), and two with dimensions \(10\times4\) (the equal sides of the triangle and the length of the prism).

So area of the three lateral faces:

- \(12\times4=48\)
- \(2\times(10\times4)=80\)
- Total lateral area: \(48 + 80=128\)

Area of the two triangular bases: \(2\times48 = 96\)

Total surface area: \(128+96 = 224\)? Wait, but let's check the net again. Wait, the net has more rectangles? Wait, no, the net in the picture: there are two triangles (top and the other side) and three rectangles? Wait, no, the bottom part has two rectangles of \(10\times4\) and one of \(12\times4\), and then the middle rectangle? Wait, I think I made a mistake earlier.

Wait, looking at the net:

- Two triangles (area \(2\times\frac{1}{2}\times12\times8 = 96\))
- Three rectangles: one \(12\times4\), two \(10\times4\), and two \(10\times4\)? Wait, no, the net has:

Wait, the vertical rectangles:

- Middle rectangle: \(12\times4\) (area 48)
- Left rectangle: \(10\times4\) (area 40)
- Right rectangle: \(10\times4\) (area 40)
- Bottom two rectangles: \(10\times4\) (area 40 each)? No, that's not right. Wait, the net is drawn as:

Top part: two triangles (left and right) and a middle rectangle (12x4). Then below the middle rectangle, there are two rectangles (each 10x4) and another? Wait, no, the net is for a polyhedron that when assembled, the rectangles are the lateral faces. Wait, maybe the length of the prism is 4 in, and the triangular base has sides 12, 10, 10.

Wait, let's calculate the area of each face:

1. Triangular faces: 2 faces, each with area \(\frac{1}{2}\times12\times8 = 48\). So \(2\times48 = 96\).

2. Rectangular faces:

- Face 1: 12 in (length) and 4 in (width): area \(12\times4 = 48\).

- Face 2: 10 in (length) and 4 in (width): area \(10\times4 = 40\).

- Face 3: 10 in (length) and 4 in (width): area \(10\times4 = 40\).

- Face 4: 10 in (length) and 4 in (width): area \(10\times4 = 40\).

- Face 5: 10 in (length) and 4 in (width): area \(10\times4 = 40\). Wait, no, that's too many.

Wait, I think the correct approach is to use the formula for the surface area of a triangular prism: \(SA = 2B+Ph\), where \(B\) is the area of the triangular base, \(P\) is the perimeter of the triangular base, and \(h\) is the height (length) of the prism.

We have:

- \(B=\frac{1}{2}\times12\times8 = 48\)

- \(P = 12 + 10+10=32\)

- \(h = 4\) (the distance between the two triangular bases, which is the width of the rectangular faces)

So \(SA=2\times48+32\times4=96 + 128 = 224\). Wait, but let's check the net again. Wait, the net has:

- Two triangles (area 96)

- Three rectangles? No, a triangular prism has three rectangular lateral faces. Wait, the three rectangular lateral faces have dimensions:

- One with length 12 and width 4 (area 48)

- Two with length 10 and width 4 (area 40 each, so \(2\times40 = 80\))

So total lateral area \(48 + 80=128\), plus \(2\times48 = 96\), total \(128 + 96=224\).

Wait, but maybe I made a mistake in the base of the triangle. Let's re - check the triangle: the height is 8 in, the slant side is 10 in, so the half - base is \(\sqrt{10^{2}-8^{2}} = 6\) in, so the full base is 12 in. That's correct.

So the surface area is \(224\) square inches.

Step2: Verify the calculations

- Area of triangular base: \(\frac{1}{2}\times12\times8 = 48\). Two bases: \(2\times48 = 96\).

- Perimeter of triangular base: \(12 + 10+10 = 32\). Lateral surface area: \(32\times4 = 128\).

- Total surface area: \(96+128 = 224\).

Answer:

224