7. a 4-cm cube is attached to the top of a right triangular prism as shown. determine the surface area of the composite object, to the nearest square centimetre.
4 cm
5 cm
9 cm
15 cm
a. 298 cm²
b. 352 cm²
c. 336 cm²
d. 368 cm²
unit 2: powers and exponent laws
8. write the number of unit squares in this large square as a power.

Question

Accepted Answer

### Question 7
# Explanation:
## Step 1: Surface Area of Triangular Prism (without the area covered by the cube)
First, find the surface area of the right triangular prism. The triangular base has legs 9 cm and 12 cm (since it's a right triangle with hypotenuse 15 cm, by Pythagoras: $ \sqrt{15^2 - 9^2} = 12 $ cm), and height 5 cm? Wait, no, the prism's length is 5 cm? Wait, the diagram: the triangular base has base 9 cm, height 12 cm (wait, no, 9, 12, 15 is a Pythagorean triple: $ 9^2 + 12^2 = 81 + 144 = 225 = 15^2 $), so the triangular base is right-angled with legs 9 cm and 12 cm, hypotenuse 15 cm. The prism's length (the distance along the direction perpendicular to the triangle) is 5 cm? Wait, the cube is 4 cm, attached to the top. Wait, maybe the prism's dimensions: the triangular face has base 9 cm, height 12 cm? Wait, no, the diagram shows 9 cm, 5 cm, 15 cm. Wait, maybe I misread. Let's re-examine: the right triangular prism has a right triangle with legs 9 cm and 12 cm (since 9-12-15 is a right triangle), and the length of the prism (the side along the direction of the right angle) is 5 cm? Wait, no, the cube is 4 cm, attached to the top. So the surface area of the prism: two triangular faces, three rectangular faces. But then we attach a cube on top, so we have to subtract the area of the square where the cube is attached (since it's covered) and add the lateral surface area of the cube (since the top of the cube is exposed, but the bottom is covered). Wait, no: when you attach the cube to the prism, the area of the prism's top face that's covered by the cube is a square of 4x4, so we subtract that from the prism's surface area, and add the 5 faces of the cube (since the bottom face of the cube is attached to the prism, so not exposed). Wait, no: the total surface area is surface area of prism + surface area of cube - 2*(area of the square where they are attached) (because the prism's top face loses that area, and the cube's bottom face loses that area, so total loss is 2*A, but actually, when combining, the overlapping area is internal, so we subtract 2*A? Wait, no: surface area of composite = surface area of prism + surface area of cube - 2*A, where A is the area of the overlapping region (since the prism's top face had A, now it's covered, so we subtract A from the prism's surface area; the cube's bottom face had A, now it's covered, so we subtract A from the cube's surface area. So total is (SA_prism - A) + (SA_cube - A) = SA_prism + SA_cube - 2A. But A is the area of the square where they are attached, which is 4*4 = 16 cm².

First, calculate SA_prism:

Triangular base area: $ \frac{1}{2} \times 9 \times 12 = 54 $ cm² (since 9-12-15 is a right triangle, legs 9 and 12). There are two triangular faces: 2*54 = 108 cm².

Rectangular faces: three rectangles. The sides: one with dimensions 9 cm (base of triangle) and 5 cm (length of prism), one with 12 cm (height of triangle) and 5 cm, and one with 15 cm (hypotenuse) and 5 cm. Wait, no, maybe the length of the prism is 5 cm? Wait, the diagram shows 9 cm, 5 cm, 15 cm. Wait, maybe the triangular base has base 9 cm, height 5 cm? Wait, that can't be, because 9-5-15: 9² + 5² = 81 +25=106 ≠ 225. So that's wrong. Wait, maybe the prism's length is 5 cm, and the triangular base is 9 cm (base), 12 cm (height), 15 cm (hypotenuse), and the length of the prism (the side perpendicular to the triangle) is 5 cm. So the rectangular faces:

- One with 9 cm (base) and 5 cm (length): area = 9*5 = 45
- One with 12 cm (height) and 5 cm (length): area = 12*5 = 60
- One with 15 cm (hypotenuse) and 5 cm (length): area = 15*5 = 75

So lateral surface area of prism (the three rectangles) is 45 + 60 + 75 = 180 cm². Then total surface area of prism is 108 (two triangles) + 180 (three rectangles) = 288 cm².

Now, the cube: surface area of cube is 6*(4*4) = 96 cm². But when we attach the cube to the prism, the area where they meet is a square of 4*4 = 16 cm². So we have to subtract 2*16 = 32 cm² (because the prism's top face had 16 cm², now covered, so subtract 16; the cube's bottom face had 16 cm², now covered, so subtract 16). So total surface area of composite: 288 (prism) + 96 (cube) - 32 (overlap) = 352 cm²? Wait, but wait, maybe the prism's top face is not the triangular face, but the rectangular face? Wait, no, the cube is attached to the top of the prism, which is a right triangular prism, so the "top" is a rectangular face. Wait, maybe I messed up the prism's dimensions. Let's re-express:

Wait, the right triangular prism: the triangular base is right-angled with legs 9 cm and 12 cm (since 9-12-15), and the length of the prism (the side along the direction of the right angle) is 5 cm? No, maybe the prism's length is 5 cm, and the triangular face has base 9 cm, height 12 cm, and the other side 15 cm. Wait, no, maybe the diagram shows the prism with length 5 cm, and the triangular face has base 9 cm, height 12 cm (wait, no, 9 and 5? Wait, the diagram has 9 cm, 5 cm, 15 cm. Wait, maybe the right triangle has legs 9 cm and 5 cm? But 9² + 5² = 81 +25=106, which is not 15²=225. So that's impossible. Therefore, the right triangle must have legs 9 cm and 12 cm, hypotenuse 15 cm, and the length of the prism (the side perpendicular to the triangle) is 5 cm. Then the top face of the prism (the rectangular face) is 12 cm (height of triangle) by 5 cm? No, wait, the top face of the prism would be a rectangle with length equal to the length of the prism (5 cm) and width equal to the hypotenuse? No, no. Wait, a right triangular prism has two triangular bases (right triangles) and three rectangular lateral faces. The lateral faces are: one with area (leg1 * length), one with (leg2 * length), and one with (hypotenuse * length). So if the legs are 9 and 12, hypotenuse 15, and length of prism is 5 cm, then lateral faces are 9*5, 12*5, 15*5. Then the two triangular bases are (9*12)/2 each. Then the total surface area of the prism is 2*(9*12/2) + 9*5 + 12*5 + 15*5 = 108 + 45 + 60 + 75 = 288 cm², as before.

Now, the cube is 4 cm, attached to the top of the prism. The "top" of the prism is one of the lateral faces? Wait, no, the cube is attached to the top, which is a rectangular face. Wait, maybe the length of the prism is 5 cm, and the top face is the rectangle with dimensions 9 cm (leg) and 5 cm (length)? No, that doesn't make sense. Wait, maybe the prism's length is 5 cm, and the top face is the rectangle with dimensions 12 cm (leg) and 5 cm (length)? No, this is confusing. Wait, maybe the cube is attached to the rectangular face of the prism that has area 9*5? No, the cube is 4x4, so the area where it's attached is 4x4=16. So if the prism's top face (the one where the cube is attached) has area, say, 9*5=45, then we subtract 16 from that, and add the 5 faces of the cube (since the bottom face is attached). Wait, no, the total surface area is:

Surface area of prism: 2*(area of triangle) + area of three rectangles.

Then, when we attach the cube, we cover a square of 4x4 on one of the rectangular faces of the prism, so we subtract that area from the prism's surface area, and add the lateral surface area of the cube (since the bottom face of the cube is covered, so we add 5*4*4=80).

Wait, let's recalculate:

Area of triangular bases: 2*( (9*12)/2 ) = 108 cm².

Area of rectangular faces:

- Face 1: 9*5 = 45 cm² (let's say this is the front face)
- Face 2: 12*5 = 60 cm² (back face)
- Face 3: 15*5 = 75 cm² (top face? No, top face would be the one with hypotenuse? No, maybe the top face is 9*5, and the cube is attached to that. Wait, the cube is 4 cm, so attached to a 9*5 face. Then the area covered is 4*4=16, so we subtract 16 from the prism's surface area (because that part is now internal), and add the 5 faces of the cube (4*4*5=80).

So total surface area:

Prism SA: 108 + 45 + 60 + 75 = 288

Subtract the covered area: 288 - 16 = 272

Add cube's lateral SA: 272 + 80 = 352 cm². Which matches option b.

## Step 2: Verify with Cube and Prism
Wait, another way: surface area of composite = SA_prism + SA_cube - 2*A, where A is the area of the overlapping region (since both the prism and cube lose that area). SA_prism = 288, SA_cube = 6*16=96, A=16. So 288 + 96 - 2*16 = 384 - 32 = 352. Yes, that's correct.

# Answer:
b. $ 352 \, \text{cm}^2 $

### Question 8
# Explanation:
## Step 1: Determine the Side Length of the Large Square
Looking at the grid, it appears to be a square with 7 unit squares per side (since the grid has 7 columns and 7 rows, assuming the diagram is a 7x7 grid). Wait, no, let's count: the grid has 7 units per side? Wait, the diagram shows a square grid with, say, 7 rows and 7 columns (since it's a large square made of unit squares). Wait, no, maybe 7x7? Wait, no, the problem says "write the number of unit squares in this large square as a power". If the large square has side length $ n $, then the number of unit squares is $ n^2 $. Looking at the grid, if it's a 7x7 grid, then $ 7^2 $. Wait, but maybe 6x6? No, the diagram (as per the user's image) shows a grid with 7 columns and 7 rows? Wait, no, let's check: the grid has 7 unit squares along each side? Wait, the user's image for question 8 shows a square grid with, say, 7x7? Wait, no, maybe 7x7? Wait, the problem says "the number of unit squares in this large square as a power". So if the large square has side length $ n $, then the number of unit squares is $ n^2 $. From the diagram, it's a 7x7 grid? Wait, no, maybe 7? Wait, no, let's see: the grid has 7 columns and 7 rows, so $ 7^2 $. Wait, but maybe 6? No, the diagram (as per the user's image) shows a square with 7 unit squares per side. So the number of unit squares is $ 7^2 $ (if 7x7) or $ 6^2 $ (if 6x6). Wait, looking at the grid, it's a 7x7 grid? Wait, no, the user's image: the grid has 7 columns and 7 rows? Wait, the grid in the image for question 8 is a square with 7 unit squares along each side (since there are 7 vertical lines and 7 horizontal lines, making 6 spaces? No, no: the number of unit squares along a side is equal to the number of small squares, so if the grid has 7 small squares per side, then it's $ 7^2 $. Wait, but maybe 6? No, the diagram shows a grid with 7x7? Wait, no, let's count: the grid has 7 columns and 7 rows, so the number of unit squares is $ 7 \times 7 = 7^2 $. Wait, but maybe 6? No, the user's image: the grid is 7x7? Wait, no, the problem says "write the number of unit squares in this large square as a power". So if the large square has side length $ n $, then the number of unit squares is $ n^2 $. From the diagram, it's a 7x7 grid, so $ 7^2 $. Wait, but maybe 6? No, the grid in the image has 7 unit squares per side. So the answer is $ 7^2 $ (or $ 6^2 $ if I'm wrong, but likely 7x7). Wait, no, let's check: the grid has 7 columns and 7 rows, so 7x7=49, which is $ 7^2 $.

## Step 2: Express as a Power
The number of unit squares is the square of the side length. If the side length is 7, then it's $ 7^2 $.

# Answer:
$ 7^2 $ (or if the grid is 6x6, $ 6^2 $, but based on the diagram, likely $ 7^2 $)

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QUESTION IMAGE

Question

Explanation:

Question 7

Step 1: Surface Area of Triangular Prism (without the area covered by the cube)

Step 1: Determine the Side Length of the Large Square

Step 2: Express as a Power

Answer:

Question 8