HELP ME WITH MY MATH PLZ

Break down the figure into a cuboid and a triangle.

You have a cuboid with:

Height: 7cm

Length: 8cm

Width: 6cm

H x L = (7cmx8cm) = 56 cm^2

since there are 2 sides equal to each other (front and back) we add another 56 cm^2

Now calculate a side:

L x W = (8cmx6cm) = 48 cm^2

again, there are 2 sides equal to each other (top and bottom) we add another 48 cm^2

Finally, we calculate:

H x W = (7cmx6cm) = 42 cm^2

again, there are 2 sides equal to each other (right and left) we add another 42 cm^2

We add all up:

56 cm^2

56 cm^2

48 cm^2

48 cm^2

42 cm^2

42 cm^2

SA: 292 cm^2  (This is the surface area of the cuboid)

Now we have to calculate the surface area of the triangle prism, but first we have to break it down into 2 triangles to make it easier; since they're equal in measure, their sides are equal too:

T1:

A (Height) = 7 cm

B (Base) = 6 cm

C (Hypotenuse) = 9 cm

Formula: A = 1/2 bh

A = 1/2 (6cm)(7cm)

A = 1/2 (42cm^2)

A = 21 cm^2

Therefore, you have 2 triangles with SA: 21 cm^2 each

We're not done yet.

Now calculate the bottom of the figure in the middle of the triangles. This is a rectangle and we can use the rectangle surface area formula to calculate it. We can do this by multiplying one side by the other:

Base = 6cm

Height = 6cm

B x H = (6cmx6cm) = 36 cm^2

There's another rectangle in the back of the triangles which is collided with the cuboid, this has the following mesuarements:

Base = 6cm

Height = 7cm

B x H = (6cmx7cm) = 42 cm^2

We have one last rectangle which is the one in the front in a slope. We simply do the same:

Base = 6cm

Height = 9cm

B x H = (6cmx9cm) = 56 cm^2

To calculate the surface area of the prism, add all up:

21 cm^2

21 cm^2

36 cm^2

42 cm^2

56 cm^2

SA: 176 cm^2 (This is the surface area of the triangle prism)

The answer to your question "Find the surface area of the composite figure" is to simply add both surface areas up:

Cuboid SA: 292 cm^2

Triangle prism SA: 176 cm^2

Composite figure SA: 468 cm^2

Hope this helps, I spent half an hour on this haha.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents