Similar Figures and Surface Area Practice Complete this assessment to review what you’ve learned. It will not count toward your grade. 5 of 55 of 5 Items Question Rosie is building two dog houses, one for each of her dogs. One dog is bigger than the other, so she decides to make one house 12 the size of the other. How much smaller will the surface area of the small house be compared to that of the big house? Write your response as a fraction.(1 point) The surface area of the small house will be the size of the big house. Remaining Attempts : 3 Skip to navigation page 16 of 17 Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.

Answer 1

To compare the surface areas of the two dog houses based on their sizes, we need to understand how the changes in dimensions affect the surface area.

Let's denote the dimensions of the larger dog house as \( L \), \( W \), and \( H \), where \( L \) is the length, \( W \) is the width, and \( H \) is the height. The surface area \( A \) of a rectangular prism (which is a box shape for the dog house) is given by the formula:

\[ A = 2(LW + LH + WH) \]

If the smaller dog house is \( \frac{1}{12} \) the size of the larger one, then each linear dimension (length, width, and height) of the smaller house will be \( \frac{1}{12} \) of the corresponding dimensions of the larger house:

• Dimensions of the smaller house: \( \frac{L}{12}, \frac{W}{12}, \frac{H}{12} \)

Let's calculate the surface area of the smaller house:

\[ A_{\text{small}} = 2\left(\left(\frac{L}{12}\right)\left(\frac{W}{12}\right) + \left(\frac{L}{12}\right)\left(\frac{H}{12}\right) + \left(\frac{W}{12}\right)\left(\frac{H}{12}\right)\right) \]

This simplifies to:

\[ A_{\text{small}} = 2\left(\frac{LW}{144} + \frac{LH}{144} + \frac{WH}{144}\right) = \frac{2}{144}(LW + LH + WH) = \frac{1}{72}(LW + LH + WH) = \frac{1}{72}A_{\text{large}} \]

Now, we can find how much smaller the surface area of the smaller house is compared to that of the larger house.

The difference in surface area is:

\[ A_{\text{large}} - A_{\text{small}} = A_{\text{large}} - \frac{1}{72}A_{\text{large}} = A_{\text{large}}\left(1 - \frac{1}{72}\right) = A_{\text{large}}\left(\frac{72 - 1}{72}\right) = A_{\text{large}}\left(\frac{71}{72}\right) \]

To express the relative change in terms of surface area, we compare it as follows:

The smaller house’s surface area is \( \frac{1}{72}A_{\text{large}} \), and the surface area of the smaller house is smaller by \( A_{\text{large}} - \frac{1}{72}A_{\text{large}} \), which simplifies to:

\[ \text{Surface area of smaller house compared to larger house} = \frac{71}{72}A_{\text{large}} \]

In terms of fraction format, we can say the surface area of the smaller house will be \( \frac{1}{72} \) the size of the larger house, and the decrease is given by \( \frac{71}{72} \).

So,

\[ \text{Difference in surface area} = \frac{71}{72}A_{\text{large}} - A_{\text{small}} = A_{\text{large}} \]

Thus, the surface area of the small house will be \( \frac{1}{72} \) of the big house. The answer is:

\[ \frac{1}{72} \]

However, if you need just the factor by which it is smaller:

\[ \text{Smaller fraction} = \frac{1}{72} \]

This quantity represents what fraction of the larger dog's house’s area the smaller dog house occupies or the smaller unit's relative size.