Question

Graph triangle DEF with vertices D(2,4) E(6,4) and F(4,8) and it’s image after a dilation centered at the origin with a scale factor of 3/2. Then, find the difference between the area of the image and the area of the preimage.The graphing part I got correct. I’m just confused on how to find the difference between the area of the image and the preimage.

Answer 1

We have to find the difference between the area of the image and the pre-image.

When we scale a figure with a scale factor k, the the area of the image figure will be k² times the area of the pre-image figure.

We will prove this by calculating both areas with the traditional method: the area of a triangle is equal to half the product of the base and the height.

We can identify the base and the height in the graph of both figures as:

We then can calculate the area of the pre-image as:

$A=\frac{bh}{2}=\frac{4\cdot4}{2}=\frac{16}{2}=8$

and the area of the image as:

$A^{\prime}=\frac{b^{\prime}h^{\prime}}{2}=\frac{6\cdot6}{2}=\frac{36}{2}=18$

Then, the difference in the area between the image and the pre-image is:

$d=A^{\prime}-A=18-8=10$

NOTE: we can now test that the relation between the areas is k²:

$\begin{gathered} k=\frac{3}{2} \\ A^{\prime}=k^2\cdot A \\ A^{\prime}=(\frac{3}{2})^2\cdot8=\frac{9}{4}\cdot8=\frac{72}{4}=18 \end{gathered}$

Answer: the difference of areas between the image and the pre-image is 10 square units.