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=(bh)/(2)=(4\cdot4)/(2)=(16)/(2)=8`

and the area of the image as:

`A^(\prime)=(b^(\prime)h^(\prime))/(2)=(6\cdot6)/(2)=(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=(3)/(2) \\ A^(\prime)=k^2\cdot A \\ A^(\prime)=((3)/(2))^2\cdot8=(9)/(4)\cdot8=(72)/(4)=18 \end{gathered}`

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