Question

trapezoid ijkl is shown on the graph below. ijkl is dilated by a scale factor of 2 centered at point j to create ijkl.
what is the area of ijkl?
write your answer as a whole number or as a decimal rounded to the nearest tenth.
square units

Explanation:

Step1: Find coordinates of vertices

From the graph:
$I(1,6)$, $J(9,6)$, $L(3,3)$, $K(5,3)$

Step2: Calculate original trapezoid area

Area of trapezoid: $\frac{1}{2} \times (b_1 + b_2) \times h$
Here, $b_1 = IJ = 9-1=8$, $b_2 = LK = 5-3=2$, $h = 6-3=3$
$\text{Area}_{IJKL} = \frac{1}{2} \times (8+2) \times 3 = 15$

Step3: Relate scaled area to original

When dilated by scale factor $k$, area scales by $k^2$.
$k=2$, so $\text{Area}_{I'J'K'L'} = 15 \times 2^2$
$\text{Area}_{I'J'K'L'} = 15 \times 4 = 60$

Answer:

60