$ABCD$ is a square. Find the area of the green region.

[asy]
unitsize(3 cm);

pair A, B, C, D, P, Q;

A = (0,1);
B = (1,1);
C = (1,0);
D = (0,0);
P = (0,0.6);
Q = extension(P, P + rotate(90)*(B - P), C, D);

fill(D--P--Q--cycle,green);
draw(A--B--C--D--cycle);
draw(B--P--Q);
draw(rightanglemark(B,P,Q,2));

label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$P$", P, W);
label("$Q$", Q, S);
label("$5$", (A + P)/2, W, red);
label("$8$", (C + Q)/2, S, red);
[/asy]
	

To find the area of the green region, we first need to determine the coordinates of point $ Q $ and use it to find the area of triangle $ DPQ $.

### Step 1: Coordinates of Points
Given the square $ ABCD $ with points:
- $ A(0, 1) $
- $ B(1, 1) $
- $ C(1, 0) $
- $ D(0, 0) $
- $ P(0, 0.6) $

We need to find the coordinates of point $ Q $, where $ Q $ lies on $ CD $ and $ PQ $ is perpendicular to $ BP $.

### Step 2: Equation of Line $ BP $
The slope of $ BP $ can be calculated since $ B(1, 1) $ and $ P(0, 0.6) $ are given:
\[
\text{slope of } BP = \frac{1 - 0.6}{1 - 0} = \frac{0.4}{1} = 0.4
\]
Thus, the equation of line $ BP $ is:
\[
y - 0.6 = 0.4(x - 0)
\]
\[
y = 0.4x + 0.6
\]

### Step 3: Slope of Perpendicular Line $ PQ $
The slope of $ PQ $ must be the negative reciprocal of $ 0.4 $, which is $ -\frac{1}{0.4} = -2.5 $.

### Step 4: Equation of Line $ PQ $
The equation of line $ PQ $, passing through $ P(0, 0.6) $ with slope $ -2.5 $, is:
\[
y - 0.6 = -2.5(x - 0)
\]
\[
y = -2.5x + 0.6
\]

### Step 5: Intersection of Line $ PQ $ with Line $ CD $
The line $ CD $ is horizontal at $ y = 0 $ (since $ C(1, 0) $ and $ D(0, 0) $). Substitute $ y = 0 $ in the equation of $ PQ $:
\[
0 = -2.5x + 0.6
\]
\[
2.5x = 0.6
\]
\[
x = \frac{0.6}{2.5} = 0.24
\]
Thus, the coordinates of $ Q $ are $ (0.24, 0) $.

### Step 6: Area of Triangle $ DPQ $
The area of triangle $ DPQ $ can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
Here, the base $ DQ $ is the distance between $ D(0, 0) $ and $ Q(0.24, 0) $, which is $ 0.24 $. The height is the distance from $ P(0, 0.6) $ to line $ CD $, which is $ 0.6 $. Therefore:
\[
\text{Area} = \frac{1}{2} \times 0.24 \times 0.6 = \frac{1}{2} \times 0.144 = 0.072
\]

Thus, the area of the green region is $ \boxed{0.072} $ square units.

Would you like more details on any part of the solution or have any other questions?

Here are some related problems:
1. How would you find the area if $ P $ were at a different position on $ AD $?
2. What is the area of the triangle if $ PQ $ were parallel to $ AB $ instead of perpendicular to $ BP $?
3. How does the area of the green region change if the square is scaled up by a factor of 2?
4. What would be the coordinates of $ Q $ if $ P $

$ABCD$ is a square. Find the area of the green region.

This problem involves finding the area of a triangular region inside a square using coordinate geometry. Learn how to determine the coordinates of key points, apply the slope formula, and calculate the area of the triangle. Suitable for students in Grades 8-10.

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