Math Problem Statement

$ABCD$ 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]

Solution

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

Step 1: Coordinates of Points

Given the square ABCDABCD with points:

  • A(0,1)A(0, 1)
  • B(1,1)B(1, 1)
  • C(1,0)C(1, 0)
  • D(0,0)D(0, 0)
  • P(0,0.6)P(0, 0.6)

We need to find the coordinates of point QQ, where QQ lies on CDCD and PQPQ is perpendicular to BPBP.

Step 2: Equation of Line BPBP

The slope of BPBP can be calculated since B(1,1)B(1, 1) and P(0,0.6)P(0, 0.6) are given: slope of BP=10.610=0.41=0.4\text{slope of } BP = \frac{1 - 0.6}{1 - 0} = \frac{0.4}{1} = 0.4 Thus, the equation of line BPBP is: y0.6=0.4(x0)y - 0.6 = 0.4(x - 0) y=0.4x+0.6y = 0.4x + 0.6

Step 3: Slope of Perpendicular Line PQPQ

The slope of PQPQ must be the negative reciprocal of 0.40.4, which is 10.4=2.5-\frac{1}{0.4} = -2.5.

Step 4: Equation of Line PQPQ

The equation of line PQPQ, passing through P(0,0.6)P(0, 0.6) with slope 2.5-2.5, is: y0.6=2.5(x0)y - 0.6 = -2.5(x - 0) y=2.5x+0.6y = -2.5x + 0.6

Step 5: Intersection of Line PQPQ with Line CDCD

The line CDCD is horizontal at y=0y = 0 (since C(1,0)C(1, 0) and D(0,0)D(0, 0)). Substitute y=0y = 0 in the equation of PQPQ: 0=2.5x+0.60 = -2.5x + 0.6 2.5x=0.62.5x = 0.6 x=0.62.5=0.24x = \frac{0.6}{2.5} = 0.24 Thus, the coordinates of QQ are (0.24,0)(0.24, 0).

Step 6: Area of Triangle DPQDPQ

The area of triangle DPQDPQ can be calculated using the formula: Area=12×Base×Height\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} Here, the base DQDQ is the distance between D(0,0)D(0, 0) and Q(0.24,0)Q(0.24, 0), which is 0.240.24. The height is the distance from P(0,0.6)P(0, 0.6) to line CDCD, which is 0.60.6. Therefore: Area=12×0.24×0.6=12×0.144=0.072\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 0.072\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 PP were at a different position on ADAD?
  2. What is the area of the triangle if PQPQ were parallel to ABAB instead of perpendicular to BPBP?
  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 QQ if PP

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Triangles

Formulas

Slope formula: (y2 - y1) / (x2 - x1)
Equation of a line: y = mx + c
Area of a triangle: 1/2 * base * height

Theorems

Pythagorean Theorem (implicitly used in determining the perpendicular)

Suitable Grade Level

Grades 8-10