In the square ABCD, line XY is drawn parallel to the base as shown in the figure below. P and Q are any two arbitrary points on XY, forming triangles ABQ and CDP. What fraction of the area of the square is the total green shaded area?
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In geometry problems, look for relationships that simplify calculations. Here, the sum of the heights of the two triangles \((h + (s-h))\) is always equal to the side of the square (\(s\)). This is the key to solving the problem quickly.
Step 1: Understanding the Concept:
The problem asks for the sum of the areas of two triangles inside a square, expressed as a fraction of the square's total area. The key is that the bases of the triangles are sides of the square, and their heights are related. \
Step 2: Key Formula or Approach:
The area of a triangle is given by the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Let the side length of the square ABCD be \(s\). The area of the square is \(s^2\).
Let the line XY be at a height \(h\) from the base CD. Since XY is parallel to CD and AB, the perpendicular distance from any point on XY to CD is \(h\).
The perpendicular distance from any point on XY to the top side AB will be \(s - h\).
Step 3: Detailed Explanation:
We need to calculate the areas of the two shaded triangles, \(\triangle ABQ\) and \(\triangle CDP\). Area of \(\triangle ABQ\):
Base = AB = \(s\).
Height = Perpendicular distance from point Q (on XY) to the base AB. This height is \((s - h)\).
Area(\(\triangle ABQ\)) = \(\frac{1}{2} \times s \times (s - h)\).
\
Area of \(\triangle CDP\):
Base = CD = \(s\).
Height = Perpendicular distance from point P (on XY) to the base CD. This height is \(h\).
Area(\(\triangle CDP\)) = \(\frac{1}{2} \times s \times h\).
Total Shaded Area:
Total Area = Area(\(\triangle ABQ\)) + Area(\(\triangle CDP\))
\[ \text{Total Area} = \left( \frac{1}{2} \times s \times (s - h) \right) + \left( \frac{1}{2} \times s \times h \right) \] \
Factor out the common term \(\frac{1}{2}s\):
\[ \text{Total Area} = \frac{1}{2}s [ (s - h) + h ] \]
\[ \text{Total Area} = \frac{1}{2}s [ s ] = \frac{1}{2}s^2 \] Fraction of the area of the square:
Fraction = \(\frac{\text{Total Shaded Area}}{\text{Area of Square}}\)
\[ \text{Fraction} = \frac{\frac{1}{2}s^2}{s^2} = \frac{1}{2} \]
Step 4: Final Answer:
The total green shaded area is exactly half of the area of the square. The fraction is \(\frac{1}{2}\) or 0.5. Notice that the result is independent of the position of the line XY (the value of \(h\)) and the positions of points P and Q on that line.
