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  1. Similar Triangles
  2. areas
  3. pythagoreanTheorem
  4. circles
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  6. threeDimensionalGeometry
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  9. miscellaneous
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Exercise 261
In rectangle $ABCD$, we have $A=(0,0), B=(200,20), D=(2,y)$, for some integer $y$. What is the area of rectangle $ABCD$?
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Solution
  1. $AD\bot AB$, so $\frac{20}{200}\cdot \frac{y}2=-1$, $y=-20$.
  2. $AB=\sqrt{200^2+20^2}=20\sqrt{101}, AD=\sqrt{2^2+20^2}=2\sqrt{101}$.
  3. Area$=40\cdot 101=4040$.
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