Try this beautiful problem from area of rectangle from Singapore Math Olympiad, 2012, Junior Section.
Problem – Area of Rectangle (SMO Exam)
In the diagram below , A and B (20,0) lie on the x-axis and c(0,30) lies on the y-axis such that \(\angle {ABC} = 90^\circ\).A rectangle DEFG is inscribed in triangle ABC . Given that the area of triangle CGF is 351, calculate the area of the rectangle DEFG .
- 468
- 456
- 654
- 400
Key Concepts
Area of Triangle
Area of Rectangle
2-D Geometry
Check the Answer
But try the problem first…
Answer: 468
Singapore Mathematics Olympiad,
Challenges and Thrills – Pre – College Mathematics
Try with Hints
First hint
We can try this sum from taking
OA = \(\frac {30^2}{20} = 45\)
So the area of \(\triangle {ABC} = \frac {(20+45)\times 30}{2} = 975\)
Try to do the rest of the sum………………………
Second Hint
Now lets try to find height of \(\triangle {CGF}\)
Suppose height of \(\triangle {CGF}\) be ‘h’. Then
\((\frac {h}{30})^2 = \frac {351}{975} = (\frac {3}{5})^2\)
\(\frac {h}{30 – h} = \frac {3}{2}\)
Now we have almost reach the answer . Try to find the area of Rectangle DEFG……
Final Step
Note that the rectangle DEFG has the same base as \(\triangle {CGF}\). Then its area is
\( 351 \times \frac {2}{3} \times 2 = 468 \) (Answer )
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