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AMC 10 USA Math Olympiad

Problem from Area of Rectangle | SMO 2012 | Junior Section

Try this beautiful problem from area of rectangle from Singapore Math Olympiad, 2012, Junior Section. You may use sequential hints if required.

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 .

Problem from area of rectangle
  • 468
  • 456
  • 654
  • 400

Key Concepts


Area of Triangle

Area of Rectangle

2-D Geometry

Check the Answer


Answer: 468

Singapore Mathematics Olympiad,

Challenges and Thrills – Pre – College Mathematics

Try with Hints


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………………………

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……

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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