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

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 )