 How Cheenta works to ensure student success?
Explore the Back-Story

# Problem from Area of Rectangle | SMO 2012 | Junior Section 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

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 )

## Subscribe to Cheenta at Youtube

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

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 )

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.

### Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy  