The Area of Rectangle (The More Challenging Version)

Geometry Level pending

There is a square A B C D ABCD with side length of 6 6 . Points E E and F F are on lines A B AB and B C BC respectively such that A E = 1.5 AE = 1.5 and F C = 2 FC = 2 . Points G G and H H exists such that E F G H EFGH is a rectangle and D D lies on line G H GH . Find the area of rectangle E F G H EFGH .


The answer is 33.

Hua Zhi Vee by Hua Zhi Vee , 16, Malaysia

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1 solution

Hua Zhi Vee
Jun 9, 2017

Connect DE and DF. Note that A r e a o f △ D E F = 1 2 ( A r e a o f E F G H ) = A r e a o f A B C D − A r e a o f △ A E D − A r e a o f △ B E F − A r e a o f △ C D F Area\ of \bigtriangleup DEF = \frac{1}{2}(Area\ of\ EFGH) = Area\ of\ ABCD - Area\ of \bigtriangleup AED - Area\ of \bigtriangleup BEF - Area\ of \bigtriangleup CDF .

A r e a o f △ D E F Area\ of \bigtriangleup DEF

= A r e a o f A B C D − A r e a o f △ A E D − A r e a o f △ B E F − A r e a o f △ C D F = Area\ of\ ABCD - Area\ of \bigtriangleup AED - Area\ of \bigtriangleup BEF - Area\ of \bigtriangleup CDF

= 36 − 1 2 × 1.5 × 6 − 1 2 × 4.5 × 4 − 1 2 × 2 × 6 = 36 - \frac{1}{2} \times 1.5 \times 6 - \frac{1}{2} \times 4.5 \times 4 - \frac{1}{2} \times 2 \times 6

= 36 − 4.5 − 9 − 6 = 36 - 4.5 - 9 - 6

= 16.5 = 16.5

A r e a o f E F G H Area\ of\ EFGH

= 2 × A r e a o f △ D E F = 2 \times Area\ of \bigtriangleup DEF

= 2 × 16.5 = 2 \times 16.5

= 33 = 33

∴ \therefore The area of E F G H EFGH is 3

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