Algebra Equations

Algebra Level 2

Above shows a rectangle A B C D ABCD with F F be the midpoint of A B AB , A B = 20 AB = 20 , B C = 16 BC= 16 , E D = B G = x ED = BG = x . Lines F H FH and I J IJ are parallel to each other. I C = 7 16 IC = \frac7{16} DC.

What is the difference of the area between the trapezium C G I J CGIJ and the area of the trapezium D E F H DEFH ?

Take x = A B ⋅ A D [ 2.5 ( A B ) ] x =\dfrac{AB\cdot AD}{[2.5(AB)]} .


The answer is 8.75.

Muhd Nazmi by Muhd Nazmi , 23, Malaysia

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1 solution

Muhd Nazmi
Oct 5, 2014

∣ H = ∣ I 16 10 = x ( 7 16 20 ) 1 2 ( x + 16 ) ( 20 2 ) − 1 2 ( 16 − x + 14 ) ( 7 16 20 ) = 8.75 |H\quad =\quad |I\\ \\ \frac { 16 }{ 10 } \quad =\quad \frac { x }{ (\frac { 7 }{ 16 } 20) } \\ \\ \frac { 1 }{ 2 } (x+16)(\frac { 20 }{ 2 } )\quad -\quad \frac { 1 }{ 2 } (16-x+14)(\frac { 7 }{ 16 } 20)\quad =\quad 8.75

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