Question

ABCD is a trapezium with AB= x cm, DC=( x+ 4 )cm and the distance between the parallel sides AB and DC is 1/2 x cm

Answer 1

We know that the smaller base is x cm, the greater base is x+4 cm, and that the height is x/2 cm.

Since the area of a trapezium is given by

$A=\dfrac{(b+B)h}{2}$

If we plug the expressions we have

$A=\dfrac{(x+(x+4))\frac{x}{2}}{2}=\dfrac{(2x+4)x}{4}=\dfrac{2x^2+4x}{4}$

If this must equal 84, we can solve for x as follows: start with

$\dfrac{2x^2+4x}{4}=84$

Multiply both sides by 4:

$2x^2+4x=336$

Subtract 336 from both sides:

$2x^2+4x-336=0$

We can divide both sides by 2 to make computation simpler:

$x^2+2x-168=0$

Use the quadratic formula (or any mean you prefer) to solve this equation and get the solutions

$x_1=-14,\quad x_2=12$

We can't accept the negative solution, because it would imply

$AB=-14,\quad CD=-10$

and we can't have negative side lengths. So, the answer is 12.