Question

Given the rectangle abcd shown below has a total area of 72. E is in the midpoint of bc and f is the midpoint of dc. What is the area of the inscribed triangle aef.

Answer 1

Refer to the attached image.

Given the rectangle ABCD of length 'l' and height 'h'.

Therefore, CD=AB = 'l' and BC = AD = 'h'

We have to determine the area of triangle AEF.

Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

Area of triangle ADF =
`(1)/(2)bh`

=
`(1)/(2)(DF * AD)`

=
`(1)/(2)((l)/(2) * h)`

`=(lh)/(4)`

Area of triangle ECF =
`(1)/(2)bh`

=
`(1)/(2)(CF * CE)`

=
`(1)/(2)((l)/(2) * (h)/(2))`

`=(lh)/(8)`

Area of triangle ABE =
`(1)/(2)bh`

=
`(1)/(2)(AB * BE)`

=
`(1)/(2)(l * (h)/(2))`

`=(lh)/(4)`

Now, area of triangle AEF =

Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

=
`72 - ((lh)/(4) + (lh)/(8) + (lh)/(4))`

=
`72 - ((2lh+lh+2lh)/(8))`

=
`72 - ((5lh)/(8))`

=
`72 - ((5 * 72)/(8))`

`=(72 * 8 - (5 * 72))/(8)`

= 27 units

Therefore, the area of triangle AEF is 27 units.