Trapezoid Area Problem: Finding Area When DE Divides Triangle ABC in Half

The trapezoid DECB forms part of triangle ABC.

AB = 6 cm
AC = 10 cm

Calculate the area of the trapezoid DECB, given that DE divides both AB and AC in half.

DE crosses AB and AC, that is to say:

$AD=DB=\frac{1}{2}AB=\frac{1}{2}\times6=3$

$AE=EC=\frac{1}{2}AC=\frac{1}{2}\times10=5$

Now let's look at triangle ADE, two sides of which we have already calculated.

Now we can find the third side DE using the Pythagorean theorem:

$AD^2+DE^2=AE^2$

We substitute our values into the formula:

$3^2+DE^2=5^2$

$9+DE^2=25$

$DE^2=25-9$

$DE^2=16$

We extract the root:

$DE=\sqrt{16}=4$

Now let's look at triangle ABC, two sides of which we have already calculated.

Now we can find the third side (BC) using the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute our values into the formula:

$6^2+BC^2=10^2$

$36+BC^2=100$

$BC^2=100-36$

$BC^2=64$

We extract the root:

$BC=\sqrt{64}=8$

Now we have all the data needed to calculate the area of the trapezoid DECB using the formula:

(base + base) multiplied by the height divided by 2:

Keep in mind that the height in the trapezoid is DB.

$S=\frac{(4+8)}{2}\times3$

$S=\frac{12\times3}{2}=\frac{36}{2}=18$