Question

The sides of convex quadrilateral ABCD are AB = 8, BC = 5, $CD = 5,$ and $DA = 8.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $E$ so that $BE:ED = 1:1.$ Find the area of ABCD.

Answer 1

To find the area of convex quadrilateral \(ABCD\) given that \(AB = 8\), \(BC = 5\), \(CD = 5\), and \(DA = 8\), with diagonals \(AC\) and \(BD\) intersecting at point \(E\) such that \(BE:ED = 1:1\), we can use the fact that the diagonals divide the quadrilateral into four triangles \(ABE\), \(BCE\), \(CDE\), and \(DAE\).

Since \(BE = ED\), point \(E\) is the midpoint of diagonal \(BD\). We can find the area of quadrilateral \(ABCD\) using the formula for the area of a quadrilateral when we have the lengths of the sides and the property of the diagonals.

Let \(s\) be the semiperimeter of \(ABCD\). We can calculate it as follows:

\[
s = \frac{AB + BC + CD + DA}{2} = \frac{8 + 5 + 5 + 8}{2} = 13
\]

Using Brahmagupta's formula for the area \(K\) of a cyclic quadrilateral, which can be adapted for a general quadrilateral, is given by:

\[
K = \sqrt{(s - AB)(s - BC)(s - CD)(s - DA)}
\]

Calculating each term:

\[
s - AB = 13 - 8 = 5
\]
\[
s - BC = 13 - 5 = 8
\]
\[
s - CD = 13 - 5 = 8
\]
\[
s - DA = 13 - 8 = 5
\]

Substituting these values into the formula:

\[
K = \sqrt{5 \cdot 8 \cdot 8 \cdot 5} = \sqrt{(5 \cdot 5) \cdot (8 \cdot 8)} = \sqrt{25 \cdot 64} = \sqrt{1600} = 40
\]

Thus, the area of quadrilateral \(ABCD\) is

\[
\boxed{40}
\]