Quadrilateral $ABCD$ has right angles at $B$ and $D$, and $AC=3$. If $ABCD$ has two sides with distinct integer lengths, then what is the area of $ABCD$? Express your answer in simplest radical form.

Guest Apr 15, 2018

#1**+1 **

ac is the diaganal.

thus pythagorean theorem says that ab^2 + bc^2 = ac^2

and ad^2 + dc^2 = ac^2 and ac= 3 so ac^2 =9

there are no two distinct integers when squared and then added = 9

quadleraterals have 4 sides so that means 2 will have radicals.

so lets just pick sides like 2 and \(\sqrt{}\)5 that gives us 9

then the other two sides might be 1 and \(\sqrt{}\)8

1 and 2 satisfy the two distinct integer lengths

now for area we just find the area of the two triangles and add them up

for the area of the quadrilateral. area of triangle is bh/2

(2*\(\sqrt{}\)5) /2 = \(\sqrt{}\)5

(1*\(\sqrt{}\)8) /2 = (\(\sqrt{}\)8)/2

added together for total area is \(\sqrt{}\)5 + (\(\sqrt{}\)8)/2

maybe they want those in one term?

(2\(\sqrt{}\)5 +\(\sqrt{}\)8) / 2

Guest Apr 16, 2018