Question

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 the simplest radical form.

Answer 1

Answer:

34 is the answer

Step-by-step explanation:

Answer 2

Answer:

$\sqrt{2}+\sqrt{5}$

Step-by-step explanation:

Since there are 2 diagonal angles that are 90 degrees, we can figure out the shape is made of 2 right triangles, so we can use the Pythagorean theorem to find out the length of each side.

The problem said that there must be 2 distinct sides with integer lengths, but there are no 2 integers that satisfy $x^2+y^2 = 3^2$ so we can deduce that both right triangles contain 1 integer value side and 1 non-integer value side.

There are only 2 positive integer values that would fit in $x^2+y^2 = 3^2$ for x: 1 and 2. That means the 2 triangles' equation is:

$1^2+\sqrt{8}^2 = 3^2$ and $2^2 + \sqrt{5}^2 = 3^2$.

Now, since these are right triangles, their areas are just going to be their 2 legs multiplied by 1/2.

The first triangle's area is:

$1\cdot\sqrt{8}\cdot1/2 = 1\cdot2\sqrt{2}\cdot1/2 = \sqrt{2}$

The second triangle's area is:

$2\cdot\sqrt{5}\cdot1/2 = \sqrt{5}$

The total area of the quadrilateral would be the sum of the 2 triangles, which is just $\sqrt{2}+\sqrt{5}$.

I hope this helped you.