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.

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 [tex]x^2+y^2 = 3^2[/tex] 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 [tex]x^2+y^2 = 3^2[/tex] for x: 1 and 2. That means the 2 triangles' equation is:

[tex]1^2+\sqrt{8}^2 = 3^2[/tex] and [tex]2^2 + \sqrt{5}^2 = 3^2[/tex].

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:

[tex]1\cdot\sqrt{8}\cdot1/2 = 1\cdot2\sqrt{2}\cdot1/2 = \sqrt{2}[/tex]

The second triangle's area is:

[tex]2\cdot\sqrt{5}\cdot1/2 = \sqrt{5}[/tex]

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

I hope this helped you.

keshavgandhi04 keshavgandhi04 · Answer 2 · 2021-12-28T14:35:29+01:00

The total area of the quadrilateral ABCD is [tex](\sqrt{2}+ \sqrt{5} )[/tex] and this can be determined by using the formula of the area of the triangle.

Given :

Quadrilateral ABCD has right angles at B and D, and AC=3.
ABCD has two sides with distinct integer lengths.

The following steps can be used in order to determine the area of ABCD:

Step 1 - Let the length of the segment AD be 'x' and the length of the segment CD be 'y'.

Step 2 - Now, apply the Pythagorean theorem on the triangle ACD.

[tex]x^2+y^2= 3^2[/tex]

Step 3 - According to the given data, ABCD has two sides with distinct integer lengths. So, there are two possibilities:

[tex]1^2+(\sqrt{8} )^2= 3^2[/tex]

[tex](2)^2+(\sqrt{5} )^2=3^2[/tex]

So, the possible sides of the quadrilateral will be [tex]\rm 1, \;2,\; 2\sqrt{2},\;and \;\sqrt{5}[/tex].

Step 4 - So, the area of the triangle ABC is:

[tex]A=\dfrac{1}{2}\times \sqrt{8} \times 1\\A = \sqrt{2}[/tex]

Step 5 - Now, the area of the triangle ACD is:

[tex]A'=\dfrac{1}{2}\times \sqrt{5} \times 2\\A' = \sqrt{5}[/tex]

Step 6 - So, the total area of the quadrilateral ABCD is:

[tex]{\rm Area } = \sqrt{2} + \sqrt{5}[/tex]

The total area of the quadrilateral ABCD is [tex](\sqrt{2}+ \sqrt{5} )[/tex].

For more information, refer to the link given below:

https://brainly.com/question/25240753