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Montano1993 [528]
3 years ago
6

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

at is the area of $ABCD$? Express your answer in the simplest radical form.
Mathematics
2 answers:
Jet001 [13]3 years ago
8 0

Answer:

34 is the answer

Step-by-step explanation:

GuDViN [60]3 years ago
3 0

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.

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