Mathematics, 20.08.2020 16:01 jennifer9983

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.

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Answer from: nevelle

The total area of the quadrilateral ABCD is $(\sqrt{2}+ \sqrt{5} )$ 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.

x^2+y^2= 3^2

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

$1^2+(\sqrt{8} )^2= 3^2$

$(2)^2+(\sqrt{5} )^2=3^2$

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

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

$A=\dfrac{1}{2}\times \sqrt{8} \times 1\\A = \sqrt{2}$

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

$A'=\dfrac{1}{2}\times \sqrt{5} \times 2\\A' = \sqrt{5}$

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

${\rm Area } = \sqrt{2} + \sqrt{5}$

The total area of the quadrilateral ABCD is $(\sqrt{2}+ \sqrt{5} )$ .

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Answer from: dondre54

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

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.

Answer from: Quest

1.3 million hoops i believe

Answer from: Quest

y= -2x+4, have a great day

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Quadrilateral $ABCD$ has right angles at $B$ and $D$, and $AC=3$. If $ABCD$ has two sides with disti...