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QuestionIdentify the quadrilateral formed by the points (−5,0),(0,4),(5,0),(0,−4)(-5,0),(0,4),(5,0),(0,-4). Options: A. Trapezoid B. Rhombus C. Square D. Rectangle.

Studdy Solution

STEP 1

Assumptions1. The vertices of the quadrilateral are at the points (-5,0), (0,4), (5,0), (0,-4) . We are using the Euclidean distance formula to calculate the distances between points d=(x−x1)+(y−y1)d=\sqrt{\left(x_{}-x_{1}\right)^{}+\left(y_{}-y_{1}\right)^{}}

STEP 2

First, let's plot the points on a Cartesian plane.

STEP 3

After plotting, we can see that the points form a quadrilateral with the following vertices1. Point A at (-5,0)
2. Point B at (0,)
3. Point C at (5,0) . Point D at (0,-)

STEP 4

Next, we need to calculate the distances between each pair of points. We'll start with the distance between points A and B.
AB=(0−(−))2+(4−0)2AB = \sqrt{\left(0-(-)\right)^{2}+\left(4-0\right)^{2}}

STEP 5

Calculate the distance between points A and B.
AB=(5)2+(4)2=25+16=41AB = \sqrt{(5)^{2}+(4)^{2}} = \sqrt{25+16} = \sqrt{41}

STEP 6

Next, calculate the distance between points B and C.
BC=(5−0)2+(0−4)2BC = \sqrt{\left(5-0\right)^{2}+\left(0-4\right)^{2}}

STEP 7

Calculate the distance between points B and C.
BC=(5)2+(−4)2=25+16=41BC = \sqrt{(5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41}

STEP 8

Next, calculate the distance between points C and D.
CD=(0−5)2+(−4−0)2CD = \sqrt{\left(0-5\right)^{2}+\left(-4-0\right)^{2}}

STEP 9

Calculate the distance between points C and D.
CD=(−5)2+(−4)2=25+16=41CD = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41}

STEP 10

Finally, calculate the distance between points D and A.
DA=(−5−0)2+(0−(−4))2DA = \sqrt{\left(-5-0\right)^{2}+\left(0-(-4)\right)^{2}}

STEP 11

Calculate the distance between points D and A.
DA=(−5)+(4)=25+16=41DA = \sqrt{(-5)^{}+(4)^{}} = \sqrt{25+16} = \sqrt{41}

STEP 12

Now that we have the lengths of all sides, we can see that all sides are equal in length. This means the quadrilateral is a rhombus.
The answer is B. Rhombus.

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