Lesson 14

Which one doesn’t belong?

A quadrilateral has vertices \((0,0), (4,3), (13,\text-9),\) and \((9,\text-12)\).

1. What type of quadrilateral is it? Explain or show your reasoning.
2. Find the perimeter of this quadrilateral.
3. Find the area of this quadrilateral.

Use the applet to answer the questions.

1. First, observe and describe the image:
   1. What kind of line is \(BC\) in reference to the circle?
   2. What does the measurement of angle \(BDC\) appear to be?
2. Move point \(D\) around. How does the measurement of angle \(BDC\) appear to change? Use a corner of the index card to compare the size of the angle as it moves.
3. Move point \(D\) to a location with integer coordinates. Each student in the group should choose a different spot for \(D\). Calculate the slopes of segments \(BC\) and \(BD\). What do your results tell you?
4. Now select the button labeled “slopes.” This will show the slopes of \(BD\) and \(BC\) and their product. Move point \(D\) around. What do you notice about the product of the slopes?
5. Based on the results, write a conjecture that captures what you are seeing.

What kind of shape is quadrilateral \(ABCD\)? It looks like it might be a rhombus. To check, we can calculate the length of each side. Using the Pythagorean Theorem, we find that the lengths of segments \(AB\) and \(CD\) are \(\sqrt{45}\) units, and the lengths of segments \(BC\) and \(DA\) are \(\sqrt{37}\) units. All side lengths are between 6 and 7 units long, but they are not exactly the same. So our calculations show that \(ABCD\) is not really a rhombus, even though at first glance we might think it is.

We did just show that two pairs of opposite sides of \(ABCD\) are congruent. This means that \(ABCD\) must be a parallelogram. Checking slopes confirms this. Sides \(AB\) and \(CD\) each have slope \(\frac12\). Sides \(BC\) and \(DA\) each have slope 6.

Can we find the area of triangle \(EFG\)? That seems tricky, because we don’t know the height of the triangle using \(EG\) as the base. However, angle \(EFG\) seems like it could be a right angle. In that case, we could use sides \(EF\) and \(FG\) as the base and height.

To see if \(EFG\) is a right angle, we can calculate slopes. The slope of \(EF\) is \(\frac86\) or \(\frac43\), and the slope of \(FG\) is \(\text-\frac{3}{4}\). Since the slopes are opposite reciprocals, the segments are perpendicular and angle \(EFG\) is indeed a right angle. This means that we can think of \(EF\) as the base and \(FG\) as the height. The length of \(EF\) is 10 units and the length of \(FG\) is 5 units. So the area of triangle \(EFG\) is 25 square units because \(\frac12 \boldcdot 10 \boldcdot 5 = 25\).