GEOM 2 | Lesson 4 | Practice Solution

1.  \(XY = 15\)

2.   \(D\) is the midpoint of segment \(AC\):  \((\frac{5}{2},\frac{13}{2})\)

3.   \(7x – 1 = 4x + 5\); \(x = 2\) and \(VQ = 13\)

4.  \(7x + 10 = 9x – 2\); \(x = 6\) so \(AD = 104\)

5.  Let \(z\) represent the altitude.

\(\frac{20}{z} =\frac{15}{z}\); \(z^2 = 15•30\) or \(z = 10√3\)

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