Class Sumary 02/05/09

Today we spent most of the class going over questions 24, 25 of pg. 358. After that, Mrs. Saatkamp gave us a worksheet with more problems to work on so we could practice for the quiz TOMORROW.

Explanations: (refer to diagrams on pg. 358)

24. (find EC, GC, and EF)

First you have to find which triangles are similar to each other:

Angle A is congruent to Angle BCA since they are alternate interior angles.

Angle E is congruent to Angle B since they are opposite angles in the paralellogram ECBF.

Therefore, ∆ABC~∆CEG- AA similarity. Therefore, the following proportion is true for those triangles:

By substituting the given information into the proportions and solving, the result is:

GC=4

EC=5

Finally, because BC= 6 and it is opposite, and therefore congruent to EF in the parallelogram ECBF, EF= 6.

25. (Find AE, AC, CF, CB, CD, EF):

1. First you have to find which triangles are similar to each other:

Angle CEF is congruent to angle CAB since they are corresponding angles.

Angle CFE is congruent to angle CBA since they are also corresponding angles.

Therefore, Triangle ABC ~ triangle EFC- AA Similarity. Thus, the following proportion is true for those triangles:

After adding in the given infromation, the proportion would look like this:

After solving it by first using the first and last ratios to solve for EF, and using the second and last ratio to solve for FC, the result would be:

EF= 16.7

FC or CF=13.3 (repeating).

2. Now to find a second pair of similar traingles:

Angle ADC is congruent to angle BDC, for they are both right angles.

Triangle ABC and Traingle ACD also have angle CAD in common, therefore they are similar. Thus, the following proportion is true for those triangles: