tseday871221 said:
You have two similar RIGHT triangles, triangle TAB and triangle MCB.
Angle A is the right angle in triangle TAB and angle C is the right angle in triangle MCB.
triangle 1: TA = x, AB = 125 feet, TB = ?
triangle 2: BC = 50 feet, CM = 30 feet, MB = ?
Using the staments above, how can I set up this problem to figure out how to find x?
IF you have followed the standard practice of naming similar triangles so that corresponding vertices are named in the same relative positions, then
TA/MC = AB/CB = TB/MB
Use the first two ratios, and substitute the values given for TA, AB, MC, and CB:
x/30 = 125/50
50x = 30*125
x = (30*125)/50
x = 75
If all you were supposed to do was find the value of x, you're done!
If you are also supposed to find the lengths of the remaining sides in each triangle, you can use the Pythagoean Theorem in each triangle.
(TA)<SUP>2</SUP> + (AB)<SUP>2</SUP> = (TB)<SUP>2</SUP>
75<SUP>2</SUP> + 125<SUP>2</SUP> = (TB)<SUP>2</SUP>
5625 + 15625 = (TB)<SUP>2</SUP>
21250 = (TB)<SUP>2</SUP>
sqrt(21250) = TB
25 sqrt(34) = TB
(MC)<SUP>2</SUP> + (BC)<SUP>2</SUP> = (MB)<SUP>2</SUP>
30<SUP>2</SUP> + 50<SUP>2</SUP> = (MB)<SUP>2</SUP>
900 + 2500 = (MB)<SUP>2</SUP>
3400 = (MB)<SUP>2</SUP>
sqrt(3400) = MB
10 sqrt(34) = MB
It just seemed to me that it was easier to use the given values and the ratios of corresponding sides to find the value of x first. Though I got the same lengths for TB and MB that arthur did, you'll note I ended up with a different value for x.
