In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AC=5 and BH=2, find AH and CH.
1 answer:
Answer:
AH = 1 or 4
CH = 4 or 1
Step-by-step explanation:
An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...
AH/BH = BH/CH
AH·CH = BH²
The problem statement tells us AH + CH = AC = 5, so we can write
AH·(5 -AH) = BH²
AH·(5 -AH) = 2² = 4
This gives us the quadratic ...
AH² -5AH +4 = 0 . . . . in standard form
(AH -4)(AH -1) = 0 . . . . factored
This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.
You might be interested in
Answer:
8
Step-by-step explanation:
Let's set this up to see if we can simplify it a bit:
Notice we have 4 tens on top and 3 on bottom, so we can eliminate the bottom 3 altogether and leave just one on top. And we have 2 fives on the top and 3 on bottom, so we can eliminate the 2 on the top and leave one on the bottom. Now that looks like this:
10 divided by 5 is 2, and 2 cubed is 8