Math topic in the second day of junior high school! ! ! ! ! ! !

Solution: do high AD, BE, CF

Let BD=x, then CD=BC-BD=6-x,

In the right triangle ABD, AD 2 = AB 2-BD 2 = 16-X 2 is obtained from Pythagorean theorem.

In the right-angled triangle ACD, AD 2 = AC 2-CD 2 = 25-(6-x) 2 is obtained from Pythagorean theorem.

So16-x 2 = 25-(6-x) 2,

The solution is x=9/4,

Substitute in the above formula, get, AD=5√7/4,

So △ ABC area = (1/2) * BC * ad = (1/2) * 6 * (5 √ 7/4) =15 √ 7/4,

By the area method,

△ABC area = (1/2) * ab * cf =15 √ 7/4,

The solution is CF= 15√7/8.

△ABC area = (1/2) * BC * be =15 √ 7/4,

The solution is BE=3√7/2,

So the three heights of △ABC are: AD: 5 √ 7/4, BE: 3 √ 7/2, CF: 15 √ 7/8.