In the equal femoral triangle abc with the base AC, the height ak is drawn. it is known that AB = 10 and BK = 6.

In the equal femoral triangle abc with the base AC, the height ak is drawn. it is known that AB = 10 and BK = 6. find the height of the AK and the base of the ac if the angle is sharp

1. The height of AK cuts off from the isosceles △ ABC rectangular △ AKB, in which ∠AKB = 90 °, AB = 10 is the hypotenuse (since it lies opposite the right angle), AK and BK = 6 are the legs.
Using the Pythagorean theorem, we find the length of the leg AK:
AK = √ (AB² – BK²) = √ (10² – 6²) = √ (100 – 36) = √64 = 8.
2. Since △ ABC is isosceles, then AB = BC = 10.
Point K divides side BC into two segments – BK and CK, then:
BC = BK + CK.
Substitute the known values:
6 + CK = 10;
CK = 10 – 6;
CK = 4.
3. Consider a rectangular △ AKC: ∠AKC = 90 °, AC – hypotenuse (since it lies opposite the right angle), AK = 8 and CK = 4 – legs.
By the Pythagorean theorem, we find the length of the hypotenuse AC:
AC = √ (AK² + CK²) = √ (8² + 4²) = √ (64 + 16) = √80 = 4√5.
Answer: AK = 8, AC = 4√5.