In an isosceles triangle with a side side of 13 cm, a height is drawn equal to 12 cm. Find the base of the triangle

1. If the height is drawn to the base. Let ABC be an isosceles triangle, AB = BC = 13 cm, BH = 12 cm – height.
In an isosceles triangle, the height drawn to the base is also the median and bisector. Consequently:
AH = CH = AC / 2.
Consider the triangle AHB: angle AHB = 90 degrees, since BH is the height, AB = 13 cm is the hypotenuse, since it lies opposite the right angle, BH = 12 cm and AH are the legs.
By the Pythagorean theorem, we find AN:
AH = √ (AB ^ 2 – BH ^ 2) = √ (13 ^ 2 – 12 ^ 2) = √ (169 – 144) = √25 = 5 (cm).
Since AH = AC / 2, then in proportion:
AC = 2AН = 2 * 5 = 10 (cm).
Answer: AC = 10 cm.
2. If the height is drawn to the side. Let ABC be an isosceles triangle, AB = BC = 13 cm, AH = 12 cm – height.
Consider the triangle AHB: the angle AHB = 90 degrees, AB = 13 cm is the hypotenuse, since it lies opposite the right angle, AH = 12 cm and BH are the legs.
By the Pythagorean theorem, we find the ВН:
BH = √ (AB ^ 2 – AH ^ 2) = √ (13 ^ 2 – 12 ^ 2) = √ (169 – 144) = √25 = 5 (cm).
Then:
BC = BH + CH;
CH = BC – ВН;
CH = 13 – 5 = 8 (cm).
Consider a triangle AНС: angle AНС = 90 degrees, AH = 12 cm and HC = 8 cm – legs, AC – hypotenuse.
By the Pythagorean theorem:
AC = √ (AH ^ 2 + CH ^ 2) = √ (12 ^ 2 + 8 ^ 2) = √ (144 + 64) = √208 = 4√13 (cm).
Answer: AC = 4√13 cm.