The height BH of an isosceles triangle ABC (AB = BC) refers to the AC

| April 23, 2021 | education

The height BH of an isosceles triangle ABC (AB = BC) refers to the AC side as 7:48. Find the base of the AC if the perimeter of the triangle ABC is 49cm.

BH / AC = 7/48.
P = AB + BC + AC = AC + 2x, where x = AB = BC.
AC + 2x = 49.
x = (49 – AC) / 2.
Consider a triangle ABH, BH = AC / 2.
By the Pythagorean theorem:
(BH) ^ 2 = x ^ 2 – (AC / 2) ^ 2.
Divide both parts into AC ^ 2, we get:
(BH / AC) ^ 2 = (x / AC) ^ 2 – 1/4.
Substitute x:
(BH / AC) ^ 2 = ((49 – AC) / (2 * AC)) ^ 2 – 1/4.
(7/48) ^ 2 = ((49 – AC) / (2 * AC)) ^ 2 – 1/4.
(7/12) ^ 2 = ((49 – AC) / (AC)) ^ 2 – 1.
(49 + 576) / 576 = ((49 – AC) / (AC)) ^ 2.
(AC) ^ 2 * (625) / 576 = (49 – AC) ^ 2.
Let’s remove the degree in both parts:
AC * (25/24) = ± (49 – AC).
1) AC * (25/24) = 49 – AC.
AC * (24 + 25) = (49 * 24).
AC = (49 * 24) / 49 = 24 (cm).
2) AC * (25/24) = – (49 – АС), АС * (1 – 25/24) = 49. Not suitable, since the expression (1 – 25/24) <0
Thus, AC = 24 cm.



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