In triangle ABC, Area of ABC=12/25 in^2, AC=6/5in, and BC=1 in. What is the length of AB?

Answer 1

Area = 12/25 = 48/100 = .48 in^2 = A

AC = 6/5 = 1.2

BC = 1 = AB

AB= 1

Use Heron's Formula for Area of a Triangle

.48 = square root of (s(s-a)(s-b)(s-c))

.48^2 = .2304 = 1.6(.6)(.4)(.6) = .2304

Answer 2

To make nomenclature simpler I will use a = BC, b = AC, and c = AB.

a = 1, b = 6/5, Area= 12/25, c = ?

First I will use the formula Area to find angle C.

Then I will use the Law of Cosines to find c.

Then I will check by Heron's Formula

Area = (1/2)abSinC

12/25 = (1/2)(6/5)(1) SinC

SinC = 4/5

CosC = 3/5 (Pythagoren Identity) We don't need to even know what C is because this is such a pretty problem.

c^2 = a^2 + b^2 - 2abCosC

c^2 = 1^1 + (6/5)^2 - 2(1)(6/5)(3/5)

c^2 = 1 + 36/25 - 36/25

c^2 = 1

c = 1 This is the answer you are asking for -- the length of AB

Now to check via Heron's Formula

Perimeter = a + b + c = 1 + 6/5 + 1 = 16/5

s = semiperimeter = 8/5

Area = sqrt[s(s-a)(s-b)(s-c)]

Area = sqrt[(8/5)(3/5)(2/5)(3/5)]

Area = sqrt[144/625]

Area = 12/25 Q.E.D.