Find the area of triangle ABC if AB = BC = 12 and <ABC = 120 degrees
Should approach this with SSS similarity? If yes will the answer be clear or will i need to manipulate it, and how.
You can use the law of cosines to find side AC, then, use
area=sqrt(s(s-a)(s-b)(s-c))
where s is the half perimeter.
Have you come across this area formula for the area of a triangle ?
- if two sides are a and b, and Ø is the angle between them, then the area = (1/2)(ab)sinØ
so for yours,
area = (1/2)(12)(12)sin 120°
= .....
To find the area of triangle ABC, we can use SSS similarity, also known as the Law of Sines. However, in this case, it would be more straightforward to apply the formula for the area of a triangle using Heron's formula.
Heron's formula states that the area of a triangle with side lengths a, b, and c can be found using the formula:
Area = √[s(s - a)(s - b)(s - c)]
where s represents the semiperimeter of the triangle, given by:
s = (a + b + c) / 2
In our case, AB = BC = 12, so a = b = 12. Let's substitute these values into the formula:
s = (12 + 12 + c) / 2
s = (24 + c) / 2
s = 12 + c/2
Now, we know that the angle <ABC is 120 degrees, and we have two equal sides of length 12. By using the Law of Cosines, we can find the third side length, c. The Law of Cosines is given by:
c^2 = a^2 + b^2 - 2ab * cos(<C)
Substituting the known values, we have:
c^2 = 12^2 + 12^2 - 2 * 12 * 12 * cos(120)
c^2 = 144 + 144 - 288 * (-0.5)
c^2 = 144 + 144 + 144
c^2 = 432
c = √432
Now that we have the side lengths a, b, and c, we can calculate the semiperimeter:
s = (12 + 12 + √432) / 2
s = (24 + √432) / 2
s ≈ 12 + 10.39
s ≈ 22.39
Finally, we can substitute the values of side lengths and the semiperimeter into Heron's formula to find the area of triangle ABC:
Area = √[22.39(22.39 - 12)(22.39 - 12)(22.39 - √432)]
Area = √[22.39 * 10.39 * 10.39 * (22.39 - √432)]
Now, you can calculate this expression using a calculator to find the area of triangle ABC.