How do I find the area of a triangle when I only know the three side lengths?
There are two methods to find the area: Method 1: use cosine rule to find out one angle: cosA=(b^2+c^2-a^2)/2bc, and then use area formula 1/2bcSinA to work out the area. Method 2: use Heron's formula which is Area=(s(s-a)(s-b)(s-c))^1/2, where s=(a+b+c)/2, but neither A-level nor GCSE will use Heron's formula to solve area for triangle, and I believe Heron's formula is not in mark scheme, so you don't need to learn the second method. The first method is more common. I hope you find this answer helpful.
By Herons formula , you can find the area of the triangle
Let’s say we have a triangle with lengths a, b and c. Now we gotta find an angle so we can use the equation: Area=1/2 * (length of side 1) * (length of side 2) * sin(the angle opposite side 3) To do this we use the cosine rule which states: a^2 + b^2 -2ab*cosC=c^2 Where C is the angle opposite the side length c. Now we rearrange: cosC = (a^2 + b^2 - c^2)/(2ab) Solving for C we can now use: Area = 1/2 ab*cosC
Hi, the most straightforward answer to this inquiry would be by using Heron's formula which is: Area = sqrt( p ( p - a ) ( p - b ) ( p -c ) ) where a, b, and c are the lengths of the sides of the triangle, and p is 1/2 of the perimeter (i.e. (a + b + c) / 2). This formula works for any type of triangle, as long as a, b, and c are actually valid lengths to form a triangle. Another way involving trigonometry is to use the cosine law, but that will involve having to calculate the angles for the triangle.
use the cosine rule b^2 + C^2 -A^2/ 2bc to find the cosA angle. then do cos-1 to find the angle. then use sine rule= 1/2AB sinC to find the area Hope it works
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