Find areaof triangle having three sides as 4m6m and 8m?

Answer 1

Solution 1: Heron's Forumla

Obviously, by the Pythagorean Theorem (42 + 62 =? 82) is false, (16 + 36 = 52), so we have to apply a concept known as "Heron's Formula".

The semiperimeter of this triangle is (4 + 6 + 8)/2 = 2 + 3 + 4 = 9, and we take the square root of the corresponding expression:

9(9-4)(9-6)(9-8) = 1(3)(5)(9)

sqrt(15(9)) = 3 sqrt(15) m.

The proof of Heron's Formula, while brute forced and long (a bit too long to show here), is at the site I linked below.

The formula itself is sqrt(s(s-a)(s-b)(s-c)), where a,b,c are side lengths of the triangle, s is the semiperimeter (perimeter divided by 2).

Solution 2: Brute Algebra

Let this triangle be ABC. Let BC = 8m, BA = 4m, and AC = 6m. Produce point D on BC, such that m

16 - x2 = 36 - (64 - 16x + x2)

16 - x2 = 36 - 64 + 16x - x2 = -x2 + 16x - 28

We see 2 same degree terms, so we add x2 on each side, getting us to the simple linear equation:

44 = 16x

Thus, x = 11/4. By Pythagorean theorem, sqrt(16 - 121/16) = 3 sqrt(15)/4. Therefore, by the area formula, A = [ABC] = 8(3 sqrt(15)/4)/2 = 3 sqrt(15).

Solution 3: Angles

Using the triangle model from Solution 2, we have 1/2 BA(AC) sin C, where C is the angle between BA and AC. By the Laws of Cosines, we have:

64 = 16 + 36 - 48 cos C

12 = -48 cos C

-1/4 = cos C

The simplification of arccos(-1/4) is nasty, but sin(arccos(-1/4)) = sqrt(15)/4.

So, 4(6) * sqrt(15)/3 * 1/2 = 3 sqrt(15).

There are more solutions, involving immense trigonometry, incenters, orthocenters, etc., but these are the 3 of them that can be highlighted easily.