Here are all the possible triangles with integer side lengths and perimeter = 12, which means s = 12/2 = 6. Find the areas using Heron's formula, simplifying the square roots.
a) 6, 5, 1
b) 6, 4, 2
c) 6, 3, 3
d) 5, 5, 2
e) 5, 4, 3
f) 4, 4, 4
Answers in the comments.
a) 6, 5, 1
ReplyDeleteArea = sqrt(6(6-6)(6-5)(6-1)
= sqrt(6 * 0 * 1 * 5)
= 0
b) 6, 4, 2
Area = sqrt(6(6-6)(6-4)(6-2)
= sqrt(6 * 0 * 2 * 4)
= 0
c) 6, 3, 3
Area = sqrt(6(6-6)(6-3)(6-3)
= sqrt(6 * 0 * 3 * 3)
= 0
The first three are line segments, not proper triangles, so the have no area.
d) 5, 5, 2
Area = sqrt(6(6-5)(6-5)(6-2)
= sqrt(6 * 1 * 1 * 4)
= sqrt(24) = 2sqrt(6) square units
e) 5, 4, 3
Area = sqrt(6(6-5)(6-4)(6-3)
= sqrt(6 * 1 * 2 * 3)
= sqrt(36) = 6 square units
The 3-4-5 triangle is a right triangle, and right triangles with integer sides always have integer areas.
f) 4, 4, 4
Area = sqrt(6(6-4)(6-4)(6-4)
= sqrt(6 * 2 * 2 * 2)
= sqrt(48) = 4sqrt(3) square units
For any given perimeter, the way to make the largest triangle is to make the equilateral triangle. If perimeter is not a multiple of 3, the sides won't be whole numbers.
sides 6,7,8.. 21 odd perimeter how i does Jsqrtk/4. Do the semi 21 need diveded in by two. 10.5 * 10.5-6*10.5-7*10.5-8.
ReplyDeleteThink of it in fractions instead of decimals to simplify the square root.
Deletesqrt(21/2*(21/2-12/2)*(21/2-14/2)*(21/2-16/2)) =
sqrt(21*9*7*5)/sqrt(2*2*2*2) =
sqrt(7*3*3*3*7*5)/4 =
21*sqrt(15)/4