In triangle ABC, we know the side lengths are AB=9sqrt2, BC=10sqrt2, and CA=11sqrt2. Find the height of triangle ABC from A to BC.
Acute scalene triangle.
Sides: a = 12.728 b = 14.142 c = 15.556
Area: T = 84.852
Perimeter: p = 42.426
Semiperimeter: s = 21.213
Angle ∠ A = α = 50.48° = 50°28'48″ = 0.881 rad
Angle ∠ B = β = 58.993° = 58°59'34″ = 1.03 rad
Angle ∠ C = γ = 70.527° = 70°31'38″ = 1.231 rad
Height: hA = 13.333
Height: hB = 12
Height: hC = 10.909
There are many ways to find the height of the triangle. The easiest way is from the area and base length. The area of a triangle is half of the product of the length of the base and the height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.
AB = 9√2
BC = 10√2
CA = 11√2
The semiperimeter of the triangle = 30√2 / 2 = 15√2
Using Heron's formula.....the area A is given by
√[ 15√2 (15√2 - 9√2) (15√2 - 10√2) ( 15√2 - 11√2) ] =
√[ 15√2 (6√2) ( 5√2) (4√2) ] =
√ [1800 * 4] =
√(3600 * 2 ) =
60√2
Since BC = 10√2....the altitude from A can be found as
60√2 = (1/2) (10√2) (Altitude)
60 = (1/2)(10) (Altitude )
60 = 5 (Altitude)
12 = Altitude from A to BC