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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.

 Nov 12, 2019
 #1
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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.
 

 Nov 12, 2019
 #2
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Is that right?

Can you keep the radicals and simplify it from there?

 Nov 12, 2019
 #3
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+1

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

 

 

cool cool cool

 Nov 13, 2019
 #4
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Thx CPhill

 Nov 13, 2019

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