How many units are in the sum of the lengths of the two longest altitudes in a triangle with sides \(8\), \(12\), and \(17\)?
+579 \(8+12=20\)
So the sum of the lengths of the two longest altitudes in that triangle is \(20\)
+128408 The semi-perimeter of this triangle = (8 + 12 + 17) /2 = 37/2 = 18.5
The area of this triangle = sqrt ( 18.5 * ( 18.5- 8) (18.5 -12) ( 18.5 - 17) ) =
sqrt [ 18.5 * 10.5 * 6.5 * 1.5 ] =
sqrt [ 1893. 93.75 ]
The longest two altitudes will be drawn to the shortest two sides
So, using the area formula for a triangle, we have
sqrt (1893.9375 ) = (1/2)(8) *altitude 1
sqrt(1893.9375) / 4 = altitude 1 (1)
And
sqrt (1893.9375 ) = (1/2)(12) * altitude 2
sqrt (1893.9375 ) / 6 = altitude 2 (2)
Adding (1) and (2) we get
sqrt (1893.9375) / 4 + sqrt(1893.9375) / 6 ≈ 18.133 units
