A triangle has a base 17 units long and the lengths of the other two sides are equal. If the side lenghts are integers, what is the shortest possible side length?
2007-06-27 04:24:35 · 8 answers · asked by Mo 1 in Science & Mathematics ➔ Mathematics
let the triangle be ABC with the
base BC = 17 unit
draw a perpendicular from A to BC.
Let AH be the height of this triangle
Now, look at the right triangle AHC (angle AHC = 90 degrees)
HC = BC/2 = 8.5
AC*AC = AH*AH + HC*HC
AC^2 = AH*AH + (8.5 * 8.5)
AC^2 = AH*AH + 72.25
we now have
8*8 = 64
9*9 = 81
81 is closest to 72.25
AC^2 = 81
so, AC = 9
2007-06-27 04:53:39 · answer #1 · answered by buoisang 4 · 0⤊ 0⤋
In every triangle, each side is less than the sum of the other 2. If x is the length of the 2 equal sides of your isosceles trianle, then we must have x + x = 2x > 17 => x> 8.5. Since the smallest integer greater than 8.5 is 9, it follows the shortest possible length for the 2 equal sides is 9.
2007-06-28 11:18:38 · answer #2 · answered by Steiner 7 · 0⤊ 1⤋
The length of the other sides must obey the Triangle Inequality:
s+s>=17.
The smallest integer s which satisfies this inequality is 9.
2007-06-27 11:28:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The short sides would have to be 9 units. A length of 8 units would be too short.
2007-06-27 11:27:44 · answer #4 · answered by tastywheat 4 · 0⤊ 0⤋
Let it be x.
Now sum of two sides of a triangle is greater than the third side.
So, 2x > 17 which is possible for the least integer value of x = 9.
2007-06-27 11:52:01 · answer #5 · answered by Madhukar 7 · 0⤊ 0⤋
x + x > 17
2x > 17
x > 17/2 or 9
2007-06-27 11:27:49 · answer #6 · answered by gfulton57 4 · 0⤊ 0⤋
9 units. - (The two sides combined must be longer than the base.)
2007-06-27 11:28:18 · answer #7 · answered by Irv S 7 · 0⤊ 0⤋
2a > 17
a > 17/2
a > 8.5
a = 9
2007-06-27 11:27:53 · answer #8 · answered by yeeeehaw 5 · 0⤊ 0⤋