the length of the base of an isosceles triangle is one third the length of one of its legs. If the perimetet of the triangle is 66 inches, what is the length of the base
2006-10-20 13:04:32 · 5 answers · asked by imo 1 in Science & Mathematics ➔ Mathematics
Let length of base be b"
Thus length of legs are each 3b"
So Perimeter = b + 3b + 3b
= 7b
So 7b = 66
b = 9 3/7
ie base is 9 3/7 inches long
2006-10-20 13:09:14 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
Let B be the length of the base, and L be the length of the legs. Then B+2L=66, and B=L/3, or L=3B. Therefore B+2(3B)=66, or 7B=66. So B=66/7 and L=198/7.
2006-10-20 20:11:36 · answer #2 · answered by James L 5 · 0⤊ 1⤋
an isosceles triangle has 2 sides that are equal. Let x be the length of the base.
Then
x + 3x + 3x = 66
7x = 66
x = 66/7 = 9.4
2006-10-20 20:08:25 · answer #3 · answered by animal 2 · 0⤊ 0⤋
Let the length of one of the equal sides be L. Then the perimeter can be stated as L + L + L/3
2L + L/3 = 66
6L + L = 198
7L = 198
L = 28.29
L/3 = 9.43
Approximately
2006-10-20 20:12:31 · answer #4 · answered by kindricko 7 · 0⤊ 0⤋
you can't divided by 3 because is an isosceles triangle and has different sides
2006-10-20 20:12:19 · answer #5 · answered by coolasty 2 · 0⤊ 0⤋