Two isosceles triangles have perimeters of 8 and 15 centimeters, respectively. They have the same base length, but the legs (the non base sides) of the larger triangle are twice as long as the legs of the smaller triangle. For each triangle, what is the length of a leg?
small: 8 = b+2x
large: 15 = b+2y
y=2x
b+y=8
b+2y=15
y=7
b=1
x=7/2
small: 7/2 + 7/2 + 1 = 8
large: 7+7+1=15
To find the length of the legs of each triangle, let's assign variables. Let's call the length of the base of each triangle "b" and the length of the legs of the smaller triangle "x".
In the first triangle, since the perimeter is 8, we have:
Perimeter of first triangle = 2*x + b = 8
In the second triangle, since the perimeter is 15 and the legs are twice as long as the legs of the smaller triangle, we have:
Perimeter of second triangle = 2*(2x) + b = 15
Now we have a system of two equations with two variables. We can solve this system to find the values of "x" and "b".
Let's start with the first equation:
2x + b = 8
Next, let's solve the second equation for "b":
2*(2x) + b = 15
4x + b = 15
b = 15 - 4x
Now substitute the value of "b" from the second equation into the first equation:
2x + (15 - 4x) = 8
2x + 15 - 4x = 8
-2x + 15 = 8
-2x = 8 - 15
-2x = -7
x = -7/(-2)
x = 7/2
Therefore, the length of a leg of the smaller triangle is 7/2 cm.
To find the length of the legs of the larger triangle, we substitute the value of "x" back into the equation for "b":
b = 15 - 4x
b = 15 - 4 * (7/2)
b = 15 - 28/2
b = 15 - 14
b = 1
Therefore, the length of a leg of the larger triangle is 1 cm.