Each of the two congruent sides of an isosceles triangle are 10 centimeters more than 1/2 the length of the base. If the perimeter of the triangle is to be at most 100 centimeters, what is the maximum length of the base?
2 [ (b/2)+10 ] + b </= 100
b + 20 + b </= 100
2 b </= 80
b </= 40
so 40 cm
To find the maximum length of the base of the isosceles triangle, we need to set up an equation based on the given information.
Let's assume that the length of the base is 'x' centimeters.
According to the given conditions, each of the two congruent sides is 10 centimeters more than (1/2)x.
Therefore, the length of each congruent side is (1/2)x + 10 centimeters.
The perimeter of a triangle is found by adding the lengths of all three sides.
In this case, the perimeter is at most 100 centimeters, so we can write the following equation:
x + (1/2)x + 10 + (1/2)x + 10 ≤ 100
Now, let's solve this equation to find the maximum length of the base:
Combining like terms, we get:
2x + 20 ≤ 100
Subtracting 20 from both sides:
2x ≤ 80
Dividing both sides by 2:
x ≤ 40
Therefore, the maximum length of the base of the isosceles triangle is 40 centimeters.