This isosceles triangle has two sides of equal length, a, that are longer than the length of the base, b. The perimeter of the triangle is 15.7 centimeters. The equation 2ab = 15.7 models this information. If one of the longer sides is 6.3 centimeters, which equation can be used to find the length of the base?
The Correct Answer and Explanation is:
Correct Answer:
The given equation is:2a+b=15.72a + b = 15.72a+b=15.7
You are told that one of the longer sides is 6.3 cm, which means:a=6.3a = 6.3a=6.3
Substitute that into the equation:2(6.3)+b=15.72(6.3) + b = 15.72(6.3)+b=15.712.6+b=15.712.6 + b = 15.712.6+b=15.7
To find the base bbb, subtract 12.6 from both sides:b=15.7−12.6b = 15.7 – 12.6b=15.7−12.6b=3.1b = 3.1b=3.1
So, the equation that can be used to find the length of the base is:12.6+b=15.712.6 + b = 15.712.6+b=15.7
Explanation
This problem involves an isosceles triangle. In an isosceles triangle, two sides have equal lengths while the third side, known as the base, has a different length. You are told that the triangle has two sides of length aaa that are longer than the base bbb, and that the total perimeter of the triangle is 15.7 centimeters.
The perimeter of any triangle is found by adding the lengths of all three sides. In this case, the perimeter is the sum of the two equal sides and the base. That gives you the expression:2a+b=15.72a + b = 15.72a+b=15.7
This equation represents the total length of the sides combined. You are then given that the length of one of the equal sides is 6.3 centimeters, so you can substitute a=6.3a = 6.3a=6.3 into the equation. That gives:2(6.3)+b=15.72(6.3) + b = 15.72(6.3)+b=15.7
Next, simplify the left side. Two times 6.3 is 12.6, so now the equation is:12.6+b=15.712.6 + b = 15.712.6+b=15.7
This equation can be used to find the length of the base. To solve for bbb, you subtract 12.6 from 15.7. That results in:b=3.1b = 3.1b=3.1
The important part of this process is recognizing the formula for the perimeter of an isosceles triangle and understanding how substitution works. The original equation, once simplified with the known value of one side, directly leads to a solvable equation for the base.
