Question

In triangle ABC, AC= BC. The length of AC is three times the length of AB. Find the lengths of the longest side of the triangle if the perimeter of the triangle is 35 inches.

a) AB = 5 inches, AC = 15 inches, BC = 15 inches
b) AB = 7 inches, AC = 21 inches, BC = 7 inches
c) AB = 6 inches, AC = 12 inches, BC = 12 inches
d) AB = 8 inches, AC = 16 inches, BC = 16 inches

Answer 1

In an isosceles triangle where two sides are equal and one side is three times a third side, with a perimeter of 35 inches, the shortest side is 5 inches and the other two sides are 15 inches each, making them the longest sides of the triangle.

Step-by-step explanation:

In this problem, we are given that in triangle ABC, AC equals BC which suggests that it is an isosceles triangle with AC and BC as its equal sides. Since we know that the length of AC is three times the length of AB and the perimeter is 35 inches, we can set up the following equation for the perimeter:

AB + AC + BC = 35 inches

Given AC = 3 × AB, and AC = BC, we can substitute in:

AB + 3 × AB + 3 × AB = 35 inches

Combining like terms, we get :

7 × AB = 35 inches

Dividing both sides by 7 yields:

AB = 5 inches

Therefore, AC = BC = 3 × AB = 3 × 5 inches = 15 inches.

The longest side of the triangle is AC (or BC), which is 15 inches. So the correct answer is option (a): AB = 5 inches, AC = 15 inches, BC = 15 inches.