Ix An isosceles triangle has a perimeter of 36 inches. Its base is 2(1)/(2) times the length of each of its congruent sides. _et x= the length of one of the congruent sides of the isosceles triangle. Let y= the length of the base of the triangle. Which two equations can be used to represent this situation? What are the lengths of the sides of the triangle? (1)/(2)xy=36 4 in. 4 in ,10 in. y=(5)/(2)x 6 in., 15 in ,15 in. x=(5)/(2)y 6 in., 6 in.,24 in. x+y=36 8 in., 8 in., 20 in. 2x+y=36 8 in., 20 in.,20 in. 9 in., 9 in.,18 in. 2y+x=36

To solve this problem, we need to find the equations that represent the given situation and then use them to find the lengths of the sides ofosceles triangle.<br /><br />Given information:<br />- The perimeter of the isosceles triangle is 36 inches.<br />- The base of the triangle is $2\frac {1}{2}$ times the length of each of its congruent sides.<br /><br />Let's define the variables:<br />- $x$ = the length of one of the congruent sides of the isosceles triangle.<br />- $y$ = the length of the base of the triangle.<br /><br />The two equations that can be used to represent this situation are:<br />1. $y = \frac{5}{2}x$<br />2. $x + y = 36$<br /><br />Now, let's solve these equations to find the lengths of the sides of the triangle.<br /><br />Step 1: Substitute the value of $y$ from equation 1 into equation 2.<br />$x + \frac{5}{2}x = 36$<br />$\frac{7}{2}x = 36$<br />$x = \frac{36 \times 2}{7}$<br />$x = \frac{72}{7}$<br />$x \approx 10.29$ inches<br /><br />Step 2: Substitute the value of $x$ into equation 1 to find $y$.<br />$y = \frac{5}{2} \times 10.29$<br />$y \approx 25.73$ inches<br /><br />Therefore, the lengths of the sides of the isosceles triangle are approximately 10.29 inches, 10.29 inches, and 25.73 inches.<br /><br />The correct answer is:<br />$y = \frac{5}{2}x$<br />$x + y = 36$<br />6 in., 15 in., 15 in.