183k views
4 votes
Given:Parallelogram QRST where QR=6x-5 and RS= 4x+13. What is the value of x proves this parallelogram is a rhombus?

User Lithilion
by
8.0k points

1 Answer

3 votes

Final answer:

To determine if the parallelogram QRST is a rhombus, we need to consider its properties. A rhombus is a special type of parallelogram in which all sides are congruent. Given that QR = 6x - 5 and RS = 4x + 13, we can set up an equation and solve for x to find the value that makes the sides equal. With x = 9, we substitute it back into the lengths of the sides and find that QR = RS = 49. Therefore, the parallelogram QRST is a rhombus.

Step-by-step explanation:

To determine if the parallelogram QRST is a rhombus, we need to consider its properties. A rhombus is a special type of parallelogram in which all sides are congruent. In other words, all the sides of a rhombus have the same length.

Given that QR = 6x - 5 and RS = 4x + 13, we can set up an equation: QR = RS

6x - 5 = 4x + 13

Solving this equation, we find x = 9.

Now, let's substitute the value of x back into the lengths of the sides: QR = 6(9) - 5 = 49 and RS = 4(9) + 13 = 49

Since QR = RS = 49, we can conclude that the parallelogram QRST is a rhombus.