Because congruent diagonals prove that a parallelogram is a rectangle, what value of x proves that parallelogram ABCD is a rectangle if AC = 48 and DO = 7x + 3?

Question

a. x = 8
b. x = 4
c. x = 6
d. x = 3

Answer 1

In a parallelogram, if the diagonals are congruent (equal in length), then it is a rectangle. Given that the length of diagonal $AC$ is 48 and the length of diagonal $DO$ is given as $7x+3$, we can set up the equation:

$$7x+3=48$$

Now, solve for $x$:

1. Subtract 3 from both sides:

$$7x=48-3$$

$$7x=45$$

2. Now, divide both sides by 7:

$$x=\frac{45}{7}$$

This results in:

$$x=6.42857\dots$$

None of the provided options $a(8),b(4),c(6),d(3)$ satisfy this equation exactly. Therefore, double-checking the problem statement and numbers is necessary because the given values do not lead to any of the provided answers being true.

If you consider reasoning simply if $7x+3$ must also equal 48 in order for the diagonals to be congruent. You might want to double-check for possible misprints or ensure the parameters match your calculation.