Question

6 this shows a diagram and a proof.
u
r
f
s
given: parallelogram ursf
with \\(\overline{us} \cong \overline{rf}\\)
prove: \\(\triangle ufs \cong \triangle rsf\\)

1. parallelogram ursf, with \\(\overline{us} \cong \overline{rf}\\) 1. given
2. parallelogram usrf is a rectangle. 2. ?
3. \\(\angle ufs\\) and \\(\angle rsf\\) are right angles. 3. definition of rectangle
4. \\(\triangle ufs\\) and \\(\triangle rsf\\) are right triangles. 4. definition of a right triangle
5. \\(\overline{fs} \cong \overline{fs}\\) 5. reflexive property of congruence
6. \\(\triangle ufs \cong \triangle rsf\\) 6. hl (hypotenuse-leg) triangle congruence

what reason can be used to justify step 2 in this proof?

a. if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.
b. if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rectangle.
c. if the opposite sides of a parallelogram are congruent, then the parallelogram is a rectangle.
d. if the opposite angles of a parallelogram are congruent, then the parallelogram is a rectangle.

7 this diagram shows parallelogram tuvw.
(diagram: parallelogram tuvw with diagonals intersecting at o, to = 3x + 6, ov = 4x + 1)
what is tv?
a. 5
b. 7
c. 21
d. 42

Explanation:

Question 6:

To justify step 2 (stating parallelogram URSF is a rectangle), we analyze the options:

• Option A: A parallelogram with congruent diagonals is a rectangle. Given \(\overline{US} \cong \overline{RF}\) (diagonals), this matches.
• Option B: Perpendicular diagonals make a rhombus, not a rectangle. Eliminate.
• Option C: Opposite sides of a parallelogram are always congruent (defining property), not a reason for being a rectangle. Eliminate.
• Option D: Opposite angles of a parallelogram are always congruent (defining property), not a reason for being a rectangle. Eliminate.

So the correct reason is Option A.

Step1: Recall parallelogram diagonal property

In a parallelogram, diagonals bisect each other. So \(TO = OU\).

Step2: Set up equation

Given \(TO = 3x + 6\) and \(OU = 4x + 1\), set \(3x + 6 = 4x + 1\).

Step3: Solve for \(x\)

Subtract \(3x\) and \(1\) from both sides: \(6 - 1 = 4x - 3x\) → \(x = 5\).

Step4: Find \(TO\) and \(OU\)

Substitute \(x = 5\) into \(TO\): \(3(5) + 6 = 15 + 6 = 21\). \(OU = 4(5) + 1 = 21\).

Step5: Find \(TV\)

Since \(TV = TO + OU\) (diagonals bisect, so \(TV\) is sum of \(TO\) and \(OU\)), \(TV = 21 + 21 = 42\).

Answer:

A. If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.

Question 7: