Question

Question
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A parallelogram has vertices upper A, upper B, upper C, and upper D. Opposite vertices are connected via diagonal lines that intersect at an interior point labeled upper O. The lines formed between each vertex and upper O are labeled. Upper A and Upper C are both 15 while Upper D is 18 and Upper B has the m equation 7 +m

To prove that quadrilateral ABCD is a parallelogram, what must be the value of m?

(1 point)
Responses

15
15

36
36

18
18

11
11
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Answer 1

To show that quadrilateral ABCD is a parallelogram, we can use the property that the diagonals of a parallelogram bisect each other.

From the information given:
- The distance AO = 15
- The distance OC = 15
- The distance DO = 18
- The distance BO = 7 + m

For ABCD to be a parallelogram, we need AO + OC to equal BO + OD. Since AO = OC, we can set up the equation based on the lengths of the diagonals.

The lengths can be expressed as:
- AO = OC = 15
- DO = 18
- BO = 7 + m

Setting the equation based on the bisection property:
AO + OC = BO + OD

Substituting the values gives us:
15 + 15 = (7 + m) + 18

This simplifies to:
30 = 7 + m + 18
30 = m + 25

Now we’ll isolate m:
m = 30 - 25
m = 5

So, the value of m that makes quadrilateral ABCD a parallelogram is **5**.

However, since 5 is not one of the options provided in your responses (15, 36, 18, 11), please double-check the problem statement or the provided vertices' values to ensure clarity.