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3 years ago

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Quadrilateral ABCD is a parallelogram if its diagonals bisect each other. Prove that Quadrilateral ABCD is a parallelogram by fi

nding the value of x.
A) x = 2
B) x = 3
C) x =- 9|2
D) x = -3|4

Information that goes with the picture:
AO = 5x + 2
BO = x − 1
CO = 3x − 7
DO = 3x + 8

1 answer:

beks73 [17]3 years ago

5 0

Because this is a parallelogram, we know that the diagonals bisect. This means that AO = DO, and CO = BO

It doesn't matter which two segments we use, but I'm going to use CO = BO for this one.

3x - 7 = x - 1

2x = 6

x = 3

Option B. is the answer.

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$\bold{\huge{\orange{\underline{ Solution }}}}$

$\bold{\underline{ Given \: Rules}}$

<u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
<u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
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$\bold{\underline{ Let's \: Begin}}$

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

$\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }$

$\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}$

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

$\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}$

$\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}$

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

$\sf{ 158 + b = 166 + a, 156 + d = 166 + a }$

$\sf{ b = 166 - 158 + a, d = 166 - 156 + a }$

$\sf{ b = 8 + a, d = 10 + a ...(3)}$

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{229+a = 161+8+a+c = 168+10+a+e}$

$\sf{ 229+a = 169+a+c = 178+a+e}$

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

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$\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}$

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$\sf{ d = 33}$

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$\sf{ a = 189 - 166 }$

$\sf{ a = 23 }$

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