What values of $a$ would make the given quadrilateral a parallelogram.

5 \frac{1}{4}

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\frac{11}{2}

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5 \frac{3}{4}

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5

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Solution

The correct option is B $\frac{11}{2}$
As the diagonal of quadrilateral bisects each other; 5b-4=3b+6
2b=10
$b = 5$
So by the same rule;3b-4=2a
15-4=2a
11=2a
$a = \frac{11}{2}$

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Similar questions

5 \frac{1}{6} - [1 \frac{1}{5} + {2 \frac{3}{4} \div 5 \frac{1}{2} \div χ - (\frac{5}{6} - \frac{2}{3})}] = 2 \frac{61}{120}

[(- 5)^{3}]^{4}

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5 \frac{1}{6} - [1 \frac{1}{5} + {2 \frac{3}{4} \div 5 \frac{1}{2} \div x - (\frac{5}{6} - \frac{2}{3})}] = 2 \frac{61}{120}

A (- 2, 5), B (4, - 1), C (9, 1), D (3, 7)

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