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if GJ = 11a and IJ = a + 10, find the value of a that makes quadrilateral FGHI a parallelogram​

Sagot :

Note: Consider J is the intersection point of diagonals of quadrilateral FGHI.

Given:

[tex]GJ=11a[/tex]

[tex]IJ=a+10[/tex]

To find:

The value of a that makes quadrilateral FGHI a parallelogram​.

Solution:

We know that the diagonals of a parallelogram bisect each other.

If FGHI is a parallelogram​ and J is the intersection point of diagonals, then

[tex]GJ=IJ[/tex]

[tex]11a=a+10[/tex]

[tex]11a-a=10[/tex]

[tex]10a=10[/tex]

Divide both sides by 10.

[tex]a=\dfrac{10}{10}[/tex]

[tex]a=1[/tex]

Therefore, the required value of a is 1 units.

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