CK-12 Geometry - Second Edition

http://www.ck12.org Chapter 6. Polygons and Quadrilaterals

Solution:Opposite sides are congruent, so we can set each pair equal to each other and solve both equations.

6 x− 7 = 2 x+ 9 y^2 + 3 = 12

4 x= 16 y^2 = 9

x= 4 y= 3 or− 3

Even thoughy=3 or -3, lengths cannot be negative, soy=3.

Diagonals in a Parallelogram

From the Parallelogram Diagonals Theorem, we know that the diagonals of a parallelogram bisect each other.

Example 3:Show that the diagonals ofF GHJbisect each other.

Solution:The easiest way to show this is to find the midpoint of each diagonal. If it is the same point, you know
they intersect at each other’s midpoint and, by definition, cuts a line in half.

Midpoint ofF H:

(

− 4 + 6

2

,

5 − 4

2

)

= ( 1 , 0. 5 )

Midpoint ofGJ:

(

3 − 1

2

,

3 − 2

2

)

= ( 1 , 0. 5 )

Example 4:Algebra ConnectionSANDis a parallelogram andSY= 4 x−11 andY N=x+10. Solve forx.