McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
4. Rectangles
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Exercise 50 Page 511

The diagonals of a rectangle are congruent.

A

Practice makes perfect

Let's analyze the given quadrilateral to find the variables x and y. Keep in mind, we have been told that the figure is a rectangle.

By the Segment Addition Postulate, we can express each diagonal as the sum of its smaller segments. FH= FM+MH and GJ=GM+MJ

Additionally, because a rectangle is a parallelogram, we know that the diagonals bisect each other. FM= MH and GM= MJ We can use this fact to rewrite the Segment Addition Postulate equations in terms of the given expressions, FM= 3x+y and GM= 13.

Equation FH=FM+MH GJ=GM+MJ
Substitution FH=FM+ FM GJ=GM+ GM
Simplification FH=2FM GJ=2GM
Substitution FH=2( 3x+y) GJ=2( 13)
Recall that the diagonals of a rectangle are congruent. Therefore, their lengths are equal. FH=GJ ⇔ 2(3x+y)=2(13) Because our quadrilateral is a rectangle, both opposite sides are congruent. We can use the given expressions to write one more equation. GH=FJ ⇔ 11=- 3x + 5y Let's create a system of equations using both equations above. 2(3x+y)=2(13) 11=-3x+5y To solve it we will use the Substitution Method.
2(3x+y)=2(13) & (I) 11=-3x+5y & (II)
3x+y=13 & (I) 11=-3x+5y & (II)
y=13-3x & (I) 11=-3x+5y & (II)
y=13-3x & (I) 11=-3x+5( 13-3x) & (II)
â–Ľ
(II): Solve for x
y=13-3x & (I) 11=-3x+65-15x & (II)
y=13-3x & (I) 11=-18x+65 & (II)
y=13-3x & (I) 18x+11=65 & (II)
y=13-3x & (I) 18x=54 & (II)
y=13-3x & (I) x=3 & (II)
Now that we have found x, we can substitute it in the first equation to find y.
y=13-3x & (I) x=3 & (II)
y=13-3( 3) & (I) x=3 & (II)
y=13-9 & (I) x=3 & (II)
y=4 x=3
Therefore, the correct answer is A.