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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

4. Rectangles

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Exercise 15 Page 508

The diagonals of a rectangle are congruent.

ZP=43

Practice makes perfect

Let's analyze the given rectangle to find the length ZP.

Finding the Value of x

First, we need to discover the value of x. Since every rectangle is a parallelogram, recall the following theorem.

Parallelogram Diagonals Theorem
In a parallelogram, the diagonals bisect each other.

Therefore, we can state the following equations. PY= WP and PX= ZP We can use this fact to substitute the given expressions, PY= 3x-5 and WP= 2x+11, into the first equation above. PY= WP ⇒ 3x-5= 2x+11 Let's solve it!

3x-5=2x+11

LHS+5=RHS+5

3x=2x+16

LHS-2x=RHS-2x

x=16

Finding the Value of ZP

Now that we know the value of x, we want to find an expression for ZP. Recall the following theorem.

Rectangle Diagonals Theorem
The diagonals of a rectangle are congruent.

This means that ZX and WY are congruent. We know that diagonals bisect each other because the figure is a parallelogram, therefore ZX = 2ZP, and WY = 2PY. With this in mind, let's find an expression for ZP.

Equation	ZX=WY
Substitution	2ZP= 2PY
Simplification	ZP=PY
Substitution	ZP= 3x-5

Finally, let's substitute the value of x that we have found out and simplify.

ZP=3x-5

x= 16

ZP=3( 16)-5

Multiply

ZP=48-5

Subtract term

ZP=43

Subchapter links

Exercises p.508-511