body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

3. Tests for Parallelograms

Continue to next subchapter

Exercise 52 Page 503

Use the Polygon Exterior Angles Sum Theorem.

x=55

Practice makes perfect

We were asked to find the value of x in the given polygon.

Recall the Polygon Exterior Angles Sum Theorem.

Polygon Exterior Angles Sum Theorem
The sum of the measures of the exterior angles of a polygon, one at each vertex, is 360.

We can write an equation where the sum of the expressions is equal to 360. (x+3)+48+2x+82+(x+7)= 360 Let's solve the above equation to find x.

(x+3)+48+2x+82+(x+7)=360

▼

Solve for x

Remove parentheses

x+3+48+2x+82+x+7=360

CommutativePropAdd

Commutative Property of Addition

2x+x+x+82+48+7+3=360

Add terms

4x+140=360

LHS-140=RHS-140

4x=220

.LHS /4.=.RHS /4.

x=55

Extra

Measure of the Angles

Now that we know the value of x, we can find the values of the exterior angles of the pentagon.

Measure of exterior angle	x=55	Simplified
x+3	55+3	58
2x	2( 55)	110
x+7	55+7	62

Let's complete our diagram with the obtained information.

Subchapter links

Exercises p.499-503