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

6. Trapezoids and Kites

Continue to next subchapter

Exercise 77 Page 530

The diagonals of a rhombus bisect each other.

9

Practice makes perfect

We want to find the MG for the given rhombus.

For all rhombi, the diagonals bisect each other at right angles. Therefore, the lengths of DM and MG are equal. Firstly let's find the length of DM, which we can calculate from Pythagorean Theorem, because we are given HM= 12 and HD= 15 and already know △ HMD is the right triangle. HD^2= DM^2+ HM^2 Let's solve the equation.

HD^2=DM^2+HM^2

HD= 15, HM= 12

15^2=DM^2+ 12^2

Calculate power

225=DM^2+144

LHS-144=RHS-144

81=DM^2

sqrt(LHS)=sqrt(RHS)

9=DM

Rearrange equation

DM=9

Finally, we can find MG. As previously mentioned, we know that the lengths of DM and MG are equal. the length of MG=9.

Subchapter links

Exercises p.526-530