Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 253: 28

Answer

$MB=\sqrt 73$

Work Step by Step

Using the Pythagorean Theorem we can find the length of $AC$. $a^{2}+b^{2}=c^{2}$ Remember, the hypotenuse is always the longest side, or the side opposite the right angle, in this case $AB$. $AC^{2}+CB^{2}=AB^{2}$ $a^{2}+ 8^{2}=10^{2}$ Simplify... $a^{2}+64=100$ Subtract 64 from both sides... $a^{2}=36$ Square root both sides... $a=6$ OR $AC=6$ Since $M$ is the midpoint of $AC$, $AM=3$ and $MC=3$ Using the Pythagorean Theorem (again) we can find the length of $MB$. $MC^{2}+CB^{2}=MB^{2}$ $3^{2}+ 8^{2}=c^{2}$ Simplify... $9+64=c^{2}$ Combine like terms... $73=c^{2}$ Square root both sides... $c=\sqrt 73$ OR $MB=\sqrt 73$
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