Chapter 5.5, Problem 16E

Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

In Exercises 15 to 19, create drawings as needed. Given: Right ∆ M N P with M P = P N and M N = 10 2 Find: P M and P N

To determine

To find:

PM and PN for a right MNP with MP=PN and MN=102.

Explanation

Approach:

For a right triangle for which the measure of the interior angles 45Â°, 45Â°, and 90Â°, if â€˜aâ€™ is the length of measure of one of the leg; opposite to the angle 45Â°, then the length of the other two sides is given by

Length of the other leg =a

Length of the hypotenuse =a2

In general

Length of the legs are equal.

Length of the hypotenuse =2Ã— (Length of one of the legs)

Calculation:

Given,

A right âˆ†MNP with MP=PN and MN=102.

Since, two of the sides of the right triangle are congruent; the angles opposite to the sides must be equal and should be 45Â° each.

Thus, the âˆ†MNP become,

Here, MN is the hypotenuse of the âˆ†MNP.

By the application of 45Â°-45Â°-90Â°.

45Â°-45Â°-90Â° theorem.

In a right triangle whose angle measures 45Â°, 45Â°, and 90Â°, the legs are congruent and the hypotenuse has a length equal to the product of 2 and the length of either leg

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