Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 5 - Section 5.5 - Special Right Triangles - Exercises - Page 259: 34

Answer

28

Work Step by Step

Since all trapezoids add up to 360 degrees, the angles of the base of the trapezoid are: $ \frac{360 - 240}{2} = 60 ^{\circ}$ This means that the red arc is given by: $\frac{60}{2} = 30^{\circ}$ This means that triangle MQT is isosceles, so MT is equal to MQ. We form a 30-60-90 right triangle that divides QTP into two equal triangles. We call h the height of the trapezoid. From this, we obtain: $h = .5\times 12 \times \frac{\sqrt{3}}{3} \\ h = 2\sqrt{3}$ Thus, we find MQ: $ MQ = \frac{2}{\sqrt3} \times h \\ MQ = 4$ Since it is isosceles, we obtain that NP equals 4. In addition, we obtained that MT, which is half of MN, is equal to MQ above. Thus. MT equals 8. Therefore: perimeter = $ 4 +4+8+12 = 28$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.