Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 102: 33

Answer

Quadrilateral ABCD is a trapezoid Area of quadrilateral ABCD = 9

Work Step by Step

ABCD is a trapezoid because it has two bases that are not congruent to each other and 2 sides that are congruent Area of trapezoid = $ \frac{a+b}{2} \times height $ a and b are the two bases ^ base AB = $D =\sqrt (x_{2}-x_{1})^2 + (y_{2}-y_{1})^2$ $D =\sqrt (1-5)^2 + (0-0)^2$ $D =\sqrt (-4)^2 + (0)^2$ $D =\sqrt (16 + 0)$ $D =\sqrt 16$ $D =4$ therefore segment $AB =4$ base DC = $D =\sqrt (x_{2}-x_{1})^2 + (y_{2}-y_{1})^2$ $D =\sqrt (4-2)^2 + (3-3)^2$ $D =\sqrt (2)^2 + (0)^2$ $D =\sqrt (4 + 0)$ $D =\sqrt 4$ $D =2$ therefore segment DC = 2 to find the height you do find the distance from b to point (4,0) and the distance from B to C or the distance from A to D and the distance from point (2,3) to point A where both final 2 distances are equal to each other. CB = $D =\sqrt (x_{2}-x_{1})^2 + (y_{2}-y_{1})^2$ $D =\sqrt (5-4)^2 + (0-3)^2$ $D =\sqrt (1)^2 + (-3)^2$ $D =\sqrt (1+9)$ $D =\sqrt 10$ therefore segment CB = $\sqrt 10$ $height^2 +1^2 = \sqrt 10^2$ $height^2 +1 = 10$ $height^2 = 9$ height = 3 Area of trapezoid = $ \frac{a+b}{2} \times height $ Area of trapezoid = $ \frac{4+2}{2} \times 3 $ Area of trapezoid = $ \frac{6}{2} \times 3 $ Area of trapezoid = $ 3 \times 3 $ Area of trapezoid = 9
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.