Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10:
ISBN 13:

Chapter 8 - Geometry - 8.5 Triangles - 8.5 Exercises - Page 565: 13


Area of the shaded portion = 1664 m$^{2}$

Work Step by Step

To find the area of the shaded portion we simply find the area of the entire rectangle and subtract the area of the un-shaded triangle. Area of a Rectangle: $A$$_{rect}$ = $l$$w$ Therefore: $A$$_{rect}$ = (52 m)(37 m) $A$$_{rect}$ = 1924 m$^{2}$ Area of a Triangle: $A$$_{tri}$ = $\frac{1}{2}$$b$$h$ Therefore: $A$$_{tri}$ = $\frac{1}{2}$(52 m)(10 m) $A$$_{tri}$ = 260 m$^{2}$ $A$$_{total}$ = $A$$_{rect}$ - $A$$_{tri}$ Therefore: $A$$_{total}$ = 1924 m$^{2}$ - 260 m$^{2}$ = 1664 m$^{2}$
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