Elementary Geometry for College Students (5th Edition)

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

Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 373: 51

Answer

$h = 2.4$ units

Work Step by Step

1. Find the area of the larger $\triangle$ Let $A = $ area of the larger $\triangle$ $A = \frac{b \times h}{2}$ $= \frac{8 \times 6}{2}$ $= \frac{48}{2}$ $= 24$ unit$^{2}$ 2. Find the area of the smaller $\triangle$ Let $S =$ area of the smaller unshaded $\triangle$ $S = \frac{4 \times 3}{2}$ (We know that $A$ and $B$ are midpoints so therefore the lengths of both the base and height is halved) $= \frac{12}{2}$ $= 6$ unit$^{2}$ 3. Find the area of the trapezoid $=(Larger \triangle)$ $- (Smaller \triangle)$ $= 24 - 6$ $= 18$ unit$^{2}$ 4. Find the hypotenuse for both the small and large $\triangle$'s 4.1 Larger $\triangle$ $a^{2} + b^{2} = c^{2}$ $(8)^{2} + (6)^{2} = c^{2}$ $64 + 36 = c^{2}$ $100 = c^{2}$ $c = ±\sqrt {100}$ $c = 10$ units 4.2 Smaller $\triangle$ $a^{2} + b^{2} = c^{2}$ $(4)^{2} + (3)^{2} = c^{2}$ $16 + 9 = c^{2}$ $c^{2} = 25$ $c = ±\sqrt {25}$ $c = 5$ units Using the formula for a trapezium, find the height of the trapezoid Let $T =$ area of the trapezium ($a = 5$ and $b = 10$, the hypotenuse of the small and large $\triangle$'s) $T = \frac{1}{2}(a+b)h$ $18 = (0.5)(5+10)h$ $18 = (0.5)(15)h$ $18 = \frac{15h}{2}$ $36 = 15h$ $h = 2.4$ units
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