Simplify: -20w³ + 11w³ Enter the original expression if it cannot be simplified.
1 answer:
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Answer:
174 units²
Step-by-step explanation:
There are a few different ways you can find the altitude of the trapezoid. Consider the attached figure with some points and lines added. SQ' is parallel to TQ, so Q'Q = 4 and PQ' = 21. Based on the side length of PS = 13, you can <u>guess</u> that the height is 12. (5-12-13 is a commonly-used Pythagorean triple.) This would make PP' = 5, P'Q' = 16, and triangle SP'Q' have side lengths 12, 16, and 20, corresponding to a 3-4-5 right triangle multiplied by 4.
Another way to find the height is to use Heron's formula for the area of triangle PSQ'. The side lengths are 13, 20, 21, so the half-perimeter is 27 and the area is √(27(27-13)(27-20)(27-21)) = √(9²·14²) = 126. The base of the triangle, PQ', is 21, so the height is ...
... h = 2A/b = 2·126/21 = 12
The area of parallelogram Q'STQ is then ...
... A = bh = 4·12 = 48
and the total area is the triangle area plus the parallelogram area:
trapezoid area = 126 + 48 = 174 . . . . units²
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Of course, with the height known, the usual formula for the area of a trapezoid can be used:
A = (1/2)(b1 +b2)h
A = (1/2)(25 +4)·12 = 29·6 = 174 . . . . units²
