Question

discussion 6: find area
find the area of the trapezoid in the figure. explain and show work. simplify the answer.
hint: since this graph doesnt claim its drawn to - scale, one should assume its not to - scale. inspect carefully to verify first that there is no right angle in this trapezoid.
use the insert math equation tool (√x icon) to typeset math equations and expressions.
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Explanation:

Step1: Simplify the radicals

Simplify \(2\sqrt{63}\), \(2\sqrt{27}\) and \(6\sqrt{3}\), \(7\sqrt{7}\)

• \(2\sqrt{63}=2\sqrt{9\times7}=2\times3\sqrt{7} = 6\sqrt{7}\)
• \(2\sqrt{27}=2\sqrt{9\times3}=2\times3\sqrt{3}=6\sqrt{3}\)

Step2: Use the area formula for a trapezoid

The area formula for a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the lengths of the parallel sides and \(h\) is the height.
Here, \(a = 6\sqrt{7}\), \(b=7\sqrt{7}\), \(h = 6\sqrt{3}\)

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Step3: Calculate the value

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Answer:

\(39\sqrt{21}\text{ m}^2\)