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Nesterboy [21]
3 years ago
8

ABDC is a trapezoid. Line AB = 2.5, and Line EF = 4.5. What is the length of Line CD ?

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0
The line CD is the median of thre trapezoid, therefore its measures is shown below:
 You must sum the largest base and the shortest base and divide both of them by 2, as following:
 CD=(4.5+2.5)/2
 CD=7/2
 CD=3.5
 Therefore, as you can see, the measure of the line CD is 3.5
 The answer is: 3.5
mezya [45]3 years ago
5 0
The correct answer is 3.5

A trapezoid is a 4-sided flat shape with straight sides. This shape has a pair of opposite sides parallel. So according to the figure we know the following:

\overline{AC}=\overline{CE}

And it is also true that:

\overline{BD}=\overline{EF}

We know that the lengths of the parallel sides are given by:

\overline{AB}=2.5 \\ \\ \overline{EF}=4.5

So the goal is to find the length of Line CD. This line is in fact the <em>median</em> of the trapezoid <span>(also called a midline or midsegment) that is a line segment half-way between the two bases and it can be found as follows:
</span>
m=\frac{\overline{AB}+\overline{EF}}{2}=\frac{2.5+4.5}{2}=\frac{7}{2} \\ \\ \therefore \boxed{\overline{CD}=m=3.5}
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<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

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Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]

Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

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