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Mathematics 5 Online
OpenStudy (anonymous):

The area of the trapezoid is 75 cm2. What is the height of the trapezoid? cm Trapezoid with base one equal to nine centimeters and base two equal to sixteen centimeters

OpenStudy (calculusxy):

|dw:1456189622938:dw|

OpenStudy (calculusxy):

Formula for finding the area of trapezoid: \(\large h(\frac{a + b}{2})\)

OpenStudy (jdoe0001):

|dw:1456190617820:dw| \(\bf \textit{area of a trapezoid}=\cfrac{h}{2}(base1+base2)\)

OpenStudy (anonymous):

h=12?

OpenStudy (calculusxy):

we know that a can equal to 9 and b can equal to 16 (the placement doesn't really matter). the total area is 75cm^2. so after substituting the values to the equation, we can find out what h is equal to.

OpenStudy (calculusxy):

\[h \times \frac{9+16}{2} = 75\text{cm}^2\]

OpenStudy (anonymous):

ok so h=75cm^2

OpenStudy (anonymous):

thanks @calculusxy

jimthompson5910 (jim_thompson5910):

`ok so h=75cm^2` that's false

jimthompson5910 (jim_thompson5910):

what does \(\Large \frac{9+16}{2}\) evaluate to?

OpenStudy (anonymous):

12 1/2?

jimthompson5910 (jim_thompson5910):

or 12.5, correct

jimthompson5910 (jim_thompson5910):

so, \[\Large h \times \frac{9+16}{2} = 75\] turns into \[\Large h \times 12.5 = 75\] do you see how to isolate h from here?

OpenStudy (anonymous):

subtract 12.5 from 75. Right?

jimthompson5910 (jim_thompson5910):

you need to undo the multiplication what is the opposite of multiplication?

OpenStudy (anonymous):

so I divide 12.5 from 75?

jimthompson5910 (jim_thompson5910):

you would divide both sides by 12.5 \[\Large h \times 12.5 = 75\] \[\Large \frac{h \times 12.5}{12.5} = \frac{75}{12.5}\] \[\Large \frac{h \times \cancel{12.5}}{\cancel{12.5}} = \frac{75}{12.5}\] \[\Large h = \frac{75}{12.5} = ???\]

OpenStudy (anonymous):

=6

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

so 6 is my answer

jimthompson5910 (jim_thompson5910):

yep `6 cm` is the height

OpenStudy (anonymous):

ok thank you so much you are a lifesaver!!!

jimthompson5910 (jim_thompson5910):

you're welcome

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