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# find the area of the trapezoid by composing into

## Tags: Mathematik

15:42 Uhr, 13.05.2022

Tag!
dear tutors and supporters:
I am attaching the problemm.

I draw a line and I can see I have a rectangle and a triangle now.
is this a good way to find the area of this trapezoid?
Danke im Voraus.

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16:00 Uhr, 13.05.2022

I am going to post some work here and see if it is good
I am gonna find the area of the triangle first.
I have two sides
the hypotenuse and a side
hypothenuse= $10$ cm
$a\phantom{\rule{0.12em}{0ex}}=7$
$b\phantom{\rule{0.12em}{0ex}}=$?

${c\phantom{\rule{0.12em}{0ex}}}^{2}={a\phantom{\rule{0.12em}{0ex}}}^{2}+{b\phantom{\rule{0.12em}{0ex}}}^{2}$
${10}^{2}={7}^{2}+{b\phantom{\rule{0.12em}{0ex}}}^{2}$

$100=49+{b\phantom{\rule{0.12em}{0ex}}}^{2}$

$100-49=49-49+{b\phantom{\rule{0.12em}{0ex}}}^{2}$

$51={b\phantom{\rule{0.12em}{0ex}}}^{2}$
√51=√b
$b\phantom{\rule{0.12em}{0ex}}=7.14$

this is the area of the triangle
now I will find the area of the rectangle and add the two together. is this right?

16:22 Uhr, 13.05.2022

Decomposing the trapezoid into a rectangle and a triangle, as you did, is certainly a good way to determine the total area.
Unfortunately, the creator of the problem has made a serious mistake, because he has given too many dimensions. The side of your triangle, which you called $b\phantom{\rule{0.12em}{0ex}},$ must be $6c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$ (difference of the lengths $18c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$ and $12c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}\right)$. From this we can calculate the length of the hypotenuse with $\sqrt{85}c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}\approx 9.22c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$ (and not $10$ cm).

So depending on which of the four dimensions given you omit in your calculation you will arrive at a different total area.
I guess its best to omit the $10$ cm which leads to a total area of $105c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$.

16:28 Uhr, 13.05.2022

The picture has a mistake
if the lower side is 12cm and the upper side 18cm the difference, your side $b\phantom{\rule{0.12em}{0ex}}$ should be 6cm but your calculation with Pythagoras for $b\phantom{\rule{0.12em}{0ex}}$ is also right .
having $b\phantom{\rule{0.12em}{0ex}}$ the area is not $b\phantom{\rule{0.12em}{0ex}}$ like you suggest but $\frac{a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}}{2}$ so if the 18cm are right it would be 6*7cm^2 and then the rectangle with Pythagoras you find 7*7,14cm^2
so either the 18cm or the $10$ cm are wrong,
calculating with the formula for trapezoids you get the area with b=6cm
ledum

16:51 Uhr, 13.05.2022

but I thought the hypothenuse side $c\phantom{\rule{0.12em}{0ex}}$ =10cm. why is it not the hypothenuse? I'm in doubt here.

17:05 Uhr, 13.05.2022

The side you call $c\phantom{\rule{0.12em}{0ex}}$ IS the hypotenuse, but if you rely on the given measures $18c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}},7c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$ and $12c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}},$ its length can't be $10c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$.
So EITHER of the four given length $7c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}},10c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}},12c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$ or $18c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}$ must be omitted and it seems that its best to omit the $10$ cm.

17:17 Uhr, 13.05.2022

Verständigungs-Stütze:

17:22 Uhr, 13.05.2022

I get it now. I solved for $c\phantom{\rule{0.12em}{0ex}}$ and got $9.22$ cm like you.

okay, now $i\phantom{\rule{0.12em}{0ex}}$ have $c\phantom{\rule{0.12em}{0ex}}=9.22$ cm
$a\phantom{\rule{0.12em}{0ex}}=7$ cm
$b\phantom{\rule{0.12em}{0ex}}=6$ cm

I am adding all this up to find the area
$9.22+7+6$

Area of the triangle $=22.22$ cm^2

area of the rectangle
$A\phantom{\rule{0.12em}{0ex}}=L\phantom{\rule{0.12em}{0ex}}\cdot W\phantom{\rule{0.12em}{0ex}}$
$=7\cdot 12$
$=84$ cm^2

now, $22.22+84$ (cm^2)

Area of the whole trapezoid $=106.22$ cm^2

is this correct?

thanks ledum as well. If I Did not reply directly to you was becasue your comment was similar to Roman

17:25 Uhr, 13.05.2022

$>$ Area of the triangle $=22.22$ cm^2
No!
Forget about the side $c\phantom{\rule{0.12em}{0ex}}!$ You don't need it. Look at the sketch I attached at the end of my last post.
Since you have given both cathets, you can simply calculate the area with $\frac{6c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}\cdot 7c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}}{2}=21c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$

BTW, another way to get the total area of the trapezoid would be to duplicate the whole trapezoid and compose a rectangle from the two pieces with the side lenghts 7 cm and $30c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}\left(=12c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}+18c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}\right)$.
You sure can calculate the area of this rectangular and divide it by 2 ;-)

17:40 Uhr, 13.05.2022

oh okay, I see it.

$A\phantom{\rule{0.12em}{0ex}}=\frac{b\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}}{2}$

$A\phantom{\rule{0.12em}{0ex}}=21$ cm^2

now to find the area of the rectangle
$A\phantom{\rule{0.12em}{0ex}}=l\phantom{\rule{0.12em}{0ex}}\cdot w\phantom{\rule{0.12em}{0ex}}$
$A\phantom{\rule{0.12em}{0ex}}=12\cdot 7$
$A\phantom{\rule{0.12em}{0ex}}=44$ cm^2

so, The trapezoid's are is $=21$ cm^2 $+44$ cm^2
$=65$ cm^2

right?.

thanks, Nuele, too.

17:42 Uhr, 13.05.2022

thanks, N8eule!.

17:44 Uhr, 13.05.2022

$>$ right?.
No!
Now your area of the rectangle is wrong. But you already knew the correct result of 7 times $12$ at $17:22$ :-)

Combining the correct results should given you the $105c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$ already mentioned above.

BTW, if you type a space between "c" and "m" when typing "7 cm^2" in text mode, you get the desired $7c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$

18:28 Uhr, 13.05.2022

I like that way best!. thanks.

18:28 Uhr, 13.05.2022

I like that way best!. thanks.

18:28 Uhr, 13.05.2022

I like that way best!. thanks.

18:42 Uhr, 13.05.2022

Roman you said the area of the rectangle is wrong. can you explain why? I did not get that.

19:03 Uhr, 13.05.2022

oh $i\phantom{\rule{0.12em}{0ex}}$ see it. it is the wrong calculation
$7\cdot 12=84c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$