

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. Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert): "Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen." 

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 $a\phantom{\rule{0.12em}{0ex}}=7$ $b\phantom{\rule{0.12em}{0ex}}=$? $c\phantom{\rule{0.12em}{0ex}}}^{2}={a\phantom{\rule{0.12em}{0ex}}}^{2}+{b\phantom{\rule{0.12em}{0ex}}}^{2$ $10}^{2}={7}^{2}+{b\phantom{\rule{0.12em}{0ex}}}^{2$ $100=49+{b\phantom{\rule{0.12em}{0ex}}}^{2}$ $10049=4949+{b\phantom{\rule{0.12em}{0ex}}}^{2}$ $51={b\phantom{\rule{0.12em}{0ex}}}^{2}$ √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? 

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 $12c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}})$. 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 $105c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$. 

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 

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

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. 

VerständigungsStütze: 

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 

$>$ 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 $30c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}(=12c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}+18c\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}})$. You sure can calculate the area of this rectangular and divide it by 2 ;) 

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. 

thanks, N8eule!. 

$>$ 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 $7c\phantom{\rule{0.12em}{0ex}}{m\phantom{\rule{0.12em}{0ex}}}^{2}$ 

I like that way best!. thanks. 

I like that way best!. thanks. 

I like that way best!. thanks. 

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

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