Finding the perimeter of a triangle with a quadrilateral cut out of a vertice

I have attached a file with the problem drawn on it. This is a question relating to the Math Praxis 0061 exam.

I am unable to figure out what avenue to take to calculate the perimeter.

Thanks,
karies4083(Happy)

My LibreOffice gives the error "The file 'geometry-questions-triangle-1.doc' is corrupt and therefore cannot be opened." In general, you should not attach .doc files because not everybody has the proper software and opening such files carries a risk (e.g., due to malicious macros). You can describe the problem with words or attach a picture file if necessary.

Sorry about that . . .I thought a picture would help in the visualization. I tried to find a picture to upload, but could not find one. So I tried to draw a picture using the key board keys .. hope it helps!

I have a triangle .. .labeled as follows. The only angle measurement given is the right angle identified in the picuture (just below the 2). The only segment measurements given are identified in the picture: 10, 8, and 2. What is the perimeter? Diagram not to scale. So basically the figure is a triangle with quadrilateral taken out of one of the verticies. You are given measurements for 3 out of the five sides of the figure.

Thanks,
karies4083

l \
l \
l \
l \ 10
l_2_ \
l_l \
l_____\
8

Okay .. . the picture I tried to make with the keys did not work, very sorry. ..

Imagine that the corner between the legs of the image is cut out in the form of a square. . Since the corner is cut out between the two legs, two additional sides are formed and a right angle is formed. The right angle is in the upper right corner, and the top section of the cut out has a measurement of 2. The green leg (with the cutout) is 8 and the hypotenus is 10. VWhat is the perimeter of the figure.

Hope this helps,
Thanks,
karies4083

To preserve spaces, you can wrap it in the [code]...[/code] tags. E.g.:

Code:

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|\
| \
|  \ 10
|  \
|____\

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How do you wrap in the code????

I saved the image as a pdf .. maybe that will be okay to open.

I wouldn't be so persistent with the questions, except no one here can answer it. It is a toughy for us(Smirk)

thanks,
karies4083

By "it" I meant the picture. Click on the "Reply With Quote" button under post #5 to see how the post was typed.

Anyway, knowing the hypotenuse and one of the legs you can find the other leg using the Pythagoras theorem. Cutting out the square (or rectangle) does not seem to change the perimeter: you remove two segments but add two of the same segments.

According the information in the problem, if I add 2 to the legnth of 8, then I have two sides with the same measurement. And if a triangle is a right angle, then the hypotenus cannot be the same legnth as one of the sides . . ?????

Sorry for all the confusion, this is such a hard problem to explain .. I wish I could draw it for you.

Thanks,
karies4083

Attachment 24215

Quote:

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Originally Posted by karies4083

According the information in the problem, if I add 2 to the legnth of 8, then I have two sides with the same measurement. And if a triangle is a right angle, then the hypotenus cannot be the same legnth as one of the sides

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I agree. Therefore, 8 must be the length of BC, not BD. Knowing BC and AB, you can find AC. Now, AC + BC = AF + EF + ED + BD regardless of the lengths of EF and ED.

But can you assume that angle FCD is a right angle, just because angle FED is a right angle???

If you can assume angle FCD is a right angle, and 8 is BC not BD ...... then according to the pythagorean theorm AC will equal 6

And using your formual:

6 + 8+10=AF+FE+ED+DB+BA=24!!!!

Quote:

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Originally Posted by karies4083

But can you assume that angle FCD is a right angle, just because angle FED is a right angle???

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I think so. Angles C, AFE and BDE are not marked as right, but if they are arbitrary, then there is no way to find the perimeter (i.e., there are figures with the given information but different perimeters).

Quote:

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Originally Posted by karies4083

If you can assume angle FCD is a right angle, and 8 is BC not BD ...... then according to the pythagorean theorm AC will equal 6

And using your formual:

6 + 8+10=AF+FE+ED+DB+BA=24!!!!

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I agree.