Similarity and Indirect Measurement

A

Akmony2000

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Hello. Here is what I worked out . I am not sure if I did this problem correctly . Thank you for your help
 

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Are you asking about problem 3 or 5?

For #3, I would definitely use fractions rather than decimals; but your answer is correct. I would have written the work as [MATH]\frac{20}{3}\times 2 = \frac{40}{3} = 13\frac{1}{3}[/MATH].

For #5, I have no idea what you are thinking. Are you following some "magical" formula that you were taught, or thinking in some way I just can't see? Please explain.

In any case, your answer is wrong.

I would first find the segment above the "x", using the two similar triangles, and then find x by subtraction. I see several other more creative ways.
 
Akmony2000 said:
Thank you for your reply. We were just given this worksheet.
Click to expand...
So, can you answer my questions? What does your work for #5 mean? (It may be more right than I know.) Can you see how to follow my suggestion? Can you identify the similar triangles? Your correct work for #3 suggests you have a reasonable understanding of them.
 
No, it's incorrect. Please explain your work (in the other thread), so we can try to help you correct it.
 
For #5 you have 160(50) and 160(40). Where did the "50" and "40" come from? Are they half of the 100 and 80 bases?

Why half?

You have two similar triangles. One has height 160 and base length 100. The other has height 160- x and base length 80. Since they are similar we have \(\displaystyle \frac{160}{100}= \frac{160- x}{80}\). Solve that equation for x.

(I spent a few seconds wondering what "ibox" meant!)
 
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