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Mars2501 [29]
3 years ago
5

Kathy stood on the middle rung of a ladder. She climbed up 4 ​rungs, moved down 3 ​rungs, and then climbed up 9 rungs. Then she

climbed up the remaining 3 rungs to the top of the ladder. How many rungs are there in the whole​ ladder?
Mathematics
1 answer:
stealth61 [152]3 years ago
4 0

<u>Answer: </u>

Kathy stood on the middle rung of a ladder There are 20 rungs in the ladder.

<u>Solution: </u>

Let there be ‘x’ rungs in the ladder.

It is given that in the beginning she is standing in the middle of the rung. Let the middle rung be  \frac{x}{2}

Now it is given that She climbed up 4 rungs which means she climbed \frac{x}{2} + 4 rungs.

Again it is given that she moved down 3 rungs which means she moved down   \left[\left(\frac{x}{2}+4\right)-3\right] rungs.

Then she moved down \left(\frac{x}{2}+1\right) rungs

Again she climbed 9 rungs up and reaches the top which means she climbed \left[\left(\frac{x}{2}+1\right)+9\right]=\left(\frac{x}{2}+10\right) rungs

Now we can use the following equation to find the number of rungs in the ladder

\begin{array}{l}{\frac{x}{2}+10=x} \\\\ {\frac{x+20}{2}=x}\end{array}

x+20 = 2x

20 = 2x-x

x = 20

Thus the number of rungs in the ladder is 20

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Step One
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Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
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Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
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28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
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sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
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Step three
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h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


7 0
3 years ago
Can I please have help?
givi [52]
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