Question

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A rod is marked with 5 equally spaced points as shown. It is rotated thrice through 180 but with different centres if it is first rotated about A then about B, and finally. about E. which point will be at the same position as it was initially?​

Answer 1

Answer:

The diagram shows a rod with five equally spaced points A, B, C, D and E marked on it. The rod is rotated three times through 180 degrees, first about A, then about B and finally about E. Which point finishes in the same position as it was at the start?

a.A

b.B

c.C

d.D

e.E

f.All of the points will finish in their original position.

g.None of the points will finish in their original position. d.D

Explanation :

Here's your rod

|-----|-----|-----|-----|A|==B==C==D==E-----|

The === is the rod

The ----- is just a virtual line

The | are just markers at the same distance of each other (the exact same distance as A from B)

The | around the A is just to mark the original position of A.

1st Rotation : around A

E==D==C==B==|A|-----|-----|

2nd Rotation : around B

A==B==|C|==D==E-----|-----|

3rd Rotation : around E

-----||-----|-----E==D==C==B==A

Original position

-----|A|==B==C==D==E-----|-----|

More Mathematical explanation :

Let's say that L is the distance between each points on the rod and is oriented from left to right.

AB = L, AC = 2L, AD = 3L, etc.

And the original positions of each point is xA0, xB0, xC0, xD0 and xE0.

Everytime the point goes to the left, its coordinate diminishes, and going right makes it go right

1st Rotation : around A

xA1 = xA0 (doesn't move)

xB1 = xB0 + 2 BA = xB0 - 2 AB = xB0 - 2L

xC1 = xC0 + 2 CA = xC0 - 2 AC = xC0 - 4L

xD1 = xD0 + 2 DA = xD0 - 2 AD = xD0 - 6L

xE1 = xE0 + 2 EA = xE0 - 2 AE = xD0 -8L

2nd Rotation : around B

xA2 = xA1 + 2 AB = xA0 - 2L (here, AB = -L)

xB2 = xB1 = xB0 - 2L (doesn't move)

xC2 = xC1 + 2 CB = xC0 - 4L + 2L = xC0 - 2L

xD2 = xD1 + 2 DB = xD0 - 6L + 4L = xD0 - 2L

xE2 = xE1 + 2 EB = xE0 - 8L + 6L = xE0 - 2L

Here, the rod just moved 2L to the left from the original position.

3rd Rotation : around E

xA3 = xA2 + 2 AE = xA0 - 2L + 8L = xA0 + 6L

xB3 = xB2 + 2 BE = xB0 - 2L + 6L = xB0 +4L

xC3 = xC2 + 2 CE = xC0 - 2L + 4L = xC0 + 2L

xD3 = xD2 + 2 DE = xD0 - 2L + 2L = xD0

xE3 = xE2 = xE0 - 2L

As you can see, xD3 = xD0, or in other words, it has the same position as at the beginning. Q.E.D.

Step-by-step explanation:

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Answer 2

Answer:

It's answer is D

Step-by-step explanation: