Tuesday 23 July 2019

Question of the week..

The figure below shows a hinged structure made up of 12 sticks. The structure can be elongated and compressed by changing the angle ABR. As shown in the figure, the distance between B and Q is 150√6 (one hundred and fifty times root six) units when the angle ABR is 90 degrees. What will this distance be when the angle ABR is changed to 60 degrees?
A. 150B. 300C. 400D. 450E. 600

4 comments:

  1. Ahh... another good ol' math related mental ability question.

    By the way "iamjayanth" your answer is wrong. The correct answer is 450(Option D).

    Now, the question is how can one get it right?

    Just join all the diagonals between the tilted squares(diamonds) between the 150sqrt(6) markings. For each square, we have the diagonal length as 30sqrt(6) (just look at figure and you will find that the lengths of BE, EH, HK, etc. are the same so we have 5 such repetitions so just divide by 5).

    From that square again divide the diagonal into half. Considering we have selected the first square/diamond BDEC... Let the mid point be O.

    We have BO=OE=15sqrt(6)

    Complete the triangle BOD. From there it can be seen clearly that angle OBD is 45(now don't ask me how... I'm not going to teach you everything... best advice is to just brush up your math because these simple stuff have been completed in earlier classes).

    Now considering triangle BOD calculate BD using Pythagoras theorem.(You know the length of OD because it is an inverted square/diamond, which is the same as BO(property of diagonals of a square))

    Now after reducing the angle between ABR to 60 degrees, the angle OBD becomes 30 degrees.

    Apply trigonometry here,

    Cos30=BO/BD
    BO=(BD)cos30= 45 units

    So, the total length BE=2(BO)= 90 units

    Again 5 such repetitions occur as explained earlier.
    So for length BQ, just multiply it by 5.

    BQ = 5(BE) = 5*90 = 450 units

    That's the answer to your question.

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