Solved Measurement Problems TCS Placement Tests
Let the side of the square ABCD be a.
Then Diagonal BD = AC = a x sqrt 2 (You will get this formula by applying Pythagoras theorem)
We have to find this diagonal length as follows:
Given, he had drawn 5 circles on diagonal BD.
After sketching 5 circles according to the question, we see that some gap between corner of square and extreme circles. Now, let us calculate that space:
In the below diagram, Here, Y is the centre of left extreme circle, BXYZ is a small square with side r and diagonal r sqrt2.
Similar observation can be made on the other extreme circle as well.
Then the required ratio = r : 8r + 2r sqrt2 = r : r(8 + 2sqrt2)
We can remove r on both sides as it is common. Therefore, we get,
= 1 : 8+2 sqrt2
= 1 : 8+2(1.414) = 1:10.828
Hence the required ratio is 1 : 10.828.
Let the side of the square be a.
Then Diagonal = a(sqrt2)
By applying a similar logic as that of problem 1, we will get a diagram as follows.
a(sqrt2) = diagonal = r sqrt2 + r + 2r + r + r sqrt2 = 4r + 2r sqrt2
a = (4r + 2r sqrt2) / sqrt2
a = 4r/sqrt2 + 2r sqrt2 / sqrt2
a = (2 x 2r) / sqrt2 + 2r
a = 2r sqrt2 + 2r = 2r(sqrt2+1)
Then perimeter of the square = 4a = 4 x 2r(sqrt2+1) = 8r(sqrt2+1)
= 8r(1.414+1) = 8r(2.414) = r(19.312)
Hence 19.312 time radius equals the perimeter of the square.
Applying similar logic to that of first and second questions, we will get the below diagram corresponding to this question :
The length of the diagonal BD = r sqrt2 + r + 10r + r + r sqrt2
= 2 rsqrt2 + 12r
Given that r = 2 cm
Then BD = 2(2)sqrt2 + 12(2) = 4 sqrt2 + 24 = 4(1.414) + 24 = 29.65 cm.
Hence the required answer is 29.65cm