## Solved Measurement Problems TCS Placement Tests

Below are three problems dealing with parameters of circles and squares.

**Question 1**

Let ABCD be a square. Alex wants to draw 5 circles of equal radius 'r' with their centres on BD such that the two extreme circles touch two sides AB, BC and AD, CD of the square respectively and each middle circle touches two circles on either side. Find the ratio of r to that of BD.

a) 2 : 11.828 b) 5 : 19.214 c) 1 : 10.828 d) 1 : 12.515

**Answer :**c) 1 : 10.828

Solution :

As per the statements in question, Alex would have drawn a diagram as shown below.

Let the radius of each circle be r.

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:

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:

Draw perpendicular lines from sides AB and BC to extreme circle. Now you get a small square of side r and of diagonal r x sqrt2.

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.

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.

Summing up our observations in the main diagram, we get,

Hence the diagonal of big square = r sqrt2 + r + 6r + r + r sqrt2 = 8r + 2r sqrt2

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.

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.

**Question 2**

In an exam, students were asked to find the perimeter of a square which contains three circles such that their centres on the diagonal of the square, the middle circle touches two extreme circles on either side and two extreme circles touch two sides of the square and the radius of the three are equal. And at what times the radius equals the perimeter?

a) 19.312 b) 12.312 c) 21.312 d) 17.312

**Answer :**a) 19.312

Solution:

Let the radius of circle be r.

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

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

But we know, 2/sqrt2 = sqrt2. Therefore, above equation becomes,

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.

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.

**Question 3**

Find the length of diagonal BD of a square ABCD when 7 circles of radius 2 cm are located in ABCD such that their centres on BD, two extreme circles touch two sides of ABCD and each middle circle touches two circles on either side.

a)32.98cm b)29.65cm c)31.12cm d)28.56cm

**Answer :**b)29.65cm

Solution :

Given that, ABCD is a square where AB, BC, CD and AD are sides of square and BD, AC are diagonals 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

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