As S is the circumcentre of ΔABC, it is equidistant from A,B and C. So AS = BS = CS = radius of its circumcircle. As AD is the perpendicular bisector of BC \(BD=\cfrac { 1 }{ 2 } \times BC=\cfrac { 1 }{ 2 } \times 72=36cm\) 
In right-angled triangle BDS, by Pythagoras theorem
BS² = BD² + SD² = 36² + 15² = 1521 = 39² ⇒ BS = 39cm.
ஃThe radius of the circumcircle of ΔABC is 39 cm.

ABC is a triangle and G is its centroid
If, \(AD=12cm\Rightarrow GD=\cfrac { 1 }{ 3 } \) of \(AD=\cfrac { 1 }{ 3 } (12)=4cm\) and
\(BE=9cm\Rightarrow BG=\cfrac { 2 }{ 3 } \) of  \(BE=\cfrac { 2 }{ 3 } (9)=6cm\) 
Also D is a midpoint of \(BC\Rightarrow BD=\cfrac { 1 }{ 2 } \) of \(BC=\cfrac { 1 }{ 2 } (8)=4c\) 
ஃ The perimeter of ΔBDG = BD + GD + BG = 4 + 4 + 6 = 14 cm

AD = DB, since D is the midpoint of AB. (Why?)
Therefore, 9x − 13 = 4x + 2. Solving this (try!) simple equation, we get x = 3.
Hence, AB = 2(9x − 13) = 2(9(3) − 13)=28 units.

Since they all work for the same number of days, the ratio in which they share the money is equal to the ratio of their work done per day.
That is,\(\cfrac { 1 }{ 4 } :\cfrac { 1 }{ 6 } :\cfrac { 1 }{ 10 } =\cfrac { 15 }{ 60 } :\cfrac { 10 }{ 60 } :\cfrac { 6 }{ 60 } =15:10:6\) 
Here, the total parts = 15 + 10 + 6 = 31
Hence, A’ s share =\(\cfrac { 15 }{ 31 } \times 3100=Rs.1500\), B; share = \(\cfrac { 10 }{ 31 } \times 3100=Rs.1000\) and
C’ s share is Rs.3100 - (Rs.1500 + Rs.1000) = Rs.600

P’s 1 day’s work = \(\cfrac { 1 }{ 20 } \) and Q’s 1 day’s work = \(\cfrac { 1 }{ 30 } \) 
P’s work for 5 days = \(\cfrac { 1 }{ 20 } \times 5=\cfrac { 5 }{ 20 } =\cfrac { 1 }{ 4 } \) 
Therefore, the remaining work = \(1-\cfrac { 1 }{ 4 } =\cfrac { 3 }{ 4 } \) (Total work is always 1)
This remaining work was done by both P and Q. 
Work done by P and Q in a day = \(\cfrac { 1 }{ 20 } +\cfrac { 1 }{ 30 } =\cfrac { 5 }{ 60 } =\cfrac { 1 }{ 12 } \) 
Therefore, the number of days they worked together =\(\cfrac { 3/4 }{ 1/12 } =\cfrac { 3 }{ 4 } \times \cfrac { 12 }{ 1 } =9days.\)
So, Q left after 9 day from the days the work started.

(A+ B) ’s 1 day’s work = \(\cfrac { 1 }{ 16 } \) 
A’s 1 day’s work = \(\cfrac { 1 }{ 48 } \) 
ஃ B’s 1 day’s work = \(\cfrac { 1 }{ 16 } -\cfrac { 1 }{ 48 } \) 
= \(\cfrac { 3-1 }{ 48 } =\cfrac { 2 }{ 48 } =\cfrac { 1 }{ 24 } \)
ஃB alone can complete the work in 24 days

If 15 men can complete the work in 40 days, then the work measured in terms of person days = 15 × 40 = 600. person days.
If the same work is to be done by 30 (15 + 15) men, then the number of days they will take is \(\cfrac { 600 }{ 30 } =20days\)

Multiplicative Factor Method:

Step 1:
Here, less days means more students. So, it is an inverse variation.
ஃ The multiplying factor is \(\cfrac { 56 }{ 27 } \) 
Step 3:
\(\therefore x=81\times \cfrac { 56 }{ 27 } \times \cfrac { 160 }{ 448 } \) 
x = 60 students.
Formula Method:
Here,P₁ = 81,D₁ = 56 and W₁ = 448
P₂ = x, D₂ = 27 and W₂ = 160
Using the formula,\(\cfrac { { P }_{ 1 }\times { D }_{ 1 } }{ { W }_{ 1 } } =\cfrac { { P }_{ 2 }\times { D }_{ 2 } }{ { W }_{ 2 } } \) 
We have,\(\cfrac { 81\times 56 }{ 448 } =\cfrac { x\times 27 }{ 160 } \) 
\(\Rightarrow x=\cfrac { 81\times 56 }{ 448 } \times \cfrac { 160 }{ 27 } \)
x = 60 students.

Proportion Method:

Here, the unknown is the payment (x). It is to be compared with the workers and the days.
Step 1:
Here, less days means less payment. So, it is a direct proportion.
ஃ The proportion is 20 : 12 :: 6 : x   ⟶1
Step 2:
Also, less workers means less payment. So, it is a direct proportion again.
ஃ The proportion is 15 : 5 :: 6 : x  ⟶2
Step 3:
Combining 1 and 2
\(\begin{matrix} 20:12 \\ 15:5 \end{matrix}\} ::6:x\)
We know that the product of the extremes = the product of the means
So,20 x 15 = 12 x 6 x 5 ⇒ \(x=\cfrac { 12\times 6\times 5 }{ 20\times 15 } =Rs.1.2lakh\) 
Multiplicative Factor Method

Here, the unknown is the payment (x). It is to be compared with the workers and the days.
Step 1:
Here, less days means less payment. So, it is a direct proportion.
ஃ The multiplying factor is \(\cfrac { 12 }{ 20 } \) (take the reciprocal).
Step 2:
Also, less workers means less payment. So, it is a direct proportion again.
ஃ The multiplying factor is \(\cfrac { 5 }{ 15 } \) (take the reciprocal)
Step 3:
\(\therefore x=6\times \cfrac { 12 }{ 20 } \times \cfrac { 5 }{ 15 } \)
x = Rs.1.2 lakh
Formula Method
Here, P₁= 15D, D₁= 20 and W₁= 6
P₂ = 5, D₂ = 12 and W₂ = x
Using the formula,\(\cfrac { { P }_{ 1 }\times { D }_{ 1 } }{ { W }_{ 1 } } =\cfrac { { P }_{ 2 }\times { D }_{ 2 } }{ { W }_{ 2 } } \) 
We have,\(\cfrac { 15\times 20 }{ 6 } =\cfrac { 5\times 12 }{ x } \) 
\(\Rightarrow x=\cfrac { 5\times 12\times 6 }{ 15\times 20 } =Rs.12lakh\)

\(\sqrt [ 3 ]{ \cfrac { 9261 }{ 8000 } } =\cfrac { \sqrt [ 3 ]{ 9261 } }{ \sqrt [ 3 ]{ 8000 } } =\cfrac { \left( 21\times 21\times 21 \right) ^{ 1/3 } }{ \left( 20\times 20\times 20 \right) ^{ 1/3 } } =\cfrac { 21 }{ 20 } \)

We find that, 675 = 3 x 3 x 3 x 5 x 5 …………….(i)
Grouping the prime factors of 675 as triplets, we are leftover with 5 x 5.
We need one more 5 to make it a perfect cube
To make 675 a perfect cube, multiply both sides of (i) by 5.
675 x 5 = (3 x 3 x 3 x 5 x 5) x 5
3375 = 3 x 3 x 3 x 5 x 5 x 5
Now, 3375 is a perfect cube. Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5.

\(\sqrt { 256 } =\sqrt { 16\times 16 } =\sqrt { 16 } \times \sqrt { 16 } =4\times 4=16\) or \(\sqrt { 256 } =\sqrt { 64\times 4 } =\sqrt { 64 } \times \sqrt { 4 } =8\times 2=16\)
Try to fill up the following table of similar problems using \(\sqrt { a\times b } =\sqrt { a } \times \sqrt { b } \)

Here, 108 = 2 x 2 x 3 x 3 x 3
= 2² x 3² x 3²
Here, the prime factor 3 does not have a second pair. Hence, 108 is not a perfect square number.

We proceed as follows:
Step 1: 64 – 1 = 63
Step 2: 63 – 3 = 60
Step 3: 60 – 5 = 55
Step 4: 55 – 7 = 48
Step 5: 48 – 9 = 39
Step 6: 39 – 11 = 28
Step 7: 28 – 13 = 15
Step 8: 15 – 15 = 0
We have subtracted successive odd numbers (starting from 1) repeatedly from the given number 64. We get zero in the 8^th step.
Therefore \(\sqrt { 64 } =8\). If we don't get zero, then the given number is not a perfect square.
Given that 685584 is a perfect square number. Is it possible for us to fi nd out its square
root by the method of successive subtraction of odd numbers?
Yes, we can, but it will be time consuming and tedious. So we need to consider other methods of fi nding the root of square numbers.

Create a 3 × 3 square below and to the left of 345 (red). Make a diagonal in each of the 9 squares created. Place the product value, say, 5 × 3 = 15, 1 in the upper part and 5 in the lower part of the square. After filling all the squares, add the numbers through coloured diagonals. The arrow indicated numbers from the left 119025 square of 345.
ஃ345² = 119025.