Question

Find the square of 35.

Answer 1

Hint: Square of a number means product of number twice to itself. For example:
Let the number a, square of a is equal to
\[\mathop {\left( a \right)}\nolimits^2 = a \times a\]
But for a large number, multiplication can be a little tedious.
The number (ab), tens place value = a, ones place value = b, can be written as (10a + b).

Complete step-by-step solution:
Square of 35 $ = \mathop {\left( {35} \right)}\nolimits^2 $
                        $ = 35 \times 35$
Number 35 can also be representing as
$35 = \left( {30 + 5} \right)$
Squaring both sides
$\mathop {\left( {35} \right)}\nolimits^2 = \mathop {\left( {30 + 5} \right)}\nolimits^2 $
$
   \Rightarrow \left( {30 + 5} \right)\left( {30 + 5} \right) \\
   \Rightarrow 30\left( {30 + 5} \right) + 5\left( {30 + 5} \right) \\
   \Rightarrow 900 + 150 + 150 + 25 \\
   \Rightarrow 1225 \\
 $
Thus, $\mathop {\left( {35} \right)}\nolimits^2 = 1225$

Therefore, the square of 35 is 1225.

Additional information: Following table of some numbers and their squares:

Number	Square	Number	Square	Number	Square
1	1	11	121	21	441
2	4	12	144	22	484
3	9	13	169	23	529
4	16	14	196	24	576
5	25	15	225	25	625
6	36	16	256	30	900
7	49	17	289	35	1225
8	64	18	342	40	1600
9	81	19	361	45	2025
10	100	20	400	50	2500

Note:
There are many alternate methods and short tricks to find the square of a two-digit number. Some of them are stated below:
i) Short trick for square of two-digit number.
Calculate product of tens digit, ones digit and 2.
Write squares of tens and ones together, add the calculated product to it, leaving the last digit of squares. Represents the square of tens and ones digit in two-digits number.
Calculation shown below:
Square of 35:
$
  \dfrac{{35}}{
  \left. {3 \times 5 \times 2} \right|30 \\
   \downarrow {\text{ }} \downarrow \\
 }{\text{ }} \\
  \dfrac{
  092\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{5} \\
   + 30 \\
 }{{1225}}{\text{ }} \\
 $

Square of 21:
$
  \dfrac{{21}}{
  \left. {2 \times 1 \times 2} \right|04 \\
   \downarrow {\text{ }} \downarrow \\
 }{\text{ }} \\
  \dfrac{
  040\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{1} \\
   + 04 \\
 }{{0441}}{\text{ }} \\
 $

ii) Consider a number with unit digit 5, i.e., $a5$
\[
  \mathop {\left( {a5} \right)}\nolimits^2 = \mathop {\left( {10a + 5} \right)}\nolimits^2 \\
  {\text{ = }}10a\left( {10a + 5} \right) + 5\left( {10a + 5} \right) \\
  {\text{ }} = 100\mathop a\nolimits^2 + 50a + 50a + 25 \\
  {\text{ }} = 100a\left( {a + 1} \right) + 25 \\
  {\text{ }} = a\left( {a + 1} \right){\text{hundred}} + 25 \\
 \]
Square of 35:
a = 3
\[
  \mathop {\left( {35} \right)}\nolimits^2 = 3\left( {3 + 1} \right){\text{hundred}} + 25 \\
  {\text{ = }}3\left( 4 \right){\text{hundred}} + 25 \\
  {\text{ }} = 1200 + 25 \\
  {\text{ }} = 1225 \\
 \]

iii) Adding odd numbers: Consider the following:
1 [one odd number] $ = 1 = \mathop 1\nolimits^2 $
1 + 3 [sum of first two odd numbers] $ = 4 = \mathop 2\nolimits^2 $
1 + 3 + 5 [sum of first three odd numbers] $ = 9 = \mathop 3\nolimits^2 $
1 + 3 + 5 + 7 [...] $ = 16 = \mathop 4\nolimits^2 $
1 + 3 + 5 + 7 + 9 [...] $ = 25 = \mathop 5\nolimits^2 $
1 + 3 + 5 + 7 + 9 + 11 [...] $ = 36 = \mathop 6\nolimits^2 $
So we can say that the sum of first $n$ odd natural numbers is $\mathop n\nolimits^2 $
Square of 21:
1 + 3 + 5 + 7 + 9 + 11 + … + 41 $ = 441 = \mathop {21}\nolimits^2 $