Question

Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication:
(i) 25 (ii)37
(iii)54 (iv)71
(v)96

Answer 1

(i) $(25{)}^2$
Here a= 2, b= 5
$\begin{array}{ccc}{a}^2& 2ab& b^2\\ (2{)}^2& 2\times 2\times 5& (5{)}^2\\ =4& =20& =25\\ \frac{+2}{6}& \frac{+2}{22}& \end{array}$

$(25{)}^2=25\times 25=625$
$(25{)}^2=625$

(ii) $(37{)}^2$
Here a= 3 , b= 7
$\begin{array}{ccc}{a}^2& 2ab& b^2\\ (3{)}^2& 2\times 3\times 7& (7{)}^2\\ =9& =42& =49\\ +4& +4& \\ 13& 46& \end{array}$
$(37{)}^2=37\times 37=1369$
$\therefore (37{)}^2=1369$

(iii) $(54{)}^2$
$\begin{array}{ccc}{a}^2& 2ab& b^2\\ (5{)}^2& 2\times 5\times 4& (4{)}^2\\ =25& =40& =16\\ +4& +1& \\ 29& 41& \end{array}$
$(54{)}^2=54\times 54=2916$
$\therefore (54{)}^2=2916$

(iv) $71{)}^2$
Here a= 7, b= 1
$\begin{array}{ccc}{a}^2& 2ab& b^2\\ (7{)}^2& 2\times 7\times 1& (1{)}^2\\ =49& =14& =1\\ +1& & \\ 50& & \end{array}$

$(71{)}^2=71\times 71=5041$
$\therefore (71{)}^2=5041$

(v) $(96{)}^2$
Here a= 9, b = 6
$\begin{array}{ccc}{a}^2& 2ab& b^2\\ (9{)}^2& 2\times 9\times 6& (6{)}^2\\ =81& =108& =36\\ +11& +3& \\ 92& 111& \end{array}$

$(96{)}^2=96\times 96=9216$
$\therefore (96{)}^2=9216$