Evaluate $(41)^{2}$ by using the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$

Open in App

Solution

Using $(a + b)^{2}$ = $a^{2} + 2 a b + b^{2}$
Writing $(41)^{2}$ as $(40 + 1)^{2}$
$=$ $40^{2} + 2 \times 40 + 1 + 1^{2}$
$=$ $1600 + 80 + 1$
The solution is $1681$

Suggest Corrections

Similar questions

(a + b)^{2} = (a^{2} + 2 a b + b^{2})

(609)^{2}

(725)^{2}

Evaluate using the identity:

(i) (49)²; (ii) (9.8)²; (iii) (59)²; (iv) (198)².

209^{2}

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

\sqrt{\frac{a^{2} + 2 a b + b^{2}}{a^{2} - 2 a b + b^{2}}}

(a + b)^{2} = a^{2} + 2 a b + b^{2}

(a + 6)^{2}

Join BYJU'S Learning Program