Question

Simplify the following using identities.
a). ${\left(109\right)}^2+{\left(91\right)}^2$
b).${\left(200\right)}^2-{\left(100\right)}^2$
c).${\left(1025\right)}^2-{\left(975\right)}^2$
d). ${\left(10\right)}^2+{\left(20\right)}^2$

Answer 1

$a\left)\right(109{)}^2+(91{)}^2$

$=(100+9{)}^2+(100-9{)}^2$

By $(a+b{)}^2+(a^2-b{)}^2=2[a^2+b^2]$

Therefore $(100+9{)}^2+(100-9{)}^2$

$=2[{100}^2+9^2]$

$=2[10000+81]$

$=2\left[10081\right]$

$=20162$

$b\left)\right(200{)}^2-(100{)}^2$

By identity $a^2-b^2=(a+b)(a-b)$

$\therefore (200{)}^2-(100{)}^2=(200+100)(200-100)$

$=300\times 100$

$=30000$

$c\left)\right(1025{)}^2-(975{)}^2$

By $(a+b{)}^2-(a^2-b{)}^2=4ab$

$=(1000+25{)}^2-(1000-25{)}^2$

$=4\left(1000\right)\left(25\right)$

$=1000\times 100$

$=100000$

$d\left)\right(10{)}^2+(20{)}^2$

Now $a^2+b^2=(a+b{)}^2-2ab$

$\therefore (10+20{)}^2-2\times 10\times 20$

$\Rightarrow (30{)}^2-400$

$\Rightarrow 900-400$

$\Rightarrow 500$