Question

The length of a rectangle is increased by 10%.express the increase in area as a percent

Answer 1

hey friend
here is ur answer
HOPE IT HELPS

Answer 2

Let, the original Length = l

Then , Area :

$l \times b$


after increasing :

$l \times \frac{110}{100} \\ \\ = > \frac{11l}{10}$

Now, Area :

$\frac{11l}{10} \times b$

Now, changed in percentage :

$\frac{ \frac{11lb}{10} - lb }{lb} \times 100 \\ \\ = > \frac{ \frac{lb}{10} }{lb} \times 100 \\ \\ = > \frac{1}{10} \times 100 \\ \\ = > 10$

So, 10% increased in area