The length of a rectangle exceeds twice its breadth by 3 cm, and its perimeter is 42 cm. Its area is:

Let the breadth be $b$, so the length is $2b+3$. The perimeter gives $l+b=21$, i.e. $3b+3=21$, so $b=6$ cm and $l=15$ cm. The area is $15\times 6=90{\text{cm}}^2$ (and the perimeter $2(15+6)=42$ cm checks out).