Question

The length of the rectangle is $(4ab+6)m$ and breadth of the rectangle is $(6b^2+2ab)m$. Find the area of the rectangle.

• A. $(24a{b}^3+18a^2{b}^2+36b^2+12ab)m^2$.
• B. $(24a{b}^3+6a^2{b}^2+6b^2+12ab)m^2$
• C. $(24a{b}^3+8a^2{b}^2+36b^2+12ab)m^2$
• D. $(8a^2{b}^2+36b^2+ab)m^2$.

Answer 1

The correct option is C $(24a{b}^3+8a^2{b}^2+36b^2+12ab)m^2$

Given:
Length of the rectangle
$=(4ab+6)m$

Breadth of the rectangle
$=(6b^2+2ab)m$

Area of a rectangle = length $\times$ breadth
$=(4ab+6)\times (6b^2+2ab)$
$=4ab\times 6b^2+4ab\times 2ab+6\times 6b^2+6\times 2ab)$
$=(24a{b}^3+8a^2{b}^2+36b^2+12ab)m^2$