# The area of a rectangle is sqrt24 cm. lf its breadth is (3-sqrt6) cm, how do you find its length in the form (a+bsqrt6)  cm where a and b are integers?

The length is $4 + 2 \sqrt{6}$
$L = \frac{A}{B} \mathmr{and} L = \frac{\sqrt{24}}{3 - \sqrt{6}} \mathmr{and} L = \frac{\sqrt{24} \cdot \left(3 + \sqrt{6}\right)}{\left(3 - \sqrt{6}\right) \cdot \left(3 + \sqrt{6}\right)}$
or $L = \frac{\sqrt{24} \cdot \left(3 + \sqrt{6}\right)}{9 - 6} = \frac{3 \sqrt{24} + \sqrt{24} \cdot \sqrt{6}}{3} = \frac{3 \sqrt{24} + 3 \sqrt{16}}{3} = \frac{\cancel{3} \left(2 \sqrt{6} + 4\right)}{\cancel{3}} = 4 + 2 \sqrt{6}$[Ans]