The area of a rectangle is $\sqrt{24}$ $sqrt24$ cm. lf its breadth is $(3 - \sqrt{6})$ $(3-sqrt6)$ cm, how do you find its length in the form $(a + b \sqrt{6})$ $(a+bsqrt6)$ cm where a and b are integers?

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The area of a rectangle is $\sqrt{24}$ $sqrt24$ cm. lf its breadth is $(3 - \sqrt{6})$ $(3-sqrt6)$ cm, how do you find its length in the form $(a + b \sqrt{6})$ $(a+bsqrt6)$ cm where a and b are integers?

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Answer 1 · 2016-11-17T11:14:14

The length is $4 + 2 \sqrt{6}$ $4+2sqrt6$

Explanation:

Area of rectangle (A) = length (L) * breadth(B). So length = area/breadth.

$L = \frac{A}{B} or L = \frac{\sqrt{24}}{3 - \sqrt{6}} or L = \frac{\sqrt{24} \cdot (3 + \sqrt{6})}{(3 - \sqrt{6}) \cdot (3 + \sqrt{6})}$ $L= A/B or L= sqrt24/(3-sqrt6) or L= (sqrt24*(3+sqrt6))/((3-sqrt6)*(3+sqrt6))$

or $L= (sqrt24*(3+sqrt6))/(9-6) = (3sqrt24+sqrt24*sqrt6)/3 =(3sqrt24+3sqrt16)/3 = (cancel3(2sqrt6+4))/cancel3= 4+2sqrt6$ [Ans]

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The area of a rectangle is $\sqrt{24}$ $sqrt24$ cm. lf its breadth is $(3 - \sqrt{6})$ $(3-sqrt6)$ cm, how do you find its length in the form $(a + b \sqrt{6})$ $(a+bsqrt6)$ cm where a and b are integers?

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Explanation:

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The area of a rectangle is sqrt24√24sqrt24 cm. lf its breadth is (3-sqrt6)(3−√6)(3-sqrt6) cm, how do you find its length in the form (a+bsqrt6) (a+b√6)(a+bsqrt6) cm where a and b are integers?

1 Answer

Explanation:

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The area of a rectangle is $\sqrt{24}$ $sqrt24$ cm. lf its breadth is $(3 - \sqrt{6})$ $(3-sqrt6)$ cm, how do you find its length in the form $(a + b \sqrt{6})$ $(a+bsqrt6)$ cm where a and b are integers?