Logarithm & Surds: rect. has length of (5+√3) cm, area of (5 + 2√3) cm^2

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IshaanM8

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Logarithm & Surds: rect. has length of (5+√3) cm, area of (5 + 2√3) cm^2

Well, thanks for opening this thread which may have taken you interest.
I'm stuck on this question:

"A rectangle of length of (5+√3)cm and has an area of (5 + 2√3) cm^2.
Calculate the exact width of the rectange with an integer denominator."

I am not very familiar with dividing surds with surds and don't know what an "integer denominator" means.
Please help in any way you can, I deeply appreciate your efforts.

Thanks!
 
The best place to start any math problem is usually by making sure you understand the definitions of all the terms used in the problem text. In this case, you need to know the definitions/formulas of "width [of a rectangle]," "area [of a rectangle]," "integer," "denominator", and "integer denominator."

Hopefully, you already know how to find the area of a rectangle when given its length. Likewise, you should already know what an integer is and what a denominator is. That leaves the only possibly unknown part as "integer denominator." But, that's nothing more than putting the concepts together. When you solve the problem to find the width, you'll end up with a fraction that has a numerator and a denominator. What you want to do is then find a way to rewrite said fraction so that the denominator is an integer (aka a whole number).

Let's start out by creating an equation we can solve for the width. Here, I'll let w stand for the width of the rectangle.

\(\displaystyle \left( 5+\sqrt{3} \right) w = 5+2\sqrt{3}\)

Next, what would happen if we isolated w by dividing both sides by \(\displaystyle 5 + \sqrt{3}\)?

\(\displaystyle w = \dfrac{5+2\sqrt{3}}{5+\sqrt{3}}\)

Where do you think you'd go from here? What do you know about rationalizing the denominatorhttps://www.mathsisfun.com/algebra/rationalize-denominator.html? Have you, perhaps, heard of multiplying by the conjugatehttps://www.mathsisfun.com/algebra/conjugate.html?

Go ahead and give the problem your best shot. If you get stuck again, that's okay, but when you reply back please include any and all work you've done on this problem. The more specific you can be about exactly where things are getting bogged down for you, the better help we can provide. Thank you.
 
IshaanM8 said:
Well, thanks for opening this thread which may have taken you interest.
I'm stuck on this question:

"A rectangle of length of (5+√3)cm and has an area of (5 + 2√3) cm^2.
Calculate the exact width of the rectange with an integer denominator."

I am not very familiar with dividing surds with surds and don't know what an "integer denominator" means.
Please help in any way you can, I deeply appreciate your efforts.

Thanks!
Click to expand...
As ksdhart has said, this is primarily an exercise in "rationalizing the denominator," which is a rather misleading term.

Back when you were studying fractions in arithmetics, you learned that it is conventional to reduce a rational number to "lowest terms." For example, you do not say

\(\displaystyle \dfrac{64}{1024}\) but rather \(\displaystyle \dfrac{1}{64}\)

because the latter is easier to understand even though both represent the same number.

A similar convention applies to fractions with square roots in the denominator. Such fractions are usually converted into a fraction of equal numeric value but with an integer in the denominator because the latter is somewhat more comprehensible than the former.
 
w = width, u = sqrt(3)

w(5 + u) = 5 + 2u : width * length = area

w = (5 + 2u) / (5 + u)

Multiply numerator and denominator by 5 - u
 
Thanks for the help everyone.
I've done the rationalising and simplifying.

I've got 19 + 5√3
------------------22


I multiplied by it's conjugate and then expanded the "brackets"
i would add a picture of my working out, but it keeps giving me error.
Please let me know if it is correct?
Thanks for your time!
 
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