# area of a square

• Jan 16th 2010, 10:08 AM
andyboy179
area of a square
i am i right in thinking that this Attachment 14854would = Attachment 14855
• Jan 16th 2010, 10:14 AM
e^(i*pi)
Quote:

Originally Posted by andyboy179
i am i right in thinking that this Attachment 14854would = Attachment 14855

Nope, the area is the product of the two sides

$\displaystyle A = \sqrt{5}(2+\sqrt{10}) = 2\sqrt{5} + 5\sqrt{2}$

In case you were wondering about the last term:

$\displaystyle \sqrt{5}\sqrt{10} = \sqrt{50} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$

Alternatively

$\displaystyle \sqrt{5}\sqrt{10} = \sqrt{5}\sqrt{5}\sqrt{2} = 5\sqrt{2}$
• Jan 16th 2010, 10:19 AM
andyboy179
Quote:

Originally Posted by e^(i*pi)
Nope, the area is the product of the two sides

$\displaystyle A = \sqrt{5}(2+\sqrt{10}) = 2\sqrt{5} + 5\sqrt{2}$

In case you were wondering about the last term:

$\displaystyle \sqrt{5}\sqrt{10} = \sqrt{50} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$

Alternatively

$\displaystyle \sqrt{5}\sqrt{10} = \sqrt{5}\sqrt{5}\sqrt{2} = 5\sqrt{2}$

how did you work out $\displaystyle A = \sqrt{5}(2+\sqrt{10}) = 2\sqrt{5} + 5\sqrt{2}$?
• Jan 16th 2010, 10:45 AM
andyboy179
..., could you explain it in more detail please?!?!?
• Jan 16th 2010, 11:16 AM
andyboy179
sorry, i'm confusing myself i think. so $\displaystyle 2\sqrt{5} + 5\sqrt{2}$ is the answer?
• Jan 16th 2010, 12:24 PM
e^(i*pi)
Quote:

Originally Posted by andyboy179
sorry, i'm confusing myself i think. so $\displaystyle 2\sqrt{5} + 5\sqrt{2}$ is the answer?

Yeah that's right.

To solve it I used the fact that the area of a rectangle is $\displaystyle length \times width$ and I also used the distributive property which says $\displaystyle a(b+c) = ab+ac$

As the sides are $\displaystyle 2+\sqrt{10}$ and $\displaystyle \sqrt{5}$ to find the area I multiplied them.

Surds act like any other number when multiplied and using the distributive property above for this case: $\displaystyle a = \sqrt{5}$ and $\displaystyle b= 2$ and $\displaystyle c=\sqrt{10}$

Therefore the area is $\displaystyle ab+ac = 2\sqrt{5} + \sqrt{5}\sqrt{10}$

As a law of surds, for positive p,q: $\displaystyle \sqrt{p}\sqrt{q} = \sqrt{pq} and so \sqrt{5}\sqrt{10} = \sqrt{50}$

You could leave the answer as $\displaystyle 2\sqrt{5}+\sqrt{50}$ but it is more usual to simplify.

To simplify a surd get the prime factors of the number inside it- in this case 50.

$\displaystyle 50 = 2 \times 5 \times 5$

Using the rule directly above $\displaystyle \sqrt{50} = \sqrt{2}\sqrt{5}\sqrt{5}$

because $\displaystyle \sqrt{a}\sqrt{a} = a$ we can simplify the above to give $\displaystyle \sqrt{2}\sqrt{5}\sqrt{5} = 5\sqrt{2}$ which is fully simplified.

I then combined this with the first part (ab) and got the final answer of

$\displaystyle A = 2\sqrt{5}+5\sqrt{2}$