area of a square

**andyboy179** · January 16th 2010, 10:08 AM

i am i right in thinking that this $area of a square-untitled1.jpg$ would = $area of a square-untitled2.jpg$

**e^(i*pi)** · January 16th 2010, 10:14 AM

Originally Posted by andyboy179

i am i right in thinking that this $Click image for larger version. Name: untitled1.jpg Views: 5 Size: 14.9 KB ID: 14854$ would = $Click image for larger version. Name: untitled2.jpg Views: 5 Size: 17.9 KB ID: 14855$

Nope, the area is the product of the two sides

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

In case you were wondering about the last term:

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

Alternatively

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

**andyboy179** · January 16th 2010, 10:19 AM

Originally Posted by e^(i*pi)

Nope, the area is the product of the two sides

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

In case you were wondering about the last term:

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

Alternatively

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

how did you work out $A = \sqrt{5}(2+\sqrt{10}) = 2\sqrt{5} + 5\sqrt{2}$ ?

**andyboy179** · January 16th 2010, 10:45 AM

..., could you explain it in more detail please?!?!?

**andyboy179** · January 16th 2010, 11:16 AM

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

**e^(i*pi)** · January 16th 2010, 12:24 PM

Originally Posted by andyboy179

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

Yeah that's right.

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

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

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

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

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

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

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

$50 = 2 \times 5 \times 5$

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

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

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

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

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