Maths year 11

Correct number but wrong sign Notice that you are multiplying negative by a negative there so the answer must be a positive. Other than that, you have now expanded both brackets. Refer back to the Pythagorean theorem and see what you need to do with the two answers you just got.

How's this?


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$-18\sqrt5 - 18\sqrt5$ is not $\sqrt5$. The 101 is correct.

So is that it,

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Well no, I just said one of your calculations is wrong.

I got 36root5

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Well, technically MINUS 36root5. But yeah you got there. Now continue on with the first part, nearly done.

Well, technically MINUS 36root5. But yeah you got there. Now continue on with the first part, nearly done.

Nope nope nope. You are getting confused.

You had $a^2+b^2=c^2$ written out.

You worked out what $a^2$ and $b^2$ are because you expanded the required brackets for the side lengths, so just add them together. What you've done right there is $(a^2)^2+(b^2)^2=c^2$ which doesn't make sense. Also, don't use the calculator because it will only gives out decimals which are not asked for.

Nope. $(9+4\sqrt5)+(101-36\sqrt5)=(9+101)+(4\sqrt5-36\sqrt5)$ Simplify it down,

Yep! So since the length of the hypotenuse, c, is followed by $c^2=110-32\sqrt5$, you can now find its length, and you will have it in the required form that part a asks for. State the values of a and b once you have done this.

I didn't quiet get that

State the values of a and b?

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You know that $c^2=110-32\sqrt5$ so how can we go about finding c itself? What do we do to both sides? Once you have done this, you can tell what the values of a and b are.

Divide both sides :/ not sure

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Square root both sides.

So since it is now in the form $c=\sqrt{a-b\sqrt5}$ you can state what the values of a and b are.