Attachment 30781

I need help from question 20 to 22!

Attachment 30782

And find X!

Please list down the solution! i got stuck in these two question for very long!

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- April 28th 2014, 07:59 AMxwyThe Pythagorean Theorem, involving radical expressions
Attachment 30781

I need help from question 20 to 22!

Attachment 30782

And find X!

Please list down the solution! i got stuck in these two question for very long! - April 28th 2014, 08:48 AMebainesRe: The Pythagorean Theorem, involving radical expressions
In question 20, notice that h, a, and a/2 form a right triangle. From Pythagoras you know that a^2 = (1/2)^2 + h^2, so rearrange this to get 'h' by itself. What do you get? Now use that value to calculate the areas for questions 21 and 22, remembering that the formula for area of a triangle is 1/2 base times height. As for question 28, note that the line segment to the left of 'x' is part of a right triangle, and you can find its length from the dimensions given. You can also find the total length of the base of the large triangle, and 'x' is the difference between these two lengths.

- April 28th 2014, 09:41 AMProve ItRe: The Pythagorean Theorem, involving radical expressions
I'm sure you mean $\displaystyle \begin{align*} a^2 = \left( \frac{a}{2} \right) ^2 + h^2 \end{align*}$.

- April 28th 2014, 11:30 AMebainesRe: The Pythagorean Theorem, involving radical expressions
Right - sorry for the typo.

- April 28th 2014, 11:15 PMxwyRe: The Pythagorean Theorem, involving radical expressions
Attachment 30784

Thanks for the reply! I am basically lost at squarroot(a^2-b^2) from my attempt as you can see.

Attachment 30785

I am sorry for not informing but I am looking at question 29 not 28. For question 29, I am perplexed at the same issue with qn 20 and 21, which is basically squareroot(a^2-b^2). As go on, it seems that have done something wrong somewhere. Please enlighten me.

Lastly, pardon me for my poor understanding in math. I have been trying very hard to self study a subject that I have not touched for years and poor at it. - April 28th 2014, 11:30 PMProve ItRe: The Pythagorean Theorem, involving radical expressions
$\displaystyle \begin{align*} a^2 - \left( \frac{a}{2} \right) ^2 &= \frac{4a^2}{4} - \frac{a^2}{4} \\ &= \frac{4a^2 - a^2}{4} \\ &= \frac{3a^2}{4} \end{align*}$

So when you take the square root:

$\displaystyle \begin{align*} \sqrt{\frac{3a^2}{4}} &= \frac{\sqrt{3}\sqrt{a^2}}{\sqrt{4}} \\ &= \frac{\sqrt{3}\,a}{2} \end{align*}$ - April 29th 2014, 02:46 AMxwyRe: The Pythagorean Theorem, involving radical expressions
hey prove it, could you please coach me on question 29 that I have posted as well? I would appreciate it very much! Thank you so much for your explanation above!

- April 29th 2014, 03:39 AMProve ItRe: The Pythagorean Theorem, involving radical expressions
In the large triangle, by Pythagoras we have

$\displaystyle \begin{align*} \left( \sqrt{3} \right) ^2 + 1^2 &= h^2 \\ 3 + 1 &= h^2 \\ 4 &= h^2 \\ h &= 2 \end{align*}$

So the entire length of the hypotenuse is 2 units.

Now looking at the triangle that has $\displaystyle \begin{align*} \sqrt{3} \end{align*}$ as its hypotenuse. Call the unknown length l and we have

$\displaystyle \begin{align*} x^2 + l^2 &= \left( \sqrt{3} \right) ^2 \\ x^2 + l^2 &= 3 \\ l^2 &= 3 - x^2 \\ l &= \sqrt{ 3 - x^2 } \end{align*}$

and in the triangle that has 1 as its hypotenuse, call the unknown length k and we have

$\displaystyle \begin{align*} x^2 + k^2 &= 1^2 \\ x^2 + k^2 &= 1 \\ k^2 &= 1-x^2 \\ k &= \sqrt{ 1 - x^2 } \end{align*}$

Since these two unknown lengths together make the largest hypotenuse, that gives

$\displaystyle \begin{align*} k + l &= 2 \\ \sqrt{1 - x^2} + \sqrt{ 3 - x^2 } &= 2 \end{align*}$

So now all that's left is solving this equation for x :)