1. ## The Pythagorean Theorem, involving radical expressions

I need help from question 20 to 22!

And find X!

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

2. ## Re: 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.

3. ## Re: 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*}.

4. ## Re: The Pythagorean Theorem, involving radical expressions

Right - sorry for the typo.

5. ## Re: The Pythagorean Theorem, involving radical expressions

Originally Posted by ebaines
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.

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

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.

6. ## Re: 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*}

7. ## Re: 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!

8. ## Re: 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