Results 1 to 8 of 8
Like Tree1Thanks
  • 1 Post By ebaines

Math Help - The Pythagorean Theorem, involving radical expressions

  1. #1
    xwy
    xwy is offline
    Newbie
    Joined
    Apr 2014
    From
    singapore
    Posts
    15

    The Pythagorean Theorem, involving radical expressions

    The Pythagorean Theorem, involving radical expressions-pythagoras.jpg
    I need help from question 20 to 22!

    The Pythagorean Theorem, involving radical expressions-pythagoras-2.jpg
    And find X!

    Please list down the solution! i got stuck in these two question for very long!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,050
    Thanks
    288

    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.
    Thanks from HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: The Pythagorean Theorem, involving radical expressions

    I'm sure you mean $\displaystyle \begin{align*} a^2 = \left( \frac{a}{2} \right) ^2 + h^2 \end{align*}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,050
    Thanks
    288

    Re: The Pythagorean Theorem, involving radical expressions

    Right - sorry for the typo.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    xwy
    xwy is offline
    Newbie
    Joined
    Apr 2014
    From
    singapore
    Posts
    15

    Re: The Pythagorean Theorem, involving radical expressions

    Quote Originally Posted by ebaines View Post
    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.
    The Pythagorean Theorem, involving radical expressions-question-20-21.jpg
    Thanks for the reply! I am basically lost at squarroot(a^2-b^2) from my attempt as you can see.

    The Pythagorean Theorem, involving radical expressions-question-29.jpg
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    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:

    $\displaystyle \begin{align*} \sqrt{\frac{3a^2}{4}} &= \frac{\sqrt{3}\sqrt{a^2}}{\sqrt{4}} \\ &= \frac{\sqrt{3}\,a}{2} \end{align*}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    xwy
    xwy is offline
    Newbie
    Joined
    Apr 2014
    From
    singapore
    Posts
    15

    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!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Radical Expressions in Algebra.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 29th 2013, 09:18 AM
  2. Radical Expressions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 17th 2009, 08:42 AM
  3. Rational Expressions and Radical Expressions
    Posted in the Algebra Forum
    Replies: 16
    Last Post: July 9th 2009, 09:29 PM
  4. Radical Pythagorean Theorem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 13th 2009, 04:51 PM
  5. radical expressions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 24th 2007, 05:27 PM

Search Tags


/mathhelpforum @mathhelpforum