Results 1 to 5 of 5

Math Help - Quadratics Problems

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    114

    Quadratics Problems

    Question No. 1

    In a right angled triangle, the second to longest side is 5 cm longer than the shortest side, and the hypotenuse is three times longer than the shortest side. Find exact length of the hypotenuse.

    (The following image is what I've visualized it to be - is it correct?)



    This is how I tried solving it and the answer was incorrect:

    a^2+b^2=c^2
    x^2+(x+5)^2=3x^2
    x^2+x^2+10x+25=3x^2


    Question No. 2

    A straight length of wire is 20 cm long. It is bent at the right angles to form the two shorter sides of a right angled triangle. If the triangle's area is 30cm^2, find:

    1. the length of the hypotenuse
    2. The triangles perimeter

    ^ I have no idea how to do either of those questions and am not sure how to visualize it either.


    Thank you so much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    For number 1, you forgot to square 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    x^2+ (x+5)^2 = (3x)^2

    x^2 +x^2 +10x +25 = 9x^2

    0 =7x^2-10x-25

    Use quad formula to solve for x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    792
    I'll list the given for number 2 and give you some tips:
    let a and b denote the "legs" of the right-angled triangle, c denotes the hypotenuse.

    a + b = 20~~~ [1]

    \frac{1}{2}ab = 30 \Rightarrow ab = 60

    Square equation number one, then remember the Pythagorean Theorem:
    a^2 + b^2 = c^2

    Once you have found the hypotenuse, the perimeter of the triangle is:
    P = a+b+c
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418
    Quote Originally Posted by juliak View Post
    Question No. 1

    In a right angled triangle, the second to longest side is 5 cm longer than the shortest side, and the hypotenuse is three times longer than the shortest side. Find exact length of the hypotenuse.

    (The following image is what I've visualized it to be - is it correct?)



    This is how I tried solving it and the answer was incorrect:

    a^2+b^2=c^2
    x^2+(x+5)^2=3x^2
    x^2+x^2+10x+25=3x^2


    Question No. 2

    A straight length of wire is 20 cm long. It is bent at the right angles to form the two shorter sides of a right angled triangle. If the triangle's area is 30cm^2, find:

    1. the length of the hypotenuse
    2. The triangles perimeter

    ^ I have no idea how to do either of those questions and am not sure how to visualize it either.

    Thank you so much.
    For question 2, you're only using the wire to create the two shorter sides.

    So call x the length of one of those sides. The other is therefore 20 - x.

    The area of a triangle is given by

    A=\frac{1}{2}bh.

    So we find 30=\frac{1}{2}x(20-x).

    30=10x-\frac{1}{2}x^2.

    Solve for x, then you have the two shorter lengths. Use Pythagoras to find the hypotenuse, and then add the 3 side lengths together to find the triangle's perimeter.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratics
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 25th 2011, 12:19 AM
  2. Quadratics
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: June 25th 2010, 09:54 AM
  3. Need help with 2 quadratics problems...
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 14th 2009, 04:54 PM
  4. Another quadratics!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 27th 2008, 05:37 AM
  5. Quadratics! need (more) help!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 26th 2008, 10:24 AM

Search Tags


/mathhelpforum @mathhelpforum