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Math Help - Quadratics

  1. #1
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    Quadratics

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

    (one of the problems I have is that I can't really visualize the triange. If there are two right angles the its not a triangle! O_O)
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  2. #2
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    Hi -

    I'm not quite sure what the problem is that you have in visualising this. You simply take a straight piece of wire and bend it into a right-angle somewhere along its length!

    Nothing more than that. The two parts of the wire are now at right-angles to one another. So they form the two shorter sides of a right-angled triangle. An imaginary line joining the two ends of the wire forms the hypotenuse of the triangle.

    To solve the problem, let's suppose that the 20cm length is now divided into two parts, one of length x cm and the other of length (20 - x) cm.

    Then use the area formula Area = (half base times height) to work out the area in terms of x. Put the answer equal to 30, and you will have a quadratic equation in x. Solve this, and you can work out the length of the hypotenuse and the perimeter.

    I hope that helps.
    Grandad
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  3. #3
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    I worked out

    x=10 +/- root(85)

    but i dont know how to work out the answer from that

    because there's two possible answers..

    and the answer in the book says

    i. 2 root(70)

    ii.20+ 2 root(70)
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  4. #4
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    Quote Originally Posted by juliak View Post
    I worked out

    x=10 +/- root(85)

    but i dont know how to work out the answer from that

    because there's two possible answers..

    and the answer in the book says

    i. 2 root(70)

    ii.20+ 2 root(70)
    It would help if you showed your working.

    You have \frac{x}{2} (20 - x) = 30 which after a bit of algebra becomes x^2 - 20x + 60 = 0. Solve for x. There are two possible values but it won't matter which one you choose (why?).

    Then from Pythagoras x^2 + (20 - x)^2 = h^2. Substitute your value of x and solve for h (the hypotenuse).
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  5. #5
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    Quadratics

    Hi -

    As it turns out, you don't actually need to solve the quadratic. The equation is indeed:

    <br />
x^2-20x+60=0<br />

    and in (i) you need the hypotenuse h where:

    <br />
h^2=x^2+(20-x)^2<br />

    and when you remove the brackets and simplify, you get:

    <br />
h^2=2x^2-40x+400<br />

    But if you look closely at the quadratic, you'll see that

    <br />
x^2-20x=-60<br />

    \therefore 2x^2-40x=-120

    So we can substitute this into the expression for h^2

    <br />
h^2=-120+400=280<br />

    This gives
    <br />
h=2\sqrt{70}<br />

    For (ii) - the perimeter - you simply have to add on the length of the original wire: Answer: <br />
20+2\sqrt{70}<br />
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    It would help if you showed your working.

    You have \frac{x}{2} (20 - x) = 30 which after a bit of algebra becomes x^2 - 20x + 60 = 0. Solve for x. There are two possible values but it won't matter which one you choose (why?).

    Then from Pythagoras x^2 + (20 - x)^2 = h^2. Substitute your value of x and solve for h (the hypotenuse).
    but the answer is a really wonky number
    like, i square X even if its something with lots and lots of decimals?
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  7. #7
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    Quote Originally Posted by juliak View Post
    but the answer is a really wonky number
    like, i square X even if its something with lots and lots of decimals?
    You're meant to use algebra to solve x^2 - 20x + 60 = 0 and get exact values for x.

    But Grandad has shown another approach that avoids any algebraic difficulty.
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