Question:

Geometry problem, triangles. ?

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http://img.photobucket.com/albums/v374/ashleyee100/Image1-3.jpg

That is a screen shot of the problem I need to solve, only I have no idea as to how to go about doing it. Any help is greatly appreciated!!

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  1. There is a theorem in geometry that says:

    Given a right triangle and the altitude to the hypotenuse,

        The altitude is the geometric mean of the segments into which it

        divides the hypotenuse.

    From the drawing, h (the altitude), divides the hypotenuse into 16 and 9.

    Hence, following the above theorem, we can write,

                h = sqrt[16(9)]

                h = 12  ANSWER


  2. well, the entire large triangle is a 3-4-5 right triangle. Im assuming you've already learned about those. if not, it basically means that a right triangle with its 2 legs having a length of a multiples of 3 and 4 and will have a hypotenuse that is a multiple of 5. For example, a 6-8-10 triangle is a 3-4-5 triangle b/c its lenghts are all just multiplied by 2.  Therefore you can find out what are the lengths of the remaining legs of the large triangle, which are 15 and 20.  You also know that the length of the base of the left small triangle is 9 (25 minus16). and WOW! it's another 3-4-5 triangle! and OMG, so is the triangle on the right side! so now you can find h, which is 12.  

  3. I got the same answer as everyone else, but I did it in a different way. First, I set the length of the left, slanted side equal to x. Then I call the side opposite that y. From that, and using the Pythagorean theorem, I was able to come up with the following 3 equations: (^2 means squared)

    h^2+9^2=x^2

    h^2+16^2=y^2

    x^2+y^2=25^2

    After this, I took the first side of the first two equations and plugged it into the third equation. Here is what the final equation looked like:

    h^2+9^2+h^2+16^2=25^2

    simplifiy squares:.

    h^2+81+h^2+256=625

    combine like terms:

    2h^2=288

    divide by 2:

    h^2=144

    take square root:

    h=12

    I hope I helped

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