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Homework Help: Im Stuck with Geometry Please Help

  1. Jun 16, 2008 #1
    Im trying to teach myself some geometry on my own, but im stuck since some weeks, with a couple of area problems of triangles. I ll really appreciate any help or advice about it.

    The formulas for a triangle are:

    A=1/2 bh and A=1/2 ab Sen C

    The problem says:
    -Find the area of a triangle that has two adyacent sides that are 5 and 4 ,and they include an angle that is 45º

    I thought that the formula might be A=1/2 ab Sen C

    so it would be A= 1/2 (4)(5)(sen 45) = 7.07106...

    but the result on the book was = 5 square root of 2

    Could I use "sen of 45" in that way? if not what is the correct procedure in order to do that problem?

    There are some other similar problems that I cant understand as well:

    -Find the area of a triangle that has two adjacent sides that are 8 and 12, and they include an angle that is 60º.

    -Find the area of a triangle if two sides are 13 and 15, and the hight over the third side is 12.

    Im stuck with those ones since some weeks, I 'll really appreaciate any help.

    Thanks a lot in advance.
  2. jcsd
  3. Jun 16, 2008 #2
    Your answer is correct, but just expressed differently. Note that sin(45) = sqrt(2)/2--verify with a table of trigonometric values. A calculator does not explicitly reveal the factor, square root of 2.
  4. Jun 16, 2008 #3
    Thanks a lot, I appreaciate it.

    Im trying to solve this one now:

    Find the area of a triangle if two sides are 13 and 15, and the hight over the third side is 12.

    Do I need to find the base in the problem? Should I use the pytagorean theorem to get the base in this problem? and if so what should be the procedure?

    Thanks in advance for your time and help.
  5. Jun 17, 2008 #4


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    Homework Helper

    Yes. Find the base. Use the Pythagorean theorem. You have two right triangles with hypotenuses 13 and 15 and side 12. The sum of the other two sides is the base.
  6. Jun 17, 2008 #5
    Use trigonometry my friend.
  7. Jun 17, 2008 #6
    thanks a lot
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