OT- Triangle Area Question / 6th Grade Math
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  1. #1
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    Default OT- Triangle Area Question / 6th Grade Math

    OK....it's been many decades since I was in 6th Grade, and no, it was not my final year of schooling.

    So here's my dilemma - the problem below is "Solve for the area of this triangle".

    My position is that it can't be solved as there is not enough information given.

    Am I wrong?

    triangle.jpg

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    Yes...if it is a right angle triangle. But that one is clearly not.

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    It's 1/2 of a parallelogram, so answer is 36.

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    I think it's missing an angle somewhere. You can draw the hypoteneuse of the small implied right triangle (the one you need
    to the area of to solve the overall problem) at any angle.

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    Quote Originally Posted by GregSY View Post
    Yes...if it is a right angle triangle. But that one is clearly not.
    The height is given as perpendicular to the length and that's what matters here. If you were given the actual length of that non-perpendicular side then you would need more info.

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    Quote Originally Posted by GregSY View Post

    My position is that it can't be solved as there is not enough information given.

    Am I wrong?

    triangle.jpg
    Not IMO, for without an angle or length of hypotenuse, that is nothing more than a drawn shape.

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    jancollc and BugRobotics are correct. (Height x Width)/2

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    Quote Originally Posted by BugRobotics View Post
    The height is given as perpendicular to the length and that's what matters here. If you were given the actual length of that non-perpendicular side then you would need more info.
    This answer would allow me to reach an answer....but I'm not sure it makes sense just yet!

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    36 square inches

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    Area = 1/2 x Base x Height

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    Quote Originally Posted by jancollc View Post
    It's 1/2 of a parallelogram, so answer is 36.
    Yep.

    Area of a parallelogram is base x height ie 12 x 6 = 72 in that case. The triangle is half of the parallelogram so area is 36.

    Of course you can figure out everything else about the triangles via the area of the parallelogram. Can be a very useful approach when you don't have enough information to go straight to trigonometry.

    Clive

    Need to work on my typing speed!

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    Oh well, that's something I've learned today.

    Thanks guys

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    If you had an accurate drawing to scale it could be solved by drafting methods plus trigonometry. Once divided into two triangles with one 90 degree corner the sides could be measured and each triangle would have 1/2 the area of a rectangle with the same dimensions. The unknown angles could be found using trigonometry.

    If a photo were imported as a layer into 2D cad you could draw a line from that corner tangent to the longest dimension and you would have your right angles. How you digitize the drawing is to place coordinates at the corners while zoomed in and then play connect-the-dots.

    I used to solve unknown geometry by using dividers on a photo and then transferring to paper. On one occasion I needed to "borrow" the dimensions of a motorcycle over-center stand from a photo in a magazine article. These stands are used for field maintenance such as removing wheels and have to be simple, light, and portable. The dimensions are criitical to ease of use.

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    To solve a trig problem like that one needs a combination of side lengths and angles.

    1. Two sides and one angle.

    2. Two angles and one side.


    Sent from my iPhone using Tapatalk

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    Quote Originally Posted by GregSY View Post
    This answer would allow me to reach an answer....but I'm not sure it makes sense just yet!
    Maybe this will help...

    Take your triangle (blue), make another of the same size and orient it on top of the first (red) to make a parallelogram (parallelogram is 4-sided shape formed by two pairs of parallel lines).

    Area of a parallelogram is base * height. You have 1/2 of the parallelogram therefore... 1/2 * base(12) * height(6). So you can see how you've been given all the necessary info.

    parallelogram.jpg

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    Quote Originally Posted by Clive603 View Post
    Yep.

    Area of a parallelogram is base x height ie 12 x 6 = 72 in that case. The triangle is half of the parallelogram so area is 36.

    Of course you can figure out everything else about the triangles via the area of the parallelogram. Can be a very useful approach when you don't have enough information to go straight to trigonometry.

    Clive

    Need to work on my typing speed!
    I think you can't fiqure out rest of the triangle from the information given. Area is just constant even if the angles and dimensions vary wildly.

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    I was going to go along with all the others who saw the 6" as height, not length of a side, and the answer easy.

    But then I realized this was in English units and impossible to solve, unless it were metric . . .

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    Quote Originally Posted by jancollc View Post
    It's 1/2 of a parallelogram, so answer is 36.
    That's the area of a right triangle with a 6" side. Your answer overstates the area by
    some amount.

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    Quote Originally Posted by jim rozen View Post
    That's the area of a right triangle with a 6" side. Your answer overstates the area by
    some amount.
    Um, no.

    A rectangle is a parallelogram, but a parallelogram is not necessarily a rectangle.

    The formula for area of a parallelogram, as prev. posted, is A=bh. Since the triangle is 1/2 of the parallelogram, it's area is 1/2(bh).

    Information given, you cannot solve the full triangle. But you can solve for area no problem.

    It's not a trig problem- it's basic geometry.

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    for fucks sake, if you don't get it, draw it up in cad and analyze the area. It's 36

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