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Math Help - Area of Triangles Problem

  1. #1
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    Red face Area of Triangles Problem

    Hello! I have another trigonometry problem. What I have done is divide the pond into two triangles, and calculated them separately using Heron's formula for the non-right-angle triangle and the standard formula for the area of a triangle.

    The surface of a fish pond has the shape shown in the diagram (attached). How many goldfish can the pond support if each fish requires 0.3 square metres surface area of water?


    All measurements are in metres. I keep getting an answer of 20 goldfish; the textbook says that the answer is 17 goldfish.

    Thank you in advance. Help is much needed.
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  2. #2
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    Quote Originally Posted by JadeKiara View Post
    Hello! I have another trigonometry problem. What I have done is divide the pond into two triangles, and calculated them separately using Heron's formula for the non-right-angle triangle and the standard formula for the area of a triangle.

    The surface of a fish pond has the shape shown in the diagram (attached). How many goldfish can the pond support if each fish requires 0.3 square metres surface area of water?


    All measurements are in metres. I keep getting an answer of 20 goldfish; the textbook says that the answer is 17 goldfish.

    Thank you in advance. Help is much needed.
    Hi JadeKiara,

    you can use Pythagoras' theorem to find the length of the hypotenuse
    of the right-angled triangle.

    Then you have all 3 sides of the other.
    You only need one angle and you can find one using the cosine rule.

    Then calculate both areas independently.
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  3. #3
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    According to Herons formula, area of the triangle is
    A = sqrt[5.62*3.38*1.62*0.62)
    To get total area add the area of right angled triangle..
    Now proceed.
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  4. #4
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    Quote Originally Posted by JadeKiara View Post
    Hello! I have another trigonometry problem. What I have done is divide the pond into two triangles, and calculated them separately using Heron's formula for the non-right-angle triangle and the standard formula for the area of a triangle.

    The surface of a fish pond has the shape shown in the diagram (attached). How many goldfish can the pond support if each fish requires 0.3 square metres surface area of water?


    All measurements are in metres. I keep getting an answer of 20 goldfish; the textbook says that the answer is 17 goldfish.

    Thank you in advance. Help is much needed.
    By Pythagoras' Theorem, the unknown length is \sqrt{5}.


    The area of the Right Angle Triangle is \frac{1}{2}\cdot 1\cdot 2 = 1.

    To find the other area, you need Heron's Formula.

    s = \frac{\sqrt{5} + 4 + 5}{2}

     = \frac{9 + \sqrt{5}}{2}.


    A = \sqrt{\left(\frac{9 + \sqrt{5}}{2}\right)\left(\frac{9 + \sqrt{5}}{2} - \sqrt{5}\right)\left(\frac{9 + \sqrt{5}}{2} - 4\right)\left(\frac{9 + \sqrt{5}}{2} - 5\right)}

     = \sqrt{\left(\frac{9 + \sqrt{5}}{2}\right)\left(\frac{9 - \sqrt{5}}{2}\right)\left(\frac{1 + \sqrt{5}}{2}\right)\left(\frac{-1 + \sqrt{5}}{2}\right)}

     = \sqrt{\left(\frac{81 - 5}{4}\right)\left(\frac{5 - 1}{4}\right)}

     = \sqrt{\left(\frac{76}{4}\right)\left(\frac{4}{4}\r  ight)}

     = \sqrt{\left(\frac{76}{4}\right)}

     = \frac{2\sqrt{19}}{2}

     = \sqrt{19}.


    So the total area is (1 + \sqrt{19})\,\textrm{m}^2.

    How many times does 0.3 go into 1 + \sqrt{19}?
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  5. #5
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    Hello, JadeKiara!

    What I have done is divide the pond into two triangles,
    and calculated them separately using Heron's formula for the non-right-angle triangle
    and the standard formula for the area of a triangle.

    The surface of a fish pond has the shape shown in the diagram (attached).
    How many goldfish can the pond support if each fish requires 0.3 mē surface area of water?

    All measurements are in metres.
    I keep getting an answer of 20 goldfish;
    the textbook says that the answer is 17 goldfish.
    Code:
             1
          * - - *
          |    * *
          |   * _ *
         2|  * √5  *
          | *       * 5
          |*         *
          *           *
              *        *
                4 *     *
                      *  *
                          *

    The right triangle has area 1.

    The oblique triangle has area \sqrt{19}.

    . . The total area is: . 1 + \sqrt{19} \;\approx\;5.359 mē.


    This will support \frac{5.359}{0.3} \:=\:17.863... \;\approx\;17 goldfish.

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