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Math Help - Please Help me with this..(Areas of Plane Figures)

  1. #1
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    Please Help me with this..(Areas of Plane Figures)

    Find the area of each figures;
    1. a rhombus with shorter diagonal 6cm and a 120 angle.

    2. A regular hexagon with side 12.

    3. A regular pentagon with side S and apothem 3.

    4. The ratio of the areas of two rectangles is 36 : 64. Find the scale factor and the ratio of the perimeters.

    5. Two similar polygons have scale factor 3 : 4, the area of the smaller polygon is 108. Find the area of the larger polygon.

    *Please explain a brief statement of how to find the answers... ㅠㅠ...
    Thank- U so much...
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  2. #2
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    Quote Originally Posted by sarayork View Post
    Find the area of each figures;
    1. a rhombus with shorter diagonal 6cm and a 120 angle.

    2. A regular hexagon with side 12.

    ...
    to #1:

    The shorter diagonal bisect the greater angle. Therefore you are dealing with 4 right triangles with the shorter leg of 3 cm and the longer leg is

    l = 3\ cm \cdot \tan(60^\circ) = 3 \sqrt{3} \ cm.

    The area is then A = 4 \cdot \underbrace{(3 \cdot 3\sqrt{3})}_{\text{area of one triangle}}\ cm^2 = 36 \sqrt{3}\ cm^2

    to #2:

    A regular hexagon consists of 6 equilateral triangles. If the length of one side of the triangle is a then the height of the triangle is: h = \frac12 \cdot a \cdot \sqrt{3}

    Therefore the area of an equilateral triangle is:

    A = \frac12 \cdot a \cdot \underbrace{\frac12 \cdot a \cdot \sqrt{3} }_{height}= \frac{a^2}{4} \sqrt{3}

    And finally the area of a regular hexagon is:

    A_6=6 \cdot \frac{a^2}{4} \sqrt{3} = \frac32 \cdot a^2 \sqrt{3}. Plug in the value of a (= 12) to calculate the area.
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  3. #3
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    Quote Originally Posted by sarayork View Post
    ...

    4. The ratio of the areas of two rectangles is 36 : 64. Find the scale factor and the ratio of the perimeters.

    5. Two similar polygons have scale factor 3 : 4, the area of the smaller polygon is 108. Find the area of the larger polygon.

    ...
    to #4:

    If the lengthes(?) of corresponding sides of similar plane figures have the ratio

    \frac{l_1}{l_2}=\frac ab then the ratio of the areas is \frac{A_1}{A_2}=\frac{a^2}{b^2}

    If the areas of similar plane figures have the ratio

    \frac{A_1}{A_2}=\frac pq then the ratio of the lengthes(?) of corresponding sides is \frac{l_1}{l_2}=\frac{\sqrt{p}}{\sqrt{q}}

    Because \frac{A_1}{A_2} = \frac{36}{64} the ration of the perimeters is: \frac{p_1}{p_2} = \frac{6}{8}

    to #5:

    Calculate first the ratio of the areas and then the actual value of the greater area.

    (For confirmation only: A_2 = 192)


    EDIT: I would have helped you with #3 but I couldn't find the word "apothem" in my dictionaries. Is it maybe the radius of the inscribed circle of the pentagon?
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  4. #4
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    Quote Originally Posted by earboth View Post
    EDIT: I would have helped you with #3 but I couldn't find the word "apothem" in my dictionaries. Is it maybe the radius of the inscribed circle of the pentagon?
    I don't have time to answer questions right now, but see here: apothem.

    -Dan
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  5. #5
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    Quote Originally Posted by sarayork View Post
    ...
    3. A regular pentagon with side S and apothem 3.

    ...
    A regular 5-gon consists of 5 isosceles triangles with the base s.
    The apothem is the height h perpendicular to the base s.
    The height bisect the angle at the top and the base s.
    This angle at the top is \tau = \frac{360^\circ}{5} = 72^\circ.

    To calculate the length of the base you use half of the isoscele triangle which is a right triangle:

    \tan(36^\circ) = \frac{\frac12 s}{h}~\implies~ s = 2h\cdot \tan(36^\circ) \approx 4.359...
    The area of one triangle is calculated by:

    A_{triangle} = \frac12 \cdot s \cdot h= \frac12 \cdot 2h\cdot \tan(36^\circ) \cdot h

    And therefore the area of the 5-gon is:

    A_5 = 5h^2 \cdot \tan(36^\circ) which means here: A_5=45 \cdot \tan(36^\circ) \approx 32.6944
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