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Math Help - Triangle and Trapezium

  1. #1
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    Triangle and Trapezium

    I need help in the below question.
    The figure below, not drawn to scale, is made up of a right-angle triangle ABC, and a trapezium BCDE. Find the difference between the sum of the areas of the unshaded parts in the figure and the area of the shaded trapezium.
    Attached Thumbnails Attached Thumbnails Triangle and Trapezium-trapezium.jpg  
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  2. #2
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    Re: Triangle and Trapezium

    Quote Originally Posted by kingman View Post
    I need help in the below question.
    The figure below, not drawn to scale, is made up of a right-angle triangle ABC, and a trapezium BCDE. Find the difference between the sum of the areas of the unshaded parts in the figure and the area of the shaded trapezium.
    1. Since |AB| = |BC| the length of CD is |CD| = 4

    F is the midpoint of AB, G is the midpoint of AC.

    The area of \Delta ABC has the same value as the area of the parallelogram CBEG.

    2. Therefore the 2 red triangles have the same area.

    The shaded region, the trapezium BFGC, has an area of \frac34 of the triangle ABC. And that is the half of a square.
    Attached Thumbnails Attached Thumbnails Triangle and Trapezium-trap_triangle.png  
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  3. #3
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    Re: Triangle and Trapezium

    Quote Originally Posted by kingman View Post
    I need help in the below question.
    The figure below, not drawn to scale, is made up of a right-angle triangle ABC, and a trapezium BCDE. Find the difference between the sum of the areas of the unshaded parts in the figure and the area of the shaded trapezium.
    Please explain why "Since |AB| = |BC| the length of CD is |CD| = 4

    F is the midpoint of AB, G is the midpoint of AC." when it is not given in the question.
    I tried using the idea but could not get the difference 8 cm square which is the answer
    thanks
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  4. #4
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    Re: Triangle and Trapezium

    Quote Originally Posted by kingman View Post
    Please explain why "Since |AB| = |BC| the length of CD is |CD| = 4

    F is the midpoint of AB, G is the midpoint of AC." when it is not given in the question.
    I tried using the idea but could not get the difference 8 cm square which is the answer
    thanks
    Touché!

    I've modified your sketch a little bit. (see attachment)

    1. Since |AB| = |BC| the length of CD is |CD| = |DG| = k

    The area of the shaded region (a trapezium) is

    a_{shaded\ region}=\frac{8+(8-k)}2 \cdot k = 8k-\frac12 k^2

    2. The area of the trapezium CBED is

    a_{trapezium\ CBED} = \frac{8+(8+k)}2 \cdot k = 8k+\frac12 k^2

    3. The area of the region with green border is:

    a_{trapezium\ CBED} + a_{\Delta GFA} = 8k+\frac12 k^2 + \underbrace{\frac12 (8-k)^2}_{a_{\Delta GFA}} = 32 +k^2

    4. The area of the triangle DCA is (only in case you need it to determine the difference of sums)

    a_{\Delta DCA} = \frac12 \cdot k \cdot 8 = 4k
    Attached Thumbnails Attached Thumbnails Triangle and Trapezium-trap_triangle2.png  
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  5. #5
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    Re: Triangle and Trapezium

    Quote Originally Posted by earboth View Post
    Touché!

    I've modified your sketch a little bit. (see attachment)

    1. Since |AB| = |BC| the length of CD is |CD| = |DG| = k

    The area of the shaded region (a trapezium) is

    a_{shaded\ region}=\frac{8+(8-k)}2 \cdot k = 8k-\frac12 k^2

    2. The area of the trapezium CBED is

    a_{trapezium\ CBED} = \frac{8+(8+k)}2 \cdot k = 8k+\frac12 k^2

    3. The area of the region with green border is:

    a_{trapezium\ CBED} + a_{\Delta GFA} = 8k+\frac12 k^2 + \underbrace{\frac12 (8-k)^2}_{a_{\Delta GFA}} = 32 +k^2

    4. The area of the triangle DCA is (only in case you need it to determine the difference of sums)

    a_{\Delta DCA} = \frac12 \cdot k \cdot 8 = 4k


    Everthing is blurred on the solution page ; so cannot read what is written on the page.
    Thanks
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