Results 1 to 7 of 7

Math Help - finding the area of abd

  1. #1
    Senior Member
    Joined
    Aug 2009
    Posts
    349

    finding the area of abd

    finding the area of abd-imag0029.jpg

    i understand how to find area and i understand that in an equilateral triangle all sides are equal. the trouble i am finding with this particular problem is that i do not know how to solve without using plug/chug. if there is a faster way of doing this problem i definitely would like to know...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2011
    Posts
    68

    Re: finding the area of abd

    Quote Originally Posted by slapmaxwell1 View Post
    Click image for larger version. 

Name:	IMAG0029.jpg 
Views:	29 
Size:	970.6 KB 
ID:	23064
    the trouble i am finding with this particular problem is that i do not know how to solve without using plug/chug.
    What do you mean by that? Do you already know how to solve it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Aug 2009
    Posts
    349

    Re: finding the area of abd

    Quote Originally Posted by scounged View Post
    What do you mean by that? Do you already know how to solve it?
    i thought i did, but its not working out...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2011
    Posts
    68

    Re: finding the area of abd

    Ok. In the problem it says that |AB| = 2|BC|. I'd solve this one by letting |BC| equal a, and calculate the area of the triangles ACE and ABD, in terms of a. Can you do that?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Aug 2009
    Posts
    349

    Re: finding the area of abd

    ok im still having trouble with this problem...i figured out that one side is sqrt(5), but then thats it. so the sides of the shades triangle are 1*k, 2*k and sqrt(5)*k.

    im using k because it is saying its a ratio. so what i am asking is this, how can i use this info to find the area of the triangle abd?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,110
    Thanks
    68

    Re: finding the area of abd

    Ok; to start, you do not need triangle BCF; disregard it.

    Let a = BC; then AB = 2a (ratio 1:2) ; still with me?
    Let AC = b

    Using triangle ACE:
    since sides = b, area = b^2 SQRT(3) / 4 ; since that area is given as 20:
    b^2 SQRT(3) / 4 = 20
    b = SQRT[80 / SQRT(3)] ; that'll give you b = ~6.796

    Using right triangle ABC:
    b^2 = a^2 + (2a)^2
    a = SQRT(b^2 / 5) ; that'll give you a = ~4.298

    Using triangle ABD:
    since sides = 2a, area = a^2 SQRT(3) ; that'll give you area = 16.00000000....

    Hope you followed...come back with questions if not.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,825
    Thanks
    713

    Re: finding the area of abd

    Hello, slapmaxwell1!

    Code:
                      A            E
                      o  *  *  *  o
                   *  |\         *
                *     | \   20  *
           D o       2|  \     *
                *     |   \   *
                   *  | 1  \ *
                    B o-----o C
                       *   *
                        * *
                         o
                         F
    ABC is a right triangle with BC\!:\!AB \,=\,1\!:\!2
    AC\!E, AB\!D, BC\!F are equilateral triangles.
    The area of \Delta AC\!E = 20.
    Find the area of \Delta AB\1D.

    . . (A)\;18\qquad (B)\;16 \qquad (C)\;15 \qquad (D)\;14 \qquad (E)\;12

    Pythagorus says: AC = \sqrt{5}

    The ratio of the areas of two similar polygons
    . . is the square of the ratios of their sides.


    The ratio of the sides of \Delta AB\!D and \Delta AC\!E is. 2\!:\!\sqrt{5}

    The ratio of their areas is. 2^2\!:\!(\sqrt{5})^2 \:=\:4\!:\!5

    We have: . \frac{\Delta AB\!D}{20} \:=\:\frac{4}{5} \quad\Rightarrow\quad \Delta AB\!D \:=\:16\;\hdots\;\text{answer (B)}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the area..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 13th 2010, 02:29 PM
  2. finding the area
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 18th 2009, 12:22 PM
  3. finding area
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 21st 2009, 04:20 AM
  4. Finding area
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 20th 2008, 04:39 PM
  5. Finding area
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 14th 2008, 06:46 PM

Search Tags


/mathhelpforum @mathhelpforum