Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

please help me solve this geometry problem? thanks!?

two vertiacal poles of height 2m and 8m are 10m apart. what is the height of the intersection of the lines joining the top of each pole to the foot of the other?

3 Answers

Relevance
  • 7 years ago
    Favorite Answer

    If the height of the intersection point of the two lines from the base is h m, then

    1/h = 1/2 + 1/8 = 5/8

    ==> Height h = 8/5 = 1.6 m

    Solution:

    i) Let AB be the pole of height 8 m and CD be the pole of 2 m height. The distance between them is AC, which is 10 m (This is not relevant here).

    ii) Let the lines joining AD and BC intersect at P; PQ be the height of the intersecting point P from base AC.

    iii) Triangles, AQP & ACD are similar;

    so, PQ/DC = AQ/AC ------- (1)

    Similarly from similar triangles, CQP & CAB, PQ/BA = CQ/CA ------ (2)

    Adding (1) & (2):

    PQ(1/AB + 1/CD) = (AQ + QC)/AC = AC/AC = 1

    ==> 1/PQ = 1/AB + 1/CD = 1/8 + 1/2 = 5/8

    Hence PQ = 8/5 = 1.6 m

  • 7 years ago

    You can think of this question as poles in cartesian plane with the pole with 2m is at the origin while the pole with 8m is the right of the 2m pole being 10m apart.

    That being said, this means the top of the 2m pole is at (0,2) and the bottom is at (0,0)

    On the other hand, the top of the 8m pole is at (10,8) and the bottom is at (10,0)

    Line A = (0,2) to (10,0) and Line B = (0,0) to (10,8)

    Let's get the slope of these lines

    Line A --> m = (0-2)/(10-0) = -1/5

    Line B --> m = (8-0)/(10-0) = 4/5

    With the equation (y-y1) = m(x-x1), let's get the equations of these lines

    Line A eqn --> (y-2) = -1/5(x-0) --> y = -x/5 + 2

    Line B eqn --> (y-0) = 4/5(x-0) --> y = 4x/5

    Equate these two lines -x/5 + 2 = 4x/5

    -x/5 - 4x/5 = -2

    -x-4x/5 = -2

    -x-4x = -10

    -5x = -10

    x = 2

    Now we want to get the y so substitute x to any of the line equations

    Using Line A eqn...... y = -2/5 + 2

    y = 8/5

    The height of the intersection of the lines is 8/5 meters.

    Source(s): Math II
  • David
    Lv 7
    7 years ago

    using coordinates

    (0, 2) (10, 0)

    m = (0 - 2) / (10 - 0)

    m = -2/10

    m = -1/5

    y - 2 = (-1/5)x

    5y - 10 = -x

    x + 5y = 10

    (0, 0) (10, 8)

    m = 8/10

    m = 4/5

    y = (4/5)x

    5y = 4x

    4x - 5y = 0

    the two lines intersect at

    x + 5y = 10

    4x - 5y = 0

    x = 10 - 5y

    4(10 - 5y) - 5y = 0

    40 - 20y - 5y = 0

    -25y = -40

    y = -40/-25

    y = 8/5 ft

    1.6 ft

    Attachment image
Still have questions? Get your answers by asking now.