Please help. Thanks.?

Attachment image

3 Answers

Relevance
  • Pope
    Lv 7
    1 year ago
    Favorite Answer

    I get a = 7, b = 9, so a + b = 16.

    The way I derived that is a bit sloppy, so instead let me just prove that it satisfies all conditions.

    Let line segments AC and BD intersect at P, with these five parameters:

    PA = 2

    PB = 3

    PC = 9

    PD = 6

    cos(∠BPA) = 7/9

    These similar triangles follow:

    ∆APB ~ ∆DPC, with ratio 1 : 3

    ∆APD ~ ∆BPC, with ratio 2 : 3

    Also, as a result of the similar triangles, ABCD is a cyclic quadrilateral.

    AB² = PA² + PB² - 2(PA)(PB)cos(∠BPA)

    AB² = 2² + 3² - 2(2)(3)cos(7/9)

    AB² = 11/3

    AB = √(33)/3

    By the similar triangles (∆APB ~ ∆DPC, with ratio 1 : 3),

    CD = 3AB

    BC² = PB² + PC² - 2(PB)(PC)cos(∠CPB)

    BC² = 3² + 9² - 2(3)(9)cos(-7/9)

    BC² = 132

    BC = 2√(33) = 6AB

    By the similar triangles (∆APD ~ ∆BPC, with ratio 2 : 3),

    DA = 2/3BC = 4AB

    AB : BC : CD : DA

    = AB : 6AB : 3AB : 4AB

    = 1 : 6 : 3 : 4

    Construct midpoints E, F, and G as given. These parallel lines result:

    EF || BD

    GF || AC

    ∠EFG = ∠DPC ... (opposite angles of a parallelogram)

    ∠DPC = ∠BPA ... (vertical angles)

    cos(∠EFG) = cos(∠BPA) = 7/9

    Attachment image
    • Login to reply the answers
  • Anonymous
    1 year ago

    It’s something to do with math...am I on the right track?

    • Login to reply the answers
  • Anonymous
    1 year ago

    ewuvogzl

    • Login to reply the answers
Still have questions? Get your answers by asking now.