Geometry(similarity) ?

How can I solve for PA on this problem?

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  • 4 weeks ago
    Favourite answer

    OR=sqr(12^2+5^2)=13

    Tri. AOP ~ Tri. AOR

    =>

    AP/AR=AO/OR

    =>

    AP/12=5/13

    =>

    AP=5(12)/13~4.62.

  • Anonymous
    4 weeks ago

    By areas.  Area of ROA = 30 sq units (5 x 12/2)

    base  RA = square root 169 = 13, 

    height is PA, half base = 6.5

    30/6.5 = 4.62 units

  • Philip
    Lv 6
    4 weeks ago

    PA/OA = 12/OR = 12/(12^2+5^2)^(1/2) = 12/(144+25)^(1/2) = 12/13;

    PA =(12/13)OA=(12/13)5=(60/13)=4.615384615=4.62 rounded to nearest hundredth 

  • 4 weeks ago

    Whatever the triangle, the sum of the 3 angles is always 180 °. So in the triangle SQU, you can write:

    In the right triangle (ORA):

    yellow + red + 90 = 180

    yellow = 90 - red

    In the right triangle (OPA):

    blue + yellow + 90 = 180

    blue = 90 - yellow

    blue = 90 - (90 - red)

    blue = red

    AP = AO.cos(blue) → we've just seen that: blue = red

    AP = AO.cos(red) → where: cos(red) = AR/OR

    AP = AO.(AR/OR)

    AP = 5 * (12/13)

    AP ≈ 4.615

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  • 4 weeks ago

    OR=sqrt(12^2+5^2)=13

    12/13=.9230

    .9230*5=  4.615

  • 4 weeks ago

    For rectangle BOAR:

    OA = BR = 5  and  BO = RA = 12

    In ΔROA:

    OR² = OA² + RA²  (Pythagorean theorem)

    OR = √(5² + 12²)

    OR = √13

    Area of ΔROA:

    (1/2) × PA × OR = (1/2) × OA × RA

    (1/2) × PA × 13 = (1/2) × 5 × 12

    PA = 5 × 12 / 13

    PA = 4.62

  • 4 weeks ago

    RO = 13

    The area of ∆ RAO = 30, 

    RO/2 * PA = 30 

    PA = 60/13 = 4.62

    ∆RPA is similar to  ∆PAO

    OA/RA = PA/RP

    5/12 = PA/RP

  • 4 weeks ago

    I'm guessing a bit at this one, but OA is 5 (same as BR)

    By pythagorean theorem, PO^2 + PA^2 = OA^2,

    Thus PO^2 + PA^2 = 25

    Ratio of PO to PA is same as the larger triangle (ie same as BR to BO) so 5 to 12

    Thus (PO) * 12 = (PA) * 5.

    I've gotten you 2 equations, you figure it out from there. 

  • 4 weeks ago

    Since RO cuts the rectangle in half through the diagonal, ROA and BRO are identical.

    So we need the height of the triangle with lengths 5 and 12 when the hypotenuse is the base.

    Let's do this.  If we put the triangle upright, we have the base of 5 and height of 12.  We can find the area:

    A = bh/2

    A = 5(12)/2

    A = 30 unit²

    We can use Pythagarean theorem to find the length of the hypotenuse:

    a² + b² = c²

    5² + 12² = c²

    25 + 144 = c²

    169 = c²

    13 = c

    With the same triangle on its hypotenuse for a base, the area is unchanged from when we solved for it above.  We can use that information to solve for the unknown height:

    A = bh/2

    30 = 13h/2

    60 = 13h

    h = 60/13

    h = 4.62 units (rounded to 2DP)

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