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Math Help - This problem is not so easy

  1. #1
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    Thumbs down This problem is not so easy

    Let ABC be a right triangle at A. AB=3, AC=4. MNPQ is a rectangle that is in the triangle ABC. Represent X equal AM. Calculate X for MNPQ be square.[IMG][img]C:\Documents and Settings\admin\Desktop\picture[/img][/IMG]
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    Quote Originally Posted by mpgc_ac
    Let ABC be a right triangle at A. AB=3, AC=4. MNPQ is a rectangle that is in the triangle ABC. Represent X equal AM. Calculate X for MNPQ be square.[IMG][img]C:\Documents and Settings\admin\Desktop\picture[/img][/IMG]
    Your link is faulty.

    -Dan
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  3. #3
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    Quote Originally Posted by mpgc_ac
    Let ABC be a right triangle at A. AB=3, AC=4. MNPQ is a rectangle that is in the triangle ABC. Represent X equal AM. Calculate X for MNPQ be square.[IMG][img]C:\Documents and Settings\admin\Desktop\picture[/img][/IMG]
    Cannot access the figure.
    But, just for fun, basing only on the wordings of your posted question,
    x = 36/37 = 0.973 unit long. --------answer.

    -----
    That is based on my assumption in the square MNPQ, where:
    --AM is on the leg AB of the right triangle. (AB=3 units long; AM=x)
    --NP is along the hypotenuse BC. (BC=5 units long)
    --AQ is along the leg AC. (AC=4 units long)

    Right triangle MAQ is similar to right triangle BAC, so, by proportion,
    AM/AB = MQ/BC
    Substituting values,
    x/3 = s/5
    s = 5(x/3) = 5x/3 unit long ----where s is a side of the square MNPQ.

    Using proportions with similar triangles will get the x, but using trigonometry from here on is easier.

    In rigth triangle BAC, let angle C = theta, then,
    cos(theta) = AC/BC = 4/5 -----**

    In right triangle MNB,
    MN is perpendicular to BC,
    BM is perpendicular to AC,
    therefore, angle BMN is congruent to angle BCA, by angles whose sides are perpendicular each to each are congruent.
    So, angle BMN = angle BCA = theta -----**
    Then,
    cos(theta) = MN/BM
    where MN = s = 5x/3
    So, substitutions,
    4/5 = (5x/3)/BM
    Cross multiply,
    4(BM) = 5(5x/3)
    BM = (25x/3)/4 = 25x/12 unit long.

    But AB = BM +x = 3, so,
    25x/12 +x = 3
    Multiply both sides by 12,
    25x +12x = 36
    37x = 36
    x = 36/37 = 0.973 unit long ----------answer.
    Last edited by ticbol; May 18th 2006 at 05:48 AM. Reason: some explanations.
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  4. #4
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    Smile

    Quote Originally Posted by ticbol
    Cannot access the figure.
    But, just for fun, basing only on the wordings of your posted question,
    x = 36/37 = 0.973 unit long. --------answer.

    -----
    That is based on my assumption in the square MNPQ, where:
    --AM is on the leg AB of the right triangle. (AB=3 units long; AM=x)
    --NP is along the hypotenuse BC. (BC=5 units long)
    --AQ is along the leg AC. (AC=4 units long)

    Right triangle MAQ is similar to right triangle BAC, so, by proportion,
    AM/AB = MQ/BC
    Substituting values,
    x/3 = s/5
    s = 5(x/3) = 5x/3 unit long ----where s is a side of the square MNPQ.

    Using proportions with similar triangles will get the x, but using trigonometry from here on is easier.

    In rigth triangle BAC, let angle C = theta, then,
    cos(theta) = AC/BC = 4/5 -----**

    In right triangle MNB,
    MN is perpendicular to BC,
    BM is perpendicular to AC,
    therefore, angle BMN is congruent to angle BCA, by angles whose sides are perpendicular each to each are congruent.
    So, angle BMN = angle BCA = theta -----**
    Then,
    cos(theta) = MN/BM
    where MN = s = 5x/3
    So, substitutions,
    4/5 = (5x/3)/BM
    Cross multiply,
    4(BM) = 5(5x/3)
    BM = (25x/3)/4 = 25x/12 unit long.

    But AB = BM +x = 3, so,
    25x/12 +x = 3
    Multiply both sides by 12,
    25x +12x = 36
    37x = 36
    x = 36/37 = 0.973 unit long ----------answer.
    I'm very happy to see this problem be solved . But can anybody help me to post to picture. I want to you to see this figure of this problem. I'm waiting for your answer.
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  5. #5
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    Quote Originally Posted by mpgc_ac
    I'm very happy to see this problem be solved . But can anybody help me to post to picture. I want to you to see this figure of this problem. I'm waiting for your answer.
    You have to do an attachment. See Attach Files under the Additional Options on the posting page. I tried one here.
    Attached Thumbnails Attached Thumbnails This problem is not so easy-02-t29.gif  
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  6. #6
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    Talking

    Quote Originally Posted by JakeD
    You have to do an attachment. See Attach Files under the Additional Options on the posting page. I tried one here.
    Thank you for your reply . Below is the picture of this problem
    Attached Thumbnails Attached Thumbnails This problem is not so easy-picture.png  
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  7. #7
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    Quote Originally Posted by mpgc_ac
    Thank you for your reply . Below is the picture of this problem
    Zeez! I knew I can imagine things, including figures in Geometry.

    ------
    Ooppss, the labels are not the same. I thought NP was on the hypotenuse BC.

    You labeled the right triangle in clockwise, but you labeled the square in counterclockwise. I pictured mine both in clockwise. That's why.
    Last edited by ticbol; May 19th 2006 at 12:46 AM.
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  8. #8
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    Cool

    Quote Originally Posted by ticbol
    Zeez! I knew I can imagine things, including figures in Geometry.

    ------
    Ooppss, the labels are not the same. I thought NP was on the hypotenuse BC.

    You labeled the right triangle in clockwise, but you labeled the square in counterclockwise. I pictured mine both in clockwise. That's why.
    Dear sir or miss!
    Thank you for solution!
    Please give me some exercices.
    Reply to me soon.
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  9. #9
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    Quote Originally Posted by mpgc_ac
    Dear sir or miss!
    Thank you for solution!
    Please give me some exercices.
    Reply to me soon.
    Hello. You like it, huh.
    Ah, I only give answers to questions posted here, so it's you people out there who should post exercises that we can solve, if we can. So, come on, show us some more.
    One problem per posting, if possible. I do not give partial answers. If there are 3 problems in a posting, I have to answer all three. If I cannot answer one of them, I will not touch the three questions altogether.
    Last edited by ticbol; May 20th 2006 at 02:47 AM.
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