:confused: 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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- May 18th 2006, 03:05 AMmpgc_acThis problem is not so easy
:confused: 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]

- May 18th 2006, 05:13 AMtopsquarkQuote:

Originally Posted by**mpgc_ac**

-Dan - May 18th 2006, 06:01 AMticbolQuote:

Originally Posted by**mpgc_ac**

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. - May 18th 2006, 07:48 PMmpgc_acQuote:

Originally Posted by**ticbol**

- May 18th 2006, 08:43 PMJakeDQuote:

Originally Posted by**mpgc_ac**

- May 18th 2006, 09:03 PMmpgc_acQuote:

Originally Posted by**JakeD**

- May 19th 2006, 01:39 AMticbolQuote:

Originally Posted by**mpgc_ac**

------

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. :) - May 20th 2006, 02:49 AMmpgc_acQuote:

Originally Posted by**ticbol**

Thank you for solution!

Please give me some exercices.

Reply to me soon. - May 20th 2006, 03:42 AMticbolQuote:

Originally Posted by**mpgc_ac**

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.