# This problem is not so easy

• May 18th 2006, 03:05 AM
mpgc_ac
This 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 AM
topsquark
Quote:

Originally Posted by mpgc_ac
: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]

-Dan
• May 18th 2006, 06:01 AM
ticbol
Quote:

Originally Posted by mpgc_ac
: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]

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.
• May 18th 2006, 07:48 PM
mpgc_ac
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.
• May 18th 2006, 08:43 PM
JakeD
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.
• May 18th 2006, 09:03 PM
mpgc_ac
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 :D . Below is the picture of this problem
• May 19th 2006, 01:39 AM
ticbol
Quote:

Originally Posted by mpgc_ac
Thank you for your reply :D . 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. :)
• May 20th 2006, 02:49 AM
mpgc_ac
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!