Find the length XY from the rectangular diagram

[chwala]

$8.485$ is indeed the length of the hypotenuse, but it is not $\sqrt{5.5^2+6^2}$. How did you arrive at $8.485$ ?

aaaaaaaarrgh...i made the same mistake just like the silly mistakes i make in chess,... sorry that was very silly of me ..to respond without thinking.

i used SOHCAHTOA,$cos 45^0$=$\frac {6}{h}$
$h$=$\frac {6}{cos 45^0}$=$8.485281374$

 

[MidgetDwarf]

aaaaaaaarrgh...i made the same mistake just like the silly mistakes i make in chess,... sorry that was very silly of me ..to respond without thinking.

i used SOHCAHTOA,$cos 45^0$=$\frac {6}{h}$
$h$=$\frac {6}{cos 45^0}$=$8.485281374$

The main thing that makes this problem "complicated." Is not that its a hard problem. As I believe phinds pointed out, is that the use of the "circle" marks on the diagram implies that those angles are congruent. I believe some users, including myself, are more familiar with the tick markings to illustrate angles are congruent.

Once you figure that the triangles on the bottom left and right are isosceles right angles. We know that that the side of this right triangle along the base is of length 6.

Together, the length of the side of the two right triangles along the base is 12. We know that the length of the base of of the rectangle is 11. Therefore, 12-11 = 1. No computation needed. It is not perpendicular bisector, but basic segment addition.

The reason why I proceeded the way I did in my first post, was that I was not familiar with the way congruent angles were drawn on the diagram. So I tried to find a general solution for when the sum of the upper angles did not measure to 45 degrees.

I noticed at the end, at the end, that they are in fact 45 degrees each. So I mentioned that once you know the triangle is a right triangle, it is a very short solution...

 

[hmmm27]

aaaaaaaarrgh...i made the same mistake just like the silly mistakes i make in chess,... sorry that was very silly of me ..to respond without thinking.

i used SOHCAHTOA,$cos 45^0$=$\frac {6}{h}$
$h$=$\frac {6}{cos 45^0}$=$8.485281374$

Oh, okay, now that makes sense.

The two things the other posters participating were doing (I assume), involved recognizing the two big triangles as isosceles (skipping the trig bit), and the numbers as dimensionless ("cm" isn't necessary, and actually wrong).

re: mnemonic "SOHCAHTOA" : heheh, thanks : hope I can remember that.