AMC Questions (1 Viewer)

[kurt.physics]

I would imagine that this question would be relatively easy, but i can not seam to get an answer.

1. Farmer Taylor of Burra Creek is very annoyed. A 4m wide road has been put through one of his rectangular padocks, splitting it into two. As a result, he has lost some of his land. If all dimensions in the figure are in metres, how much land, in square metres, has he lost?

And Also,

2. The walls of a castle form a quadrilateral PQRS as shown so that PQ = 40 m, QR = 45 m, RS = 20 m, SP = 20 m and angle PSR = 90 degrees. A guard must walk outside the walls so that he is always 2 m from the nearest part of the walls. The guard started walking around the castle clockwise and finally arrived at the starting point.

The length, in metres, of his walk was



Thanks in advanced

 

[conics2008]

Intresting ! where do you find questions like these.. I like questions with limited information =)...

Try using the 100x = A and 200+2x = P ..... then hopefully find the maximum value of x and so on. Is this a maxima and minima problems.. its looks like it ?

 

[kurt.physics]

Intresting ! where do you find questions like these.. I like questions with limited information =)...

Try using the 100x = A and 200+2x = P ..... then hopefully find the maximum value of x and so on. Is this a maxima and minima problems.. its looks like it ?

What question is that for?

I got these from the year 9 & 10 Intermediate Division Competition Paper from the AMC 2002.

If you are interested there is some challange questions on their website (http://www.amt.edu.au/wuamc.html)

So can you figure out the two questions?

 

[undalay]

AMC is either multiple choice or a number?

Q1 is nethier of those? unless there is information missing.

and ur diagram in Q2 is wrong, the 90degrees angle shouldn't be inside? shouldn't it be regular. aha

 

[kurt.physics]

Sorry, didnt put the multiple choice answers, here they are

Q1)

A) 120

B) 150

C) 160

D) 200

E) 250

Q2)

A) 125 + 4π

B) 121 + 5π

C) 125 + 5π

D) 121 + 6π

E) 125 + 6π

Sorry

 

[conics2008]

What question is that for?

I got these from the year 9 & 10 Intermediate Division Competition Paper from the AMC 2002.

If you are interested there is some challange questions on their website (http://www.amt.edu.au/wuamc.html)

So can you figure out the two questions?

Question one.....

I managed to cut the rectangle in half from the 71.. im trying to figure a way to find the width.. I named it x and also trying to find 2 pairs of equation ??? Still working on it =) im still doing question 1

 

[Iruka]

Q1) Use similar triangles. Don't forget the friendliest pythagorean triple of them all, (3,4,5).

Q2) You know the theorem about the sum of the exterior angles of a polygon? You can use it for non-convex polygons as well, you just have to count oriented turns (eg, count rotations in an anti clockwise direction as positive and those in a clockwise direction as negative.)

 

[kurt.physics]

Q1) Use similar triangles. Don't forget the friendliest pythagorean triple of them all, (3,4,5).

I noticed that that happens (see attached) but i dont know what to do from there. So have you figured it out Iruka?

 

[conics2008]

ok.. with all my dodgy working out i get 120 as the area that poor bugger lost !

soo tell us whats the answer ?

 

[undalay]

I find it strange there is no width...

 

[kurt.physics]

ok.. with all my dodgy working out i get 120 as the area.....

soo tell us whats the answer ?

How the hell did you get that? I dont have the answer, i was hoping yous could get it! What about Irukra, have you got it, or Undalay??? Are they the same?

 

[conics2008]

hmmmm well basically its kind of dodgy lol.. but it got me an answer that is one of the possible answers...

OK get ready to laugh lol....

Basically What i did is break it up.. Before you posted that 3+4= 5 thing i attempted to use it but had no hope lol..... i said screw everything ... well the question didn't say " NOT TO SCALE " so I assumed lol those lines intersect at angel 45 lol..... and i went on from there


i have a rectangle on top by 71 & 24.. my width becomes 24 lol..

and then i did 29 x 24 - ( 2 x 1/2 x 24 x 24 ) = 120


lol this is coming from a 4unit student..

 

[Iruka]

You use similar triangles and some of the theorems about transversals on parallel lines to work out that the width is 32m. 32:24 = 4:3

 

[kurt.physics]

You use similar triangles and some of the theorems about transversals on parallel lines to work out that the width is 32m.

Could you show us your magic?

 

[undalay]

You use similar triangles and some of the theorems about transversals on parallel lines to work out that the width is 32m.

nvm i see

 

[kurt.physics]

...please!!!

 

[conics2008]

You use similar triangles and some of the theorems about transversals on parallel lines to work out that the width is 32m. 32:24 = 4:3

IRUKA how can you assume that those two lines are parallel... Its just like what i assumed they might intersect the angel ????

 

[Iruka]

because the question said that the road was 4 metres wide and the paddock is a rectangle.

 

[kurt.physics]

Come again?

 

[undalay]

Q1 answer is 160