Area of Shaded Region: Solve without Year 9 Maths

[tony24810]

As shown in the picture, area of shaded region is required.

By setting formula on xy plane, the point of intersection can be deduced and hence calculated by 2 triangles and 1 square.

from my calculation, set left bottom corner as (0,0), then point of intersection is (7/21/47,1/23/47), and area of shaded region would be 4/37/47 cm^2 (4.7872 cm^2).

However, the question requires that no methods beyond year 9 are to be used.

I have tried mainly simultaneous equations mainly, adding subtracting etc and couldn't figure this out.

someone please help.well I'm not sure if this should be posted on homework section or other section, please do let me know if I'm making a mistake, thanks

 

[HallsofIvy]

I'm sorry but I have absolutely no idea what "7/21/47" means! I could interpret it as either (7/21)/47 or 7/(21/47) but those are very different. In any case, I get
$$\left(\frac{350}{47}, \frac{70}{47}\right)$$
as the point to intersection.

 

[tony24810]

i wrote "7/21/47", what i mean is 7 plus 21/47, like how you type in calculator, not sure what the correct notation for computer is.

anyhow, 530/47, 70/47 is exactly what i got too, same thing

but i can't do it without using xy plane. it's meant for year 7 student, olympic maths...

 

[HallsofIvy]

That's a very strange calculator!
$$7+ \frac{21}{47}= \frac{350}{47}$$
and
$$1+ \frac{23}{47}= \frac{70}{47}$$
just what I said so, yes, you have the correct point.

I'm not sure why your "year 7" (you said "year 9" before) is like but I just used the equations of straight lines.

 

[tony24810]

well I'm a private tutor of a year 7 kid and his school teacher gave their class this challenge question and states that no advance method should be used.

i was thinking that some methods similar to those shown in these 2 pictures should be used...

 

[haruspex]

Well, this sort of uses the xy plane, but not in so many words:

Let the intersection point be distance h from the bottom of the square and distance w from the RHS.
Mark the points:
ABCD are the corners of the square, starting top right and going clockwise.
The angled lines are CG (G being the intersection with AB) and AH.
I is the point of intersection of CG, AH.
E is the point on AB nearest I
F is the point on BC nearest I
By similar trangles AEI, ABH:
w = (10-h)*3/10
and using CFI, CBG:
h = (10-w)*2/10
etc.

 

[Infinitum]

Well, this sort of uses the xy plane, but not in so many words:

Let the intersection point be distance h from the bottom of the square and distance w from the RHS.
Mark the points:
ABCD are the corners of the square, starting top right and going clockwise.
The angled lines are CG (G being the intersection with AB) and AH.
I is the point of intersection of CG, AH.
E is the point on AB nearest I
F is the point on BC nearest I
By similar trangles AEI, ABH:
w = (10-h)*3/10
and using CFI, CBG:
h = (10-w)*2/10
etc.

I tried this problem the co-ordinate geometry way, like the OP, but it didn't result in much fun, even though I did find the answer after some work. This method is really very ingenious and elegant(and quite acceptable for a 9th year)!

Thanks for sharing, haruspex.

 

[tony24810]

cool!

thanks haruspex, that's exactly what I've been looking for!