Geometry Problem

infinitebrain · 2012-03-03 16:00:10

Hi I'm working on problems for UIL Calculator and I was wondering if anyone could solve this problem: (it's on the picture and it's asking for the shaded area)
And the answer is : .439
It would be great if someone could tell me how this is solved. Thanks.

Last edited by infinitebrain (2012-03-03 16:00:45)

Bob · 2012-03-03 20:52:42

hi infinitebrain,

I'm assuming that curve is part of a circle. See diagram below.

There's a property of all circles that when two chords cross each other (AB and EH, crossing at F) then

AF.FB = EF.FH

As we know three of these distances we can calculate the last (FH) (i) and therefore work out the diameter (ii) and hence the radius (iii).

You can then use trig to work out the angle of the sector AGB, (iv) , and then its area (v).

You can also work out the area of the triangle AGB (vi).

Subtracting gives the bit of the rectangle ABCD that is chopped out, shaded green (vii) , and so finally you can work out the shaded area (viii).

I've numbered the steps. That way, if you want more help with a step you can post back to me.

Bob

zetafunc. · 2012-03-03 22:42:56

Hmm... I got the area to be 0.4539. Is this a legitimate way of doing the problem?

My solution:

The area of the rectangle is 0.8109.

Suppose the curved area is from part of a quadratic curve with equation Ax² + Bx + C = 0.

This curve crosses the x-axis at -0.765 and 0.765 (this is 1.53/2; I've used the centre as the origin). The curve also crosses the y-axis at 0.35.

This means that:

(1) 0.585225A - 0.765B + C = 0 and
(2) 0.585225A + 0.765B + C = 0.

(2) - (1) yields 1.53B = 0 so B = 0

So Ax² + C = 0 is equation of the curve.

At x = 0, f(x) = 0.35, which means that C = 0.35. We then find A to be -0.598060575, so the equation is now;

-0.598060575x² + 0.35 = 0

Integrating that between -0.765 and 0.765 got me 0.4539 as my final answer... but this isn't what the OP said the correct answer was. Where is the error in my method?

zetafunc. · 2012-03-03 22:49:50

Even if you don't round my numbers you get the same answer... where did I go wrong? Can you not assume it's a parabola?

bobbym · 2012-03-04 00:24:41

Hi zetafunc;

I am pretty sure he is calling that an arc of a circle and not a parabola. This is arbitrary and you could think of that curve as a parabola but then you will not get .4393...

Bob · 2012-03-04 01:14:32

hi zetafuncc,

When the poster confirms the curve we'll know what to do.

But your method could easily be adapted by fitting a circle equation to those coords.

Be worth a try.

Bob

gAr · 2012-03-04 02:01:28

Hi,

infinitebrain · 2012-03-04 07:27:17

Hi bob bundy, thanks for responding to my post,
on step vii , when you're trying to chop out the green area of the rectangle, why would you need to find the area of the triangle, I'm just wondering how that helps you find out the shaded green area.

Bob · 2012-03-04 07:39:02

hi infinitebrain,

If A is the sector angle in degrees then the formula for a sector is

That will give the area shaded orange in the diagram below. (v)

You can get the area of the triangle shaded blue by half base x height. (vi)

Subtracting gives the area of the segment cut off by AB. (vii)

Hope that makes it clear.

Bob

infinitebrain · 2012-03-04 07:43:13

oh ok thank you sooo much bob! You've made me see the problem much clearer!

Bob · 2012-03-04 07:45:04

You're welcome!

Bob

infinitebrain · 2012-03-04 07:50:16

Ok now bob, just one more question and I promise I'll understand this fully:
how do you use trig the find the angle of the triangle?

Bob · 2012-03-04 08:17:34

hi

If you have the radius then you can calculate angle AGF by

AF is half 1.53 and AG = radius.

Then double this angle to get AGB the angle of the sector.

The triangle can be found by

half AB x FG

FG = radius - EF

Bob

bobbym · 2012-03-04 12:41:48

Hi;

Lots of nice solutions up there.

hammana · 2012-03-05 04:48:07

assuming a circle, the following calculationss made by hand or used as Liberty basic code will give the answer 0.43937252. Sorry, I am not yet familiar with the way to add a figure, but it is not difficult to un derstand what means each line
code
a1=atn((1.53/2)/0.35)
a2=(3.141593-2*a1)
h=sqr(0.35^2+(1.53/2)^2)
d=h/cos(a1)
r=d/2
sat=a2*r^2
st=(r-0.35)*1.53/2
sa=sat-st
answer=1.53*0.53-sa
print answer

bobbym · 2012-03-05 08:06:49

Hi hammana;

That works too! Welcome to the forum.

Math Is Fun Forum

#1 2012-03-03 16:00:10

Geometry Problem

#2 2012-03-03 20:52:42

Re: Geometry Problem

#3 2012-03-03 22:42:56

Re: Geometry Problem

#4 2012-03-03 22:49:50

Re: Geometry Problem

#5 2012-03-04 00:24:41

Re: Geometry Problem

#6 2012-03-04 01:14:32

Re: Geometry Problem

#7 2012-03-04 02:01:28

Re: Geometry Problem

#8 2012-03-04 07:27:17

Re: Geometry Problem

#9 2012-03-04 07:39:02

Re: Geometry Problem

#10 2012-03-04 07:43:13

Re: Geometry Problem

#11 2012-03-04 07:45:04

Re: Geometry Problem

#12 2012-03-04 07:50:16

Re: Geometry Problem

#13 2012-03-04 08:17:34

Re: Geometry Problem

#14 2012-03-04 12:41:48

Re: Geometry Problem

#15 2012-03-05 04:48:07

Re: Geometry Problem

#16 2012-03-05 08:06:49

Re: Geometry Problem

Board footer