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Areas and Ratios

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Areas and Ratios

by masuarezdl » Sat Jan 31, 2009 12:49 pm
Image

If the two regions above have the same area, what is the ratio t : s?
a) 2 : 3
b) 16 : 3
c) 4 : √3
d) 2 : √3^¼
e) 4 : √3^¼

As far as I can go, the areas are equal, so:
Area of Triangle (at) = (b*h) / 2 = (t*√3) / 2
Area of Square (as) = s²

If at = as, then (t*√3) / 2 = s². Then Im lost. Any hints?
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by truplayer256 » Sat Jan 31, 2009 1:00 pm
Area of triangle=(t^2times sqrt3)/4
Area of square= s^2

Since the two areas are equal:

s^2=(t^2 times sqrt3)/4
4s^2=t^2 times sqrt 3

Since we want the ratio of t to s:

4/(sqrt3)=(t^2)/(s^2)
take the square root of both sides

2/(3^1/4)=t/s. D
The mistake that you made was that you said that the area of the triangle was (t times the sqrt 3)/4, when it's really (t^2 times sqrt 3)/4
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by awilhelm » Sat Jan 31, 2009 1:43 pm
I have a different answer. What about this approach?

Let's assume that t = 2

That makes the area of the triangle = (2*sqrt3)/2 = sqrt3

If the area of the square is also equal to sqrt 3, then s must be equal to sqrt(sqrt3)

So we have t = 2, s = sqrt(sqrt3)

t/s = 2/sqrt(sqrt3)

In answer B, we have 16/3. Taking the sqrt of both 16 and 3 two times will give us 2/sqrt(sqrt3). So answer B should represent the correct ratio.

Thoughts?
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by masuarezdl » Sat Jan 31, 2009 6:32 pm
I am getting a bit lost. I found one mistake though. But why is the area of the triangle at = t²*√3) / 2 ?

An equilateral triangle split in two (vertical), will get us a triangle with sides:
1. t/2
2. (√3)/2
3. 2/2 = 1

Then, by the formula a = (b*h) / 2, a = (t * (√3/2)) / 2, right? Where does the t² come from?
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by truplayer256 » Sat Jan 31, 2009 6:39 pm
The area isn't t^2sqrt3/2, it's t^2sqrt3/4...

This is because the area of a triangle is 1/2bh

EQUILATERAL TRIANGLE:

Base=x
Height=xsqrt3/2

Area= [x^(2)(xsqrt3)/(2)]/(2)=x^(2)sqrt3/4
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by hardik.jadeja » Sat Jan 31, 2009 6:52 pm
awilhelm, your method is right..

you got t/s = 2/sqrt(sqrt3), which is nothing but 2 : 3^¼
but 16/3 is not equal to 2 : 3^¼... (x^2/y^2 is not equal to x/y)

btw, why is option D (2 : √3^¼) instead of (2 : 3^¼)?
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by awilhelm » Sat Jan 31, 2009 8:03 pm
Thanks, understood!
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by bido » Sat Feb 07, 2009 11:12 pm
GAINING confidence.

Now the question is simply askin ...what is t/s

They have the same area .therefore bh/2 = s^2

Notice that the triangle is equilateral i.e each angle is 60 degrees
base of triangle=t and heightof triangle =(t root3)/2
AREA=1/2 * t * (t root3)/2

since the question says that AREA OF TRIANGLE= AREA OF SQUARE
(t^2 root3)/4 = s^2
when u cross multiply u get: t^2 root3 = 4s^2
t^2/t^2 =4/ root 3
(t/s)^2 =4/ root 3
finding the square root of both sides gives: t/s= root4/ root of root3
=2/4th root 3

D should be the answer...Always try to beak the question down to suite you..
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by sureshbala » Sun Feb 08, 2009 12:56 am
masuarezdl wrote:I am getting a bit lost. I found one mistake though. But why is the area of the triangle at = t²*√3) / 2 ?

An equilateral triangle split in two (vertical), will get us a triangle with sides:
1. t/2
2. (√3)/2
3. 2/2 = 1

Then, by the formula a = (b*h) / 2, a = (t * (√3/2)) / 2, right? Where does the t² come from?
Dear masuarezd,

If the side of an equilateral triangle is t, then the height of that triangle will be √3/2 t. How can it be independent of t?. This is the mistake that you should look into.
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by mdavis » Sun Feb 08, 2009 9:19 am
I don't understand where people are getting the height as the square root of 3/2t?

Is this just a geometry rule that I need to memorize?
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by anniev2 » Wed Feb 18, 2009 8:28 pm
mdavis wrote:I don't understand where people are getting the height as the square root of 3/2t?

Is this just a geometry rule that I need to memorize?
Yes, apparently there is a formula for finding the area of an equilateral triangle. https://mathworld.wolfram.com/EquilateralTriangle.html

I got this same answer incorrect and it was because I was unable to set up a formula for the triangle's area.
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Re: Areas and Ratios

by x2suresh » Wed Feb 18, 2009 9:16 pm
masuarezdl wrote:Image

If the two regions above have the same area, what is the ratio t : s?
a) 2 : 3
b) 16 : 3
c) 4 : √3
d) 2 : √3^¼
e) 4 : √3^¼

As far as I can go, the areas are equal, so:
Area of Triangle (at) = (b*h) / 2 = (t*√3) / 2
Area of Square (as) = s²

If at = as, then (t*√3) / 2 = s². Then Im lost. Any hints?
at = 1/2*b*h = 1/2*t*sqrt(3)/2 = sqrt(3)/4 *t^2

sqrt(3)/4 *t^2 = s^2

s=t 3^(1/4)/2

t:s = t: t* 3^(1/4)/2 = 2:3^1/4

D
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by pkw209 » Mon Feb 01, 2010 4:24 pm
super
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