cwrw238
Find ratio of area of p to area of q.
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cwrw238
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|dw:1341591452831:dw|
eyust707
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well they are both just triangles minus the same little corner...
cwrw238
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The whole figure is a square
and the line bisects the right side of the square.
this question was from a maths paper for 16 year old's.
I would have struggled to do this at 16 - probably would have given up.
eyust707
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|dw:1341591736170:dw|
p = (1*2)/2 - A
q= (2*2)/2 - A
Subtracting the same amount from both wont effect the ratio, so:
p:q = 1/2
eyust707
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*i think*** lol i havent taken geometry since 10th grade
eyust707
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haha thats a cool one though!
eyust707
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What do you think does my logic seem sound?
cwrw238
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no - thats not the answer
eyust707
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hmm maybe you can tjust take off the A
eyust707
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but if we find that little A we are golden
cwrw238
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yes
eyust707
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I could do it using trig but thats a pain and I know there's an easier way...
cwrw238
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i did it differently using systems of equations
there maybe an easier way though
this is how i started
cwrw238
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|dw:1341592406267:dw|
phi
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|dw:1341592415019:dw|
we can label the sides of the square 1 for easy calculation.
similar triangles p and x, with x scaled by 1/2 means the area of x is 1/4 of p
the rest of the areas are easy to find.
cwrw238
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area x : area y = 1^2 : 2^2 = 1: 4
then i got 4 equations in x, y , p and q
and splved
- a bit messy !!
cwrw238
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yea - i think thats a bit better way than mine
eyust707
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haha phi isnt that just the same problem again... |dw:1341592696668:dw|
phi
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the altitudes are also in ratio of 1/2 so p's alt is 1/3 and x's is 1/6
as they add to 1/2
cwrw238
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yes
phi
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so find the area of p,
area of q is (1/2)^2 + (1/2)*(1/2)^2 + area(x)
cwrw238
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so q = 1/4 + 1/8 + area(x) = 3/8 + area(x)
phi
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area(x)= (1/2)*(1/2)*(1/6)= 1/24
area(p)= (1/2)*1*(1/3)= 1/6
phi
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so ratio of areas p/q = 2/5
cwrw238
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i got 2: 5
area p = 1/6
area q = 3/8 + 1/24 = 10/24 = 5/12
p/q = 1/6 * 12/5 = 12/60 = 2/5
- yea
cwrw238
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- quite a searching question for 16 year old, i thought
cwrw238
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- thanx
phi
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Back in the old days, maybe they worked harder at this stuff??
cwrw238
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yea - good point!