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eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0well they are both just triangles minus the same little corner...

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1The 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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
 2 years ago
Best ResponseYou've already chosen the best response.0*i think*** lol i havent taken geometry since 10th grade

eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0haha thats a cool one though!

eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0What do you think does my logic seem sound?

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1no  thats not the answer

eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0hmm maybe you can tjust take off the A

eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0but if we find that little A we are golden

eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0I could do it using trig but thats a pain and I know there's an easier way...

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1i did it differently using systems of equations there maybe an easier way though this is how i started

phi
 2 years ago
Best ResponseYou've already chosen the best response.1dw: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
 2 years ago
Best ResponseYou've already chosen the best response.1area 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
 2 years ago
Best ResponseYou've already chosen the best response.1yea  i think thats a bit better way than mine

eyust707
 2 years ago
Best ResponseYou've already chosen the best response.0haha phi isnt that just the same problem again... dw:1341592696668:dw

phi
 2 years ago
Best ResponseYou've already chosen the best response.1the 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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1so find the area of p, area of q is (1/2)^2 + (1/2)*(1/2)^2 + area(x)

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1so q = 1/4 + 1/8 + area(x) = 3/8 + area(x)

phi
 2 years ago
Best ResponseYou've already chosen the best response.1area(x)= (1/2)*(1/2)*(1/6)= 1/24 area(p)= (1/2)*1*(1/3)= 1/6

phi
 2 years ago
Best ResponseYou've already chosen the best response.1so ratio of areas p/q = 2/5

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1i 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
 2 years ago
Best ResponseYou've already chosen the best response.1 quite a searching question for 16 year old, i thought

phi
 2 years ago
Best ResponseYou've already chosen the best response.1Back in the old days, maybe they worked harder at this stuff??
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