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well they are both just triangles minus the same little corner...
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.
|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
*i think*** lol i havent taken geometry since 10th grade
haha thats a cool one though!
What do you think does my logic seem sound?
no - thats not the answer
hmm maybe you can tjust take off the A
but if we find that little A we are golden
I could do it using trig but thats a pain and I know there's an easier way...
i did it differently using systems of equations there maybe an easier way though this is how i started
|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.
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 !!
yea - i think thats a bit better way than mine
haha phi isnt that just the same problem again... |dw:1341592696668:dw|
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
so find the area of p, area of q is (1/2)^2 + (1/2)*(1/2)^2 + area(x)
so q = 1/4 + 1/8 + area(x) = 3/8 + area(x)
area(x)= (1/2)*(1/2)*(1/6)= 1/24 area(p)= (1/2)*1*(1/3)= 1/6
so ratio of areas p/q = 2/5
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
- quite a searching question for 16 year old, i thought
Back in the old days, maybe they worked harder at this stuff??
yea - good point!