Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

|dw:1341591452831:dw|

well they are both just triangles minus the same little corner...

*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

yes

I could do it using trig but thats a pain and I know there's an easier way...

|dw:1341592406267:dw|

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

yes

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

- thanx

Back in the old days, maybe they worked harder at this stuff??

yea - good point!