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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
It's a pretty hard question but it's interesting. In this length of 3 lines are given and you have to find ratio of Pink and Green area. For solving this question first you have to find length of 4th line. For this I will show you a trick which can be used in any this type of question.
First you have to assume point in the center as origin where four lines are intersecting. Then mark vertical and horizontal length from origin to sides of square as a,b,c,d as shown in figure. Then mark all corners as 1,2,3,4 and it's co-ordinates as (-c,a), (d,a), (-c,-b), (d,-b). I marked these length by previously marked a,b,c,d length from origin.
Now calculate distance between corners of square and origin by distance formula as shown:
1.Distance between O and 1:
\[\sqrt{a^{2}+c^{2}}=\sqrt{65}\]
or
\[a^{2}+c^{2}=65----(1)\]
Similarly,
2.Distance between O and 2:
\[a^{2}+d^{2}=85----(2)\]
3.Distance between O and 3:
\[c^{2}+b^{2}=25----(3)\]
4.Distance between O and 4:
\[b^{2}+d^{2}=45----(4)\]Now, by observing LHS of these equations, we can observe, (1) + (4) = (2) + (3)
\[a^{2}+c^{2}+b^{2}+d^{2} =a^{2}+d^{2}+c^{2}+b^{2} \]
Now compare RHS of these equations and assume length of 4th line as r, we get,
\[65+45 =85+r^{2} \]
\[r^{2}= 65+45-85\]
\[r=\sqrt{25}\]
\[r=5\]
Now we get length of 4th side
Now mark angles A and B angles at 3rd corner as shown in figure:
Note that A + B = 90
°
∴ cos(A) = sin(B)
Apply cosine rule at A and B
We get:
\[cos(A)=\frac{x^{2}+25-65}{2x\sqrt{25}}\]
\[cos(A)=\frac{x^{2}-40}{2x\sqrt{25}}\]
\[cos(B)=\frac{x^{2}+25-45}{2x\sqrt{45}}\]
\[cos(B)=\frac{x^{2}-20}{2x\sqrt{45}}\]
We know that,
\[sin(B)=\sqrt{1-cos^{2}(B)}\]
\[sin(B)=\sqrt{1-\left(\frac{x^{2}+20}{2x\sqrt{45}}\right)^{2}}\]
\[sin(B)=\frac{\sqrt{\left(2x\sqrt{25}\right)^{2}-\left(x^{2}-20\right)^{2}}}{2x\sqrt{25}}\]
\[sin(B)=\frac{\sqrt{100x^{2}-x^{4}+40x^{2}-400}}{2x\sqrt{25}}\]
\[sin(B)=\frac{\sqrt{140x^{2}-x^{4}-400}}{2x\sqrt{25}}\]
Now applying cos(A) = sin (B)
We get,
\[\frac{x^{2}-40}{2x\sqrt{25}}=\frac{\sqrt{140x^{2}-x^{4}-400}}{2x\sqrt{25}}\]
\[{x^{2}-40}=\sqrt{140x^{2}-x^{4}-400}\]
\[\left({x^{2}-40}\right)^{2}=\left(\sqrt{140x^{2}-x^{4}-400}\right)^{2}\]
\[x^{4}-80x^{2}+1600=140x^{2}-x^{4}-400\]
\[2x^{4}-220x^{2}+2000=0\]
\[x^{4}-110x^{2}+1000=0\]
\[x^{4}-10x^{2}-100x^{2}+1000=0\]
\[\left(x^2-10\right)\left(x^2-100\right)=0\]
\[x^2=10\]
It is not possible, so
\[x^2=100\]
\[x=10\]
Apply Heron's formula on Pink and Green triangle, For that we take,
\[S_{1}=\frac{10+\sqrt{25}+\sqrt{45}}{2}\]
\[S_{2}=\frac{10+\sqrt{25}+\sqrt{65}}{2}\]
Let Pink area be A_P and Green area A_G
\[\frac{A_{P}}{A_{G}}=\frac{\sqrt{S_{1}\left(S_{1}-10\right)\left(S_{1}-\sqrt{25}\right)\left(S_{1}-\sqrt{45}\right)}}{\sqrt{S_{2}\left(S_{2}-10\right)\left(S_{2}-\sqrt{25}\right)\left(S_{2}-\sqrt{65}\right)}}\]
By taking S_1 and S_2 values and then simplifying, we get:
\[\frac{A_{P}}{A_{G}}=\frac{\sqrt{\left(10+\sqrt{45}+\sqrt{25}\right)\left(\sqrt{45}+\sqrt{25}-10\right)\left(10+\sqrt{25}-\sqrt{45}\right)\left(10+\sqrt{45}-\sqrt{25}\right)}}{\sqrt{\left(10+\sqrt{65}+\sqrt{25}\right)\left(\sqrt{65}+\sqrt{25}-10\right)\left(10+\sqrt{25}-\sqrt{65}\right)\left(10+\sqrt{65}-\sqrt{25}\right)}}\]
After carefully simplifying this equation, we get:
\[\frac{A_{P}}{A_{G}}=\frac{\sqrt{25\times45-225}}{\sqrt{25\times65-25}}\]
\[\frac{A_{P}}{A_{G}}=\frac{\sqrt{25\times45-25\times9}}{\sqrt{25\times65-25}}\]
\[\frac{A_{P}}{A_{G}}=\frac{\sqrt{36}}{\sqrt{64}}\]
\[\frac{A_{P}}{A_{G}}=\sqrt{\frac{36}{64}}\]
\[\frac{A_{P}}{A_{G}}=\sqrt{\frac{9}{16}}\]
\[\frac{A_{P}}{A_{G}}=\frac{3}{4}\]
