1. Check the picture attached.
2. Triangles OAB and OEC are similar as they have congruent angles, as shown in the figure.
3. Similarity means that there is a scale factor of similarity, say k.
4.
[tex] \frac{12-t}{t}=k [/tex] (corresponding sides, each in front of angles alpha)
also
[tex] \frac{9-a}{a}=k[/tex] (corresponding sides in front of the vertec angles AOB and EOC)
so [tex]\frac{12-t}{t}=\frac{9-a}{a}[/tex]
[tex]\frac{12}{t}-1= \frac{9}{a}-1
[/tex]
[tex]\frac{12}{t}= \frac{9}{a}[/tex]
[tex]\frac{4}{t}= \frac{3}{a}[/tex]
[tex]t= \frac{4a}{3} [/tex]
5.
x is y-12, so let's find the areas x and y, in terms of a, so that we form an equation and solve for a.
x = [tex] \frac{1}{2}at= \frac{1}{2}a* \frac{4a}{3}= \frac{ 2a^{2} }{3} [/tex]
To find the area y we need to write OC in terms of a:
OC= [tex]t*k=\frac{4a}{3}* \frac{9-a}{a}= \frac{36-4a}{3} [/tex]
so y=[tex] \frac{1}{2}*(9-a)*\frac{4(9-a)}{3}= \frac{ 2(9-a)^{2} }{3} [/tex]
6.
x=y-12
[tex]\frac{ 2a^{2} }{3}=\frac{ 2(9-a)^{2} }{3} - \frac{36}{3} [/tex]
[tex]2a^{2}=2(81-18a+ a^{2} )-36[/tex]
[tex]2a^{2}=162-36a+ 2a^{2} -36[/tex]
[tex]36a=126[/tex]
a=126/36=3.5
