Because the triangles have same angles, we can use the relation:
[tex]\frac{x}{60}=\frac{y}{135}[/tex]
And from this we have the relation:
[tex]x=\frac{60}{135}y[/tex]
The area of this triangles are measured as half of the producr of the basis with the height.
[tex]90\text{ = }\frac{x\times y}{2}\text{ }\rightarrow x\times y=90\times2\text{ = 180}[/tex]
Now we have two equations with two uknown values. We can just substitute the value of x from the first into the second. Than we get:
[tex]\frac{60}{135}y\times y=\text{ 180 }\rightarrow\text{ y }^2\text{= }\frac{180}{60}\times135[/tex][tex]y^2=\text{ 405 }\rightarrow\text{ y = }\sqrt[]{405}\text{ = }\sqrt[\square]{9^2^{}\times5}\text{ = }9\times\sqrt[]{5}[/tex]
From this, we can substitute in the second relation that we got:
[tex]x\times y=180\text{ }\rightarrow\text{ x}\times9\times\sqrt[]{5\text{ }}=180[/tex][tex]x=\frac{180}{9\times\sqrt[]{5}}=\frac{20}{\sqrt[\square]{5}}=\frac{20\times\sqrt[]{5}}{\sqrt[]{5}\times\sqrt[]{5}}=\frac{20\times\sqrt[]{5}}{5}=4\times\sqrt[]{5}cm[/tex]
From this, we got:
[tex]x\text{ = 4}\times\sqrt[]{5}\text{ cm}[/tex]
and
[tex]\text{y = }9\times\sqrt[]{5}\text{ cm}[/tex]
