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In isosceles right triangle $ABC$, shown here, $AC=BC$. Point $X$ is on side $BC$ such that $CX=6$ and $XB=12$, and $Y$ is on side $AB$ such that $\overline{XY}\perp\overline{AB}$. What is the ratio of the area of triangle $BXY$ to the area of triangle $ABC$?

 

help.

 Feb 3, 2021

Best Answer 

 #1
avatar+11380 
+2

What is the ratio of the area of triangle $BXY$ to the area of triangle ABC?

 

Hello Guest!

 

\(sin(45°)=\frac{1}{2}\cdot \sqrt{2}=\frac{\overline{XY}}{12}\)

\(\overline{XY}=6\cdot \sqrt{2}\)

\(\overline{BC}=18\)

\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot \overline{XY}^2:\frac{1}{2}\cdot \overline{BC}^2\)

\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot 72:\frac{1}{2}\cdot 324\)

\(A_{BXY}:A_{ABC}=2:9\)

laugh  !

 Feb 3, 2021
 #1
avatar+11380 
+2
Best Answer

What is the ratio of the area of triangle $BXY$ to the area of triangle ABC?

 

Hello Guest!

 

\(sin(45°)=\frac{1}{2}\cdot \sqrt{2}=\frac{\overline{XY}}{12}\)

\(\overline{XY}=6\cdot \sqrt{2}\)

\(\overline{BC}=18\)

\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot \overline{XY}^2:\frac{1}{2}\cdot \overline{BC}^2\)

\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot 72:\frac{1}{2}\cdot 324\)

\(A_{BXY}:A_{ABC}=2:9\)

laugh  !

asinus Feb 3, 2021

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