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.
+15001 What is the ratio of the area of triangle $BXY$ to the area of triangle ABC?
Hello Guest!
\(sin(45°)=\frac{1}{2}\cdot \sqrt{2}=\frac{\overline{XY}}{12}\)
\(\overline{XY}=6\cdot \sqrt{2}\)
\(\overline{BC}=18\)
\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot \overline{XY}^2:\frac{1}{2}\cdot \overline{BC}^2\)
\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot 72:\frac{1}{2}\cdot 324\)
\(A_{BXY}:A_{ABC}=2:9\)
!
+15001 What is the ratio of the area of triangle $BXY$ to the area of triangle ABC?
Hello Guest!
\(sin(45°)=\frac{1}{2}\cdot \sqrt{2}=\frac{\overline{XY}}{12}\)
\(\overline{XY}=6\cdot \sqrt{2}\)
\(\overline{BC}=18\)
\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot \overline{XY}^2:\frac{1}{2}\cdot \overline{BC}^2\)
\(A_{BXY}:A_{ABC}=\frac{1}{2}\cdot 72:\frac{1}{2}\cdot 324\)
\(A_{BXY}:A_{ABC}=2:9\)
!
