Points $D$, $E$, and $F$ are the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively, of $\triangle ABC$. Points $X$, $Y$, and $Z$ are the midpoints of $\overline{EF}$, $\overline{FD}$, and $\overline{DE}$, respectively. If the area of $\triangle XYZ$ is 21, then what is the area of $\triangle CXY$?
XY = 1/4 AB
altitude of CXY is 3/4 the altitude ABC.
So, CXY has area 1/3 * 3/4 * 21 = 21/4
It's not 21/4
Its 63. Steve's logic is mostly right, except the area of triangle XYZ is 21, not the area of ABC.
To find the area of $\triangle CXY$, we can use the concept of similar triangles.
Since points $D$, $E$, and $F$ are the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively, we can say that $\triangle ABC$ is divided into four congruent triangles: $\triangle ADE$, $\triangle BDF$, $\triangle CEF$, and $\triangle DEF$.
Now, let's focus on $\triangle XYZ$. We know that points $X$, $Y$, and $Z$ are the midpoints of $\overline{EF}$, $\overline{FD}$, and $\overline{DE}$, respectively. This means that $\triangle XYZ$ is also divided into four congruent triangles: $\triangle XYZ$, $\triangle XDY$, $\triangle YFZ$, and $\triangle ZEX$.
Since $\triangle CCD$ and $\triangle XYZ$ share the same base $\overline{CD}$, and the height of $\triangle CCD$ is twice the height of $\triangle XYZ$, we can conclude that the area of $\triangle CCD$ is twice the area of $\triangle XYZ$.
Given that the area of $\triangle XYZ$ is 21, the area of $\triangle CCD$ is $2 \cdot 21 = 42$.
Therefore, the area of $\triangle CXY$ is also 42, since $\triangle CXY$ and $\triangle CCD$ share the same height and have bases in the same ratio.