SAT Geometry Practice

The triangle shown is a right triangle, where the interior angle shown with a right angle symbol has a measure of $90^{\circ }$.

It's shown that the other two interior angles measure $13^{\circ }$ and $a^{\circ }$.

The sum of the measures of the interior angles of a triangle is $180^{\circ }$; therefore, $90+13+a=180$.

Combining like terms on the left-hand side of this equation yields $103+a=180$.

Subtracting $103$ from both sides of this equation yields $a=77$.

It's given that triangle $XYZ$ is similar to triangle $TUV$. Therefore, each side of triangle $XYZ$ is $k$ times its corresponding side of triangle $TUV$, where $k$ is a constant.

It follows that the perimeter of triangle $XYZ$ is $k$ times the perimeter of triangle $TUV$. It's also given that $\overline{TU}$ corresponds to $\overline{XY}$ and the length of $TU$ is 18.

Let $x$ represent the length of $XY$. It follows that $x=18k$.

The table shows that the perimeters of triangles $TUV$ and $XYZ$ are 37 and 333, respectively. It follows that $333=37k$, or $9=k$.

Substituting 9 for $k$ in the equation $x=18k$ yields $x=189$, or $x=162$. Therefore, the length of $XY$ is 162.

The correct answer is 54. It's given that in triangle $ABC$, point $D$ on side $AB$ is connected by a line segment with point $E$ on side $AC$ such that line segment $DE$ is parallel to side $BC$. It follows that parallel segments $DE$ and $BC$ are intersected by sides $AB$ and $AC$.

If two parallel segments are intersected by a third segment, corresponding angles are congruent. Thus, corresponding angles $C$ and $AED$ are congruent and corresponding angles $B$ and $ADE$ are congruent. Since triangle $ADE$ has two angles that are each congruent to an angle in triangle $ABC$, triangle $ADE$ is similar to triangle $ABC$ by the angle-angle similarity postulate, where side $DE$ corresponds to side $BC$, and side $AE$ corresponds to side $AC$.

Since the lengths of corresponding sides in similar triangles are proportional, it follows that $\frac{DE}{BC}=\frac{AE}{AC}$. Since point $E$ lies on side $AC$, $AE+CE=AC$. It's given that $CE=2AE$.

Substituting $2AE$ for $CE$ in the equation $AE+CE=AC$ yields $AE+2AE=AC$, or $3AE=AC$. It's given that $BC=162$. Substituting 162 for $BC$ and $3AE$ for $AC$ in the equation $\frac{DE}{BC}=\frac{AE}{AC}$ yields $\frac{DE}{162}=\frac{AE}{3AE}$, or $\frac{DE}{162}=\frac{1}{3}$.

Multiplying both sides of this equation by 162 yields $DE=54$. Thus, the length of line segment $DE$ is 54.