The companion set to Workshop 10. Twelve SAT-style practice questions on angles, similar triangles, scaling, right-triangle trig, and circles. Name the fact, then use it.
SAT geometry rewards a handful of facts used at the right moment.
Angles: a straight line and a triangle each total \(180^\circ\), vertical angles are equal, and a slanted line crossing two parallel lines makes only small and big angles, with small + big \(= 180^\circ\).
Similar triangles: same angles, every side times the same factor \(k\).
Scaling: lengths \(\times k\), areas \(\times k^2\), volumes \(\times k^3\).
Right triangles: \(a^2 + b^2 = c^2\), SOH‑CAH‑TOA, and \(\sin A = \cos B\) when \(A + B = 90^\circ\).
Circles: \((x - h)^2 + (y - k)^2 = r^2\) has center \((h, k)\) and radius \(r\).
Here is how the wrong answers try to trip you:
Set Them Equal. Treats a small angle and a big angle along parallel lines as equal; they add to \(180^\circ\).
Matched by Letter. Assumes two separately named triangles match in letter order. Match the equal angles instead (an ordered similarity statement also gives the matches).
Area Scaled by k. Multiplies an area by the length factor \(k\) instead of \(k^2\).
Flipped Fraction. Puts the wrong side on top of a trig ratio; check opposite, adjacent, and hypotenuse from the angle you stand at.
Signs Not Flipped. Reads the center of \((x - 3)^2 + (y + 5)^2 = 16\) as \((-3, 5)\); the center is \((3, -5)\).
r Squared, Not r. Reports the right side of a circle equation, \(r^2\), as the radius.
Question 1 of 12
AnglesMedium
In the figure, lines $u$ and $v$ are parallel.
What is the value of $y$?
The takeaway: A line crossing two parallel lines (at a slant) makes only two sizes of angle: small and big. All the small angles are equal, all the big angles are equal, and small + big $= 180^\circ$. (If the crossing line is perpendicular, all eight angles are $90^\circ$.) Here $68^\circ$ is small and $y^\circ$ is big, so $y = 180 - 68 = 112$.
Similar TrianglesMedium
Right triangle $PQR$ is similar to right triangle $XYZ$, with the right angles at $R$ and $Z$, respectively.
If $\sin P = \dfrac{7}{25}$, what is the value of $\sin X$?
The takeaway: Similar triangles have the same angles; only the size changes, and every side is multiplied by the same factor. A trig ratio is a ratio of two sides, so the factor cancels: angle $X$ matches angle $P$, and $\sin X = \sin P = \frac{7}{25}$.
Right TrianglesMedium
In right triangle $GHK$, angle $H$ is a right angle.
One leg has a length of 28 centimeters.
The other leg has a length of 45 centimeters.
What is the length, in centimeters, of the hypotenuse?
The takeaway: Both given sides are legs, so use $a^2 + b^2 = c^2$: $c = \sqrt{28^2 + 45^2} = \sqrt{784 + 2025} = \sqrt{2809} = 53$ centimeters.
CirclesMedium
$(x - 2)^2 + (y + 11)^2 = 36$
In the $xy$-plane, the given equation defines a circle. What is the radius of the circle?
The takeaway: In $(x - h)^2 + (y - k)^2 = r^2$ the right side is $r^2$, not $r$. Here $r^2 = 36$, so $r = \sqrt{36} = 6$. (The diameter would be $12$.)
AnglesMedium
In the figure, two lines intersect.
What is the value of $x$?
The takeaway: Where two lines cross, the angles across from each other (vertical angles) are equal. So $5x - 20 = 105$, which gives $5x = 125$ and $x = 25$.
Right TrianglesMedium
In right triangle $KLM$, angle $L$ is a right angle.
The leg opposite angle $K$ has a length of 80 units.
The leg adjacent to angle $K$ has a length of 18 units.
What is the value of $\tan K$?
The takeaway: Tangent is opposite over adjacent (the TOA in SOH-CAH-TOA): $\tan K = \frac{\text{opposite}}{\text{adjacent}} = \frac{80}{18} = \frac{40}{9}$. Opposite and adjacent depend on which angle you stand at; tangent never uses the hypotenuse.
Similar TrianglesChallenge
In triangles $LMN$ and $PQR$, angle $L$ is congruent to angle $Q$, and angle $N$ is congruent to angle $R$. In triangle $LMN$, $LM = 15$, $MN = 18$, and $LN = 20$. In triangle $PQR$, $PQ = 9$.
What is the length of $\overline{QR}$?
The takeaway: The equal angles decide which vertices match, not the order of the letters. Angle $L$ matches $Q$ and $N$ matches $R$, so $M$ matches $P$. Side $\overline{PQ}$ matches $\overline{ML}$, so the scale factor is $\frac{9}{15} = \frac{3}{5}$, and side $\overline{QR}$ matches $\overline{LN}$: $QR = 20 \cdot \frac{3}{5} = 12$. (A similarity statement written in order, like $\triangle ABC \sim \triangle DEF$, also tells you the matches.)
ScalingMedium
Two similar square tiles are cut from the same sheet of glass.
Every side of the larger tile is 5 times the corresponding side of the smaller tile. The smaller tile has an area of 12 square inches. What is the area, in square inches, of the larger tile?
The takeaway: Multiply every length by $k$ and areas are multiplied by $k^2$ (volumes by $k^3$). The sides scale by $5$, so the area scales by $5^2 = 25$: $12 \cdot 25 = 300$ square inches.
Right TrianglesChallenge
In triangle $PQR$, angle $Q$ is a right angle, $PQ = 20$, and $QR = 21$.
What is the value of $\cos P - \sin R$?
The takeaway: Angle $Q$ is $90^\circ$, so $P + R = 90^\circ$, and sine and cosine trade places: $\cos P = \sin R$. Both equal $\frac{PQ}{PR}$, because $PQ$ is adjacent to $P$ and opposite $R$. So $\cos P - \sin R = 0$, and the side lengths are not needed.
CirclesMedium
$(x + 8)^2 + (y + 3)^2 = 49$
In the $xy$-plane, the given equation defines a circle. What are the coordinates of the center of the circle?
The takeaway: Standard form is $(x - h)^2 + (y - k)^2 = r^2$, so flip the signs to read the center. $(x + 8)$ is $(x - (-8))$ and $(y + 3)$ is $(y - (-3))$, so the center is $(-8, -3)$.
AnglesChallenge
In triangle $JKL$, the measure of angle $J$ is $59^\circ$, and the measure of angle $K$ is greater than $90^\circ$.
Which of the following could be the measure, in degrees, of angle $L$?
The takeaway: The three angles sum to $180^\circ$, so angle $L$ measures $180 - 59 - K = 121 - K$ degrees, where $K > 90$. That means angle $L$ must measure less than $31^\circ$ (and more than $0^\circ$). Only $24$ satisfies this.
ScalingChallenge
Two cylinders are similar. The volume of the larger cylinder is $27$ times the volume of the smaller cylinder. The surface area of the smaller cylinder is $40$ square inches.
What is the surface area, in square inches, of the larger cylinder?
The takeaway: Lengths scale by $k$, areas by $k^2$, volumes by $k^3$. Volume $\times 27$ means $k^3 = 27$, so $k = 3$, and surface area (an area) scales by $3^2 = 9$: $40 \cdot 9 = 360$ square inches.
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