Barron's Regents Geometry — Practice Exercises 1–11, Fully Solved

Answer: (3) $\overline{EA}\parallel\overline{CG}$, and $\overline{EH}$ is skew to $\overline{FB}$.

Why (3) works. $\overline{EA}$ and $\overline{CG}$ are both vertical edges of the prism, so they are parallel. $\overline{EH}$ lies in the top face and $\overline{FB}$ is a vertical edge from F (top) to B (bottom); they do not share a point and are not parallel, so by definition they are skew.

Why the others fail. (1) E, H, D, A are coplanar (they form one side face), but four points on a face are not collinear. (2) $\overline{HD}$ and $\overline{CD}$ share point D, so they cannot be skew. (4) $\overline{EA}$ is vertical and $\overline{BC}$ is a horizontal bottom edge; they never meet, so they are skew, not perpendicular. Perpendicular requires intersection.

Answer: (1) the union of two rays with a common endpoint.

Why (1) is the definition. Every angle is built from exactly two rays that share a common endpoint (the vertex). "Union" means the angle is those two rays taken together as one figure.

Why the others fail. (2) confuses measurement with definition — degrees are how we measure an angle, not what it is. (3) is a colloquialism at best; a triangle has three angles but an angle exists independently. (4) is not a defined term in geometry — lines are straight by definition.

Answer: (3) 48°.

Figure: BFA is a horizontal line through F. CF is perpendicular to it, pointing straight up. E is in the upper-right region between C and A, with $\angle CFE = 42^\circ$. D is in the lower-left region, on the ray opposite $\overrightarrow{FE}$ — so E, F, D are collinear.

Step 1 — use the perpendicular. Since $\overrightarrow{CF}\perp\overrightarrow{BFA}$, we know $\angle CFA = 90^\circ$.

Step 2 — split $\angle CFA$. $\angle CFA$ is split by ray FE into $\angle CFE$ and $\angle EFA$: $$\angle EFA = 90^\circ - 42^\circ = 48^\circ.$$

Step 3 — vertical angles. Since E, F, D are collinear and B, F, A are collinear, $\angle EFA$ and $\angle DFB$ are vertical angles. Vertical angles are congruent, so $m\angle BFD = 48^\circ.$

Answer: (4) 68°.

Step 1 — write the bisector equation. A bisector cuts the whole angle in half, so the whole equals twice the half: $$m\angle ABD = 2 \cdot m\angle ABC.$$ Substitute: $$8x - 12 = 2(3x + 4).$$

Step 2 — solve for x.

8x - 12 = 6x + 8
2x = 20
x = 10

Step 3 — find $m\angle ABD$. $$m\angle ABD = 8(10) - 12 = 80 - 12 = 68^\circ.$$

Answer: (2) 17°.

Step 1 — identify the angle pair. The two marked angles sit in the same relative position at their respective intersections (both on the upper-left of transversal $p$). Those are corresponding angles.

Step 2 — use the parallel test. Two lines cut by a transversal are parallel if and only if a pair of corresponding angles are congruent. Set them equal:

x + 28 = 4x - 23
28 + 23 = 4x - x
51 = 3x
x = 17

Answer: $FG = 14$.

Step 1 — translate each midpoint into an equal-length equation.

• G is the midpoint of $\overline{FH}$, so $FG = GH$. Therefore $GH = x^2 + 5x$.
• H is the midpoint of $\overline{GI}$, so $GH = HI$. Therefore $x^2 + 5x = 3x + 8$.

Step 2 — solve the quadratic.

x² + 5x = 3x + 8
x² + 2x - 8 = 0
(x + 4)(x - 2) = 0
x = -4  or  x = 2

A length can't be negative, so we test both. If $x = -4$: $FG = (-4)^2 + 5(-4) = 16 - 20 = -4$ — reject. If $x = 2$: $FG = 4 + 10 = 14$ — keep.

Step 3 — final answer. $FG = 14.$

Answer: (3) 134°.

Step 1 — draw an auxiliary line. Through the middle vertex, draw a line parallel to both $m$ and $n$. This slices $\angle 2$ into two smaller angles, one above the auxiliary line and one below.

Step 2 — use alternate interior angles. The lower piece of $\angle 2$ and $\angle 1$ are alternate interior angles between line $m$ and the auxiliary line, so the lower piece $= 32^\circ$. The upper piece of $\angle 2$ is then $78^\circ - 32^\circ = 46^\circ$.

Step 3 — identify $\angle 3$. The upper piece of $\angle 2$ and the angle above line $n$ (on the same side of the ray) are alternate interior angles between the auxiliary line and $n$, so that angle above the ray at the top vertex is also $46^\circ$. The angle $\angle 3$ marked in the figure is the supplement to the $46^\circ$ angle (they form a linear pair along line $n$): $$m\angle 3 = 180^\circ - 46^\circ = 134^\circ.$$

Answer: 15 sides.

Fast method — use exterior angles. Each exterior angle of a regular polygon is $180^\circ$ minus the interior angle: $$180^\circ - 156^\circ = 24^\circ.$$ The exterior angles of any convex polygon sum to $360^\circ$, so for a regular polygon: $$n = \frac{360^\circ}{\text{each exterior angle}} = \frac{360^\circ}{24^\circ} = 15.$$

Slow method — use interior angle formula. The formula $\dfrac{(n-2)\cdot 180^\circ}{n} = 156$ gives $180n - 360 = 156n$, so $24n = 360$ and $n = 15$. Same answer, more arithmetic.

Answer: (3) 45°.

Step 1 — find the interior angle of the regular octagon. Each interior angle: $\dfrac{(8-2)\cdot 180^\circ}{8} = \dfrac{1080^\circ}{8} = 135^\circ.$

Step 2 — find the angles at A and G that face Q. Q lies outside the octagon on the H-side. At vertex A, the ray $\overrightarrow{AQ}$ is opposite to $\overrightarrow{AB}$ (because $\overline{AB}$ was extended past A). So $\angle QAH$ and $\angle BAH$ form a linear pair: $$\angle QAH = 180^\circ - 135^\circ = 45^\circ.$$ Similarly at G, $\overrightarrow{GQ}$ is opposite $\overrightarrow{GF}$: $$\angle QGH = 180^\circ - 135^\circ = 45^\circ.$$

Step 3 — use the quadrilateral Q-A-H-G. The four vertices Q, A, H, G form a quadrilateral. Vertex H is on the boundary of the octagon between A and G, and the quadrilateral wraps around H on the outside of the octagon — so its interior angle at H is the reflex of the octagon's interior angle: $$\text{interior angle of quad at H} = 360^\circ - 135^\circ = 225^\circ.$$

Step 4 — sum the quadrilateral to 360°.

∠Q + ∠QAH + ∠AHG (reflex) + ∠HGQ = 360°
∠Q + 45° + 225° + 45° = 360°
∠Q + 315° = 360°
∠Q = 45°

Answer: 60°.

Step 1 — place the hexagon on a circle. A regular hexagon has its six vertices equally spaced on a circle centered at O. Label the vertices at $0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ$ around the center: A at $0^\circ$, B at $60^\circ$, C at $120^\circ$, D at $180^\circ$, E at $240^\circ$, F at $300^\circ$.

Step 2 — recognize both segments as diameters. $\overline{AD}$ connects A ($0^\circ$) and D ($180^\circ$) — opposite vertices, so it passes through the center. $\overline{FC}$ connects F ($300^\circ$) and C ($120^\circ$) — opposite vertices, so it also passes through the center. Therefore O is the center of the hexagon.

Step 3 — compute the central angle. $\angle AOF$ is the central angle between vertex A ($0^\circ$) and vertex F ($300^\circ$), measured the short way: $$\angle AOF = 360^\circ - 300^\circ = 60^\circ.$$

Answer — two-column proof:

Statements                                Reasons
--------------------------------------    -----------------------------------
1. WPX and YPZ are lines that            1. Given.
   intersect at P.
2. ∠WPY and ∠YPX form a linear pair    2. If two angles form a linear pair,
   (both share ray PY, and rays PW           they are supplementary.
   and PX form a straight line). So
   m∠WPY + m∠YPX = 180°.
3. ∠YPX and ∠XPZ form a linear pair    3. Same reason (rays PY and PZ form a
   (both share ray PX, and rays PY           straight line).
   and PZ form a straight line). So
   m∠YPX + m∠XPZ = 180°.
4. m∠WPY + m∠YPX =                    4. Transitive property of equality
   m∠YPX + m∠XPZ.                        (both equal 180°).
5. m∠WPY = m∠XPZ.                      5. Subtraction property of equality
                                              (subtract m∠YPX from both sides).
6. ∠WPY ≅ ∠ZPX.                       6. Definition of congruent angles.

Paragraph version. Because W, P, X are collinear, angles $\angle WPY$ and $\angle YPX$ form a linear pair and sum to $180^\circ$. Because Y, P, Z are collinear, angles $\angle YPX$ and $\angle XPZ$ also sum to $180^\circ$. Setting the two sums equal and subtracting the shared angle $\angle YPX$ from both sides gives $m\angle WPY = m\angle XPZ$, so $\angle WPY \cong \angle ZPX$. $\blacksquare$