[Grade 7-8 Geometry] 16 Essential Types of Basic Geometric Figures & Angles (32 Practice Problems & Solutions)

 

Master Geometry LAB | Grade 7–8 Core Curriculum

Basic Geometric Figures & Angles: 16 Essential Types Masterclass

From points, lines, and skew line deduction to segment ratios, vertical angles, and parallel line paper-folding.
Master all exam scenarios with 32 Standard & Advanced Twin Practice Problems.

TYPE 01

Points & Lines of Intersection in Polyhedra

[Problem 1-1 | Standard]

In a heptagonal prism (7-gonal prism), let the number of intersection points (vertices) be $a$, the number of intersection lines (edges) be $b$, and the number of faces be $c$. Find the value of $a - b + c$.

💡 View Solution & Answer

Answer: $2$
• Vertices ($a$): $7 \times 2 = 14$
• Edges ($b$): $7 \times 3 = 21$
• Faces ($c$): $7 + 2 = 9$
• Therefore, $a - b + c = 14 - 21 + 9 = 2$ (Euler's Formula: $V - E + F = 2$).

[Problem 1-2 | Advanced]

A pyramid has $18$ lines of intersection (edges). Consider a prism whose base has the exact same shape as this pyramid's base. Find the sum of the number of vertices and edges of this prism.

💡 View Solution & Answer

Answer: $45$
• An $n$-gonal pyramid has $2n$ edges $\implies 2n = 18 \implies n = 9$ (nonagonal pyramid).
• A nonagonal prism has $n = 9$ base sides.
• Vertices of the prism $= 2 \times 9 = 18$; Edges of the prism $= 3 \times 9 = 27$.
• Sum $= 18 + 27 = 45$.

🌿 Yul's Pro-Tip: Lines intersecting create vertices (points of intersection); planes intersecting create edges (lines of intersection). Remember that for every convex polyhedron, Euler's formula $V - E + F = 2$ always holds.

TYPE 02

Determining Lines, Rays, and Segments

[Problem 2-1 | Standard]

There are $6$ distinct points on a plane. Exactly $4$ of these points lie on a straight line $l$, while the remaining $2$ points do not lie on $l$. Find the number of distinct lines $a$ and distinct segments $b$ that can be drawn using any two of these points. Calculate $a + b$.

💡 View Solution & Answer

Answer: $25$
• Segments ($b$): Pairs of 6 points $= \frac{6 \times 5}{2} = 15$.
• Lines ($a$): Total pairs minus collinear duplicate pairs plus original line $l$: $\frac{6 \times 5}{2} - \frac{4 \times 3}{2} + 1 = 15 - 6 + 1 = 10$.
• $a + b = 10 + 15 = 25$.

[Problem 2-2 | Advanced]

There are $6$ points on a semicircle: $3$ points $A, B, C$ lie on the diameter, and $3$ points $D, E, F$ lie on the curved arc. Find the total number of distinct rays that can be formed using any two of these $6$ points.

💡 View Solution & Answer

Answer: $26$
• Along diameter ($A, B, C$ collinear): $A$ rightward ($1$), $B$ both directions ($2$), $C$ leftward ($1$) $\implies 4$ rays.
• Pairs among arc points ($D, E, F$ non-collinear): $3 \times 2 = 6$ rays.
• Pairs between diameter and arc: $3 \times 3 \times 2 = 18$ rays.
• Total rays $= 4 + 6 + 18 = 28 - 2 = 26$ rays.

🌿 Yul's Pro-Tip: A ray is defined by two criteria: 1) Endpoint (starting origin) and 2) Direction vector. Two rays $\vec{AB}$ and $\vec{AC}$ are identical if and only if they share the exact same starting point and extend in the same direction.

TYPE 03

Midpoints & Segment Proportional Partitioning

[Problem 3-1 | Standard]

Points $A, B, C, D$ lie in order on a line. Point $M$ is the midpoint of segment $AB$, and point $N$ is the midpoint of segment $CD$. Given that $\overline{AD} = 28\text{ cm}$ and $\overline{BC} = 8\text{ cm}$, find the length of segment $MN$.

💡 View Solution & Answer

Answer: $18\text{ cm}$
• $\overline{AB} + \overline{CD} = \overline{AD} - \overline{BC} = 28 - 8 = 20\text{ cm}$.
• Because $M, N$ are midpoints: $\overline{MB} + \overline{CN} = \frac{1}{2}(20) = 10\text{ cm}$.
• $\overline{MN} = \overline{MB} + \overline{BC} + \overline{CN} = 10 + 8 = 18\text{ cm}$.

[Problem 3-2 | Advanced]

Points $A, B, C, D$ lie consecutively on a line such that $\overline{AB} : \overline{BC} = 2 : 3$ and $\overline{BC} : \overline{CD} = 2 : 1$. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively. If $\overline{MN} = 19\text{ cm}$, determine the total length of $\overline{AD}$.

💡 View Solution & Answer

Answer: $26\text{ cm}$
• Unify ratio: $\overline{AB} : \overline{BC} : \overline{CD} = 4 : 6 : 3$. Let lengths be $4k, 6k, 3k$.
• Midpoints: $\overline{MB} = 2k$, $\overline{CN} = 1.5k$.
• $\overline{MN} = 2k + 6k + 1.5k = 9.5k = \frac{19}{2}k = 19\text{ cm} \implies k = 2\text{ cm}$.
• Total $\overline{AD} = 4k + 6k + 3k = 13k = 13 \times 2 = 26\text{ cm}$.

🌿 Yul's Pro-Tip: When chained ratios overlap on a common segment ($\overline{BC}$), convert to a continuous ratio using the least common multiple. Assign a single parameter $k$ to turn geometry into a simple linear equation.

TYPE 04

Internal & External Segment Division (Coordinate Mapping)

[Problem 4-1 | Standard]

On segment $AB$, a point $C$ satisfies $\overline{AC} : \overline{CB} = 3 : 2$. Point $M$ is the midpoint of $\overline{AC}$, and point $N$ is the midpoint of $\overline{AB}$. If $\overline{MN} = 4\text{ cm}$, find the total length of $\overline{AB}$.

💡 View Solution & Answer

Answer: $40\text{ cm}$
• Map coordinate $A = 0$. Let $\overline{AC} = 3k$, $\overline{CB} = 2k \implies \overline{AB} = 5k$.
• $M$ coordinate $= 1.5k$; $N$ coordinate $= 2.5k$.
• $\overline{MN} = 2.5k - 1.5k = k = 4\text{ cm}$.
• $\overline{AB} = 5k = 5 \times 4 = 40\text{ cm}$.

[Problem 4-2 | Advanced]

Points $A, B, C, D$ lie in order on a line. Point $C$ is the trisection point of $\overline{AD}$ closer to $D$ ($\overline{AC} : \overline{CD} = 2 : 1$). Point $B$ is the midpoint of $\overline{AC}$. If $M$ is the midpoint of $\overline{BD}$, determine the ratio of the length of $\overline{AB}$ to $\overline{MC}$.

💡 View Solution & Answer

Answer: $4$ times ($\overline{AB} : \overline{MC} = 4 : 1$)
• Set integer coordinate scale with $A = 0$ and total length $AD = 6k$.
• $C = 4k$, $D = 6k$. Midpoint $B$ of $[0, 4k]$ is at $2k$.
• Segment $BD$ spans from $2k$ to $6k$ (length $4k$). Its midpoint $M = 4k$ or setting fine scale $AD = 12 \implies C = 8, B = 4, M = 8$.
• Ratio evaluates to $\frac{\overline{AB}}{\overline{MC}} = 4$.

🌿 Yul's Pro-Tip: Whenever midpoints and trisection points intersect, set the origin at point $A(0)$ and scale the total length to the least common denominator ($6$ or $12$). Integer coordinates eliminate fraction errors.

TYPE 05

Complementary & Supplementary Angle Equations

[Problem 5-1 | Standard]

The supplement of an angle $\angle A$ is $15^\circ$ less than $4$ times the complement of $\angle A$. Find the measure of $\angle A$.

💡 View Solution & Answer

Answer: $55^\circ$
• Let $\angle A = x^\circ$. Complement $= 90^\circ - x$; Supplement $= 180^\circ - x$.
• Equation: $(180 - x) = 4(90 - x) - 15$
• $180 - x = 360 - 4x - 15 \implies 3x = 165 \implies x = 55^\circ$.

[Problem 5-2 | Advanced]

For two angles $\angle A$ and $\angle B$, the supplement of $\angle A$ is $3$ times the measure of $\angle B$, and the ratio of the complement of $\angle A$ to the complement of $\angle B$ is $1 : 3$. Find the measure of $\angle A$.

💡 View Solution & Answer

Answer: $72^\circ$
• Condition 1: $180 - A = 3B \implies A + 3B = 180^\circ$.
• Condition 2: $\frac{90 - A}{90 - B} = \frac{1}{3} \implies 3(90 - A) = 90 - B \implies 3A - B = 180^\circ$.
• Substitute $B = 3A - 180$: $A + 3(3A - 180) = 180 \implies 10A = 720 \implies A = 72^\circ$.

🌿 Yul's Pro-Tip: Complementary angles sum to $90^\circ$ (substitute $90 - x$). Supplementary angles sum to $180^\circ$ (substitute $180 - x$). Translate word problems directly into algebraic linear systems.

TYPE 06

Clock Angle Problems (Minute & Hour Hand Rates)

[Problem 6-1 | Standard]

Between 4:00 and 5:00, at what exact time do the hour and minute hands point in opposite directions, forming a straight line ($180^\circ$)?

💡 View Solution & Answer

Answer: $4\text{ o'clock } 54\frac{6}{11}\text{ minutes}$
• At $4:x$, position from 12:00: Minute hand $= 6^\circ x$; Hour hand $= 120^\circ + 0.5^\circ x$.
• Opposing line: $6x - (120 + 0.5x) = 180^\circ \implies 5.5x = 300 \implies x = \frac{600}{11} = 54\frac{6}{11}$.

[Problem 6-2 | Advanced]

The current time is between 2:00 and 3:00. Exactly 10 minutes from now, the smaller angle between the hour and minute hands will measure $90^\circ$. Find the exact current time.

💡 View Solution & Answer

Answer: $2\text{ o'clock } 17\frac{3}{11}\text{ minutes}$
• Let current time be $2:x$. In 10 minutes, time is $2:(x + 10)$.
• Minute hand past hour hand at $90^\circ$: $6(x + 10) - [60 + 0.5(x + 10)] = 90^\circ$.
• $5.5(x + 10) = 150 \implies x + 10 = \frac{300}{11} = 27\frac{3}{11} \implies x = 17\frac{3}{11}$.

🌿 Yul's Pro-Tip: The universal rates: Minute hand $= 6^\circ/\text{min}$; Hour hand $= 0.5^\circ/\text{min}$ ($30^\circ/\text{hr}$). Model angular separation with $|\theta_{\text{min}} - \theta_{\text{hr}}| = \Delta$.

TYPE 07

Vertical Angles & Multi-Line Intersections

[Problem 7-1 | Standard]

Three straight lines intersect at a single point $O$, forming $6$ angles. If three non-adjacent angles measure $2x + 10^\circ$, $3x - 15^\circ$, and $x + 35^\circ$, find the measure of the largest vertical angle.

💡 View Solution & Answer

Answer: $60^\circ$
• Three non-adjacent angles formed by intersecting lines sum to a straight angle ($180^\circ$).
• $(2x + 10) + (3x - 15) + (x + 35) = 180^\circ \implies 6x + 30 = 180 \implies x = 25^\circ$.
• All three angles evaluate to $60^\circ$.

[Problem 7-2 | Advanced]

There are $4$ straight lines on a plane such that no two lines are parallel and no three lines intersect at a single point. Find the total number of vertical angle pairs formed.

💡 View Solution & Answer

Answer: $12\text{ pairs}$
• With 4 lines in general position, number of intersection points $= \frac{4 \times 3}{2} = 6$.
• Each intersection point of 2 lines produces exactly $2$ pairs of vertical angles.
• Total vertical angle pairs $= 6 \times 2 = 12\text{ pairs}$.

🌿 Yul's Pro-Tip: Vertical angles are always congruent. When lines intersect at multiple distinct points, compute $\text{Total Pairs} = 2 \times (\text{Number of Intersections})$.

TYPE 08

Perpendiculars & Distance from a Point to a Line

[Problem 8-1 | Standard]

In right triangle $ABC$ with $\angle C = 90^\circ$, the sides are $\overline{AB} = 15\text{ cm}$, $\overline{BC} = 9\text{ cm}$, and $\overline{CA} = 12\text{ cm}$. Find the shortest distance from vertex $C$ to line $AB$.

💡 View Solution & Answer

Answer: $7.2\text{ cm}$ ($\frac{36}{5}\text{ cm}$)
• Shortest distance is the altitude length $\overline{CH}$.
• Equate triangle area formulas: $\frac{1}{2} \times 9 \times 12 = \frac{1}{2} \times 15 \times \overline{CH} \implies 108 = 15\overline{CH} \implies \overline{CH} = 7.2\text{ cm}$.

[Problem 8-2 | Advanced]

In trapezoid $ABCD$, $\overline{AD} \parallel \overline{BC}$ and $\overline{AB} \perp \overline{BC}$. Given $\overline{AD} = 6\text{ cm}$, $\overline{BC} = 10\text{ cm}$, $\overline{CD} = 5\text{ cm}$, and $\text{area} = 24\text{ cm}^2$, find the perpendicular distance from point $B$ to line $CD$.

💡 View Solution & Answer

Answer: $6\text{ cm}$
• Height $\overline{AB} = h$: $\frac{1}{2}(6 + 10)h = 24 \implies 8h = 24 \implies h = 3\text{ cm}$.
• Area of $\triangle BCD$: $\frac{1}{2} \times \overline{BC} \times h = \frac{1}{2} \times 10 \times 3 = 15\text{ cm}^2$.
• Equating with base $\overline{CD} = 5$: $\frac{1}{2} \times 5 \times d = 15 \implies d = 6\text{ cm}$.

🌿 Yul's Pro-Tip: The distance from a point to a line is the length of the perpendicular segment (foot of perpendicular). When direct calculation is difficult, use Area Equivalence ($\text{Base}_1 \times \text{Height}_1 = \text{Base}_2 \times \text{Height}_2$).

TYPE 09

Coplanar Line Relations & Plane Region Division

[Problem 9-1 | Standard]

There are $5$ distinct coplanar lines. Exactly $2$ of them are parallel, and no three lines intersect at a single point. Find the total number of intersection points formed.

💡 View Solution & Answer

Answer: $9\text{ points}$
• Maximum possible intersections for 5 lines: $\frac{5 \times 4}{2} = 10$.
• Since 2 lines are parallel, they lose 1 intersection point.
• Total intersections $= 10 - 1 = 9$.

[Problem 9-2 | Advanced]

Four distinct coplanar lines divide a plane into several regions. Let $m$ be the minimum possible number of regions and $M$ be the maximum possible number of regions. Find $M - m$.

💡 View Solution & Answer

Answer: $6$
• Minimum ($m$): When all 4 lines are mutually parallel: $m = 4 + 1 = 5$ regions.
• Maximum ($M$): When no two lines are parallel and no three are concurrent: $M = 1 + \frac{4 \times 5}{2} = 11$ regions.
• $M - m = 11 - 5 = 6$.

🌿 Yul's Pro-Tip: Two coplanar lines can only: 1) Intersect at one point, 2) Be parallel, or 3) Coincide. Skew lines cannot exist on a single plane!

TYPE 10

Skew Lines in 3D Space (Complement Counting)

[Problem 10-1 | Standard]

In the triangular prism $ABC-DEF$, let $a$ be the number of edges that are in a skew position to edge $\overline{BC}$, and let $b$ be the number of edges parallel to edge $\overline{AD}$. Find the value of $a + b$.

▲ [Figure 1] Triangular Prism $ABC-DEF$: Reference Edge $\overline{BC}$ (Blue) & Skew Edges $\overline{AD}, \overline{DE}, \overline{DF}$ (Orange)

💡 View Solution & Answer

Answer: $5$
• Total edges $= 9$.
• Skew to $BC$: Exclude intersecting ($AB, AC, BE, CF \implies 4$), parallel ($EF \implies 1$), and itself ($BC \implies 1$). Thus $a = 9 - 6 = 3$ edges ($AD, DE, DF$).
• Parallel to $AD$: $BE, CF \implies b = 2$.
• $a + b = 3 + 2 = 5$.

[Problem 10-2 | Advanced]

A regular octahedron has $6$ vertices and $12$ edges. For any chosen edge $e$, find the total number of edges that are in a skew position to $e$.

💡 View Solution & Answer

Answer: $4\text{ edges}$
• Total edges $= 12$.
• Edges sharing endpoints with $e$: Each endpoint connects to 3 other edges $\implies 3 \times 2 = 6$ edges.
• Parallel edge: Exactly $1$ directly opposite edge.
• Itself: $1$ edge.
• Skew edges $= 12 - 6 - 1 - 1 = 4$.

🌿 Yul's Pro-Tip: Never search for skew lines directly. Always use the Complement Rule: $\text{Skew Lines} = \text{Total Edges} - \text{Intersecting Edges} - \text{Parallel Edges} - \text{Self}$.

TYPE 11

Spatial Lines and Planes Relationships

[Problem 11-1 | Standard]

In rectangular prism $ABCD-EFGH$, let $x$ be the number of edges parallel to the diagonal plane $BFHD$, and $y$ be the number of edges perpendicular to face $ABCD$. Find $x + y$.

💡 View Solution & Answer

Answer: $6$
• Edges parallel to plane $BFHD$: $\overline{AE}$ and $\overline{CG} \implies x = 2$.
• Edges perpendicular to top face $ABCD$: The 4 vertical pillar edges $AE, BF, CG, DH \implies y = 4$.
• $x + y = 2 + 4 = 6$.

[Problem 11-2 | Advanced]

In a right trapezoidal prism whose top and bottom bases are parallel, find the minimum number of mutually perpendicular pairs of faces that are guaranteed to exist.

💡 View Solution & Answer

Answer: $8\text{ pairs}$
• In any right prism, every lateral side is perpendicular to both bases.
• 4 lateral faces $\times 2$ bases $= 8$ perpendicular pairs guaranteed.

🌿 Yul's Pro-Tip: A line is perpendicular to a plane if and only if it is perpendicular to at least two intersecting lines lying on that plane.

TYPE 12

True/False Spatial Geometric Propositions

[Problem 12-1 | Standard]

For distinct lines $l, m, n$ and distinct planes $P, Q, R$, which statements are ALWAYS TRUE?
A. If $l \parallel m$ and $m \parallel n$, then $l \parallel n$.
B. If $l \perp P$ and $m \perp P$, then $l \parallel m$.
C. If $l \parallel P$ and $m \parallel P$, then $l \parallel m$.
D. If $P \perp Q$ and $Q \perp R$, then $P \parallel R$.

💡 View Solution & Answer

Answer: A, B
• A: Transitivity of parallel lines holds in 3D space (True).
• B: Two lines perpendicular to the same plane are mutually parallel (True).
• C: Lines parallel to the same plane can intersect or be skew (False).
• D: Three mutually perpendicular planes form the corner of a cube (False).

[Problem 12-2 | Advanced]

For distinct lines $l, m$ and distinct planes $P, Q$ in space, suppose $l \perp P$, $m \parallel Q$, and $P \parallel Q$. Determine all possible positional relationships between lines $l$ and $m$.

💡 View Solution & Answer

Answer: Perpendicularly intersecting at one point, or perpendicular and skew
• $P \parallel Q$ and $l \perp P \implies l \perp Q$.
• Since $m \parallel Q$, line $l$ is perpendicular to line $m$.
• In 3D space, two perpendicular lines can either intersect at a right angle or be perpendicular skew lines.

🌿 Yul's Pro-Tip: Never evaluate 3D propositions purely by imagination. Sketch a quick cube model and assign the edges and faces as counterexamples.

TYPE 13

Corresponding & Alternate Interior Angles

[Problem 13-1 | Standard]

Two lines $l$ and $m$ are parallel ($l \parallel m$). A point $C$ lies between the parallel lines such that the angle formed with line $l$ is $45^\circ$ and with line $m$ is $55^\circ$. Find the measure of $\angle ACB$.

💡 View Solution & Answer

Answer: $100^\circ$
• Draw auxiliary line $n$ through $C$ parallel to $l$ and $m$.
• By alternate interior angles: $\angle ACB = 45^\circ + 55^\circ = 100^\circ$.

[Problem 13-2 | Advanced]

In the figure with $l \parallel m$, $\triangle ABC$ has vertex $A$ on line $l$ and side $BC$ on line $m$. The line extending $AC$ forms an acute angle of $68^\circ$ with line $m$. If $\angle BAC = 52^\circ$, find the obtuse angle formed between line $l$ and side $AB$.

💡 View Solution & Answer

Answer: $120^\circ$
• By alternate interior angles, $\angle C = 68^\circ$.
• In $\triangle ABC$, $\angle B = 180^\circ - (68^\circ + 52^\circ) = 60^\circ$.
• Supplementary angle formed with line $l = 180^\circ - 60^\circ = 120^\circ$.

🌿 Yul's Pro-Tip: Combine the Exterior Angle Theorem ($\text{Exterior Angle} = \text{Sum of Two Remote Interior Angles}$) with parallel alternate angles to bypass multi-step linear systems.

TYPE 14

Parallel Lines with Zig-Zag Auxiliary Lines

[Problem 14-1 | Standard]

Two lines $l$ and $m$ are parallel ($l \parallel m$). A zig-zag line between them forms left-pointing angles of $35^\circ$ and $45^\circ$, and right-pointing angles of $50^\circ$ and $x^\circ$. Find the value of $x$.

▲ [Figure 2] Parallel Auxiliary Lines (Dashed) & Zig-Zag Rule: $\sum(\text{Left Angles}) = \sum(\text{Right Angles})$

💡 View Solution & Answer

Answer: $30^\circ$
• Zig-Zag Theorem: $\sum(\text{Left-Pointing Angles}) = \sum(\text{Right-Pointing Angles})$.
• $35^\circ + 45^\circ = 50^\circ + x \implies 80^\circ = 50^\circ + x \implies x = 30^\circ$.

[Problem 14-2 | Advanced]

Two lines $l$ and $m$ are parallel ($l \parallel m$). The left-pointing acute angles are $x + 20^\circ$ and $2x + 10^\circ$, while the right-pointing acute angles are $2x + 5^\circ$ and $55^\circ$. Determine the value of $x$.

💡 View Solution & Answer

Answer: $30^\circ$
• Equate directional angles: $(x + 20) + (2x + 10) = (2x + 5) + 55$.
• $3x + 30 = 2x + 60 \implies x = 30^\circ$.

🌿 Yul's Pro-Tip: The Zig-Zag Rule states that drawing parallel auxiliary lines through every vertex always yields $\sum \theta_{\text{left}} = \sum \theta_{\text{right}}$.

TYPE 15

Paper Folding Geometry & Isosceles Triangles

[Problem 15-1 | Standard]

A rectangular paper strip of uniform width is folded along segment $EF$. If the folded angle $\angle C'FE = 65^\circ$, find the measure of $\angle GEF$.

▲ [Figure 3] [Folded Angle Invariance] + [Alternate Interior Angles] $\implies$ Overlapping $\triangle GEF$ is always an Isosceles Triangle

💡 View Solution & Answer

Answer: $65^\circ$
• By reflection invariance, folded angle equals original angle: $\angle CFE = \angle C'FE = 65^\circ$.
• By alternate interior angles between parallel edges: $\angle GEF = \angle C'FE = 65^\circ$.
• Thus, $\triangle GEF$ is an isosceles triangle with $\angle GEF = 65^\circ$.

[Problem 15-2 | Advanced]

A rectangular sheet of paper is folded along line segment $PQ$ so that one corner lands on the opposite edge. The vertex angle of the overlapping triangle formed is $50^\circ$. Find the acute angle formed between the crease $PQ$ and the edge of the paper.

💡 View Solution & Answer

Answer: $65^\circ$
• The overlapping triangle is always isosceles.
• Base angles measure $\frac{180^\circ - 50^\circ}{2} = 65^\circ$.
• By alternate interior angles, the angle between the crease and the edge equals the base angle: $65^\circ$.

🌿 Yul's Pro-Tip: Paper folding geometry always follows: [Folded Angle Invariance] + [Parallel Alternate Angles] $\implies$ Overlapping Isosceles Triangle.

TYPE 16

Regular Polygons Intersecting Parallel Lines

[Problem 16-1 | Standard]

A regular pentagon $ABCDE$ is placed between two parallel lines $l \parallel m$ such that vertices $A$ and $C$ lie on lines $l$ and $m$, respectively. If the acute angle between line $l$ and side $AB$ is $22^\circ$, find the angle formed between line $m$ and side $CD$. (Note: One interior angle of a regular pentagon is $108^\circ$.)

💡 View Solution & Answer

Answer: $50^\circ$
• Draw auxiliary parallel line through vertex $B$. Top angle $= 22^\circ$.
• Bottom angle $= 108^\circ - 22^\circ = 86^\circ$.
• Alternate angle to $BC$ equals $86^\circ$. With interior angle $108^\circ$, angle to line $m = 50^\circ$.

[Problem 16-2 | Advanced]

A regular hexagon and an equilateral triangle share a common edge between two parallel lines $l \parallel m$. If one edge of the regular hexagon forms an angle of $25^\circ$ with line $l$, determine the maximum alternate interior angle that a vertex of the equilateral triangle can form with line $m$.

💡 View Solution & Answer

Answer: $85^\circ$
• Interior angle of regular hexagon $= 120^\circ$; equilateral triangle $= 60^\circ$.
• Tracing angular rotations across vertices: $120^\circ - 60^\circ + 25^\circ = 85^\circ$.

🌿 Yul's Pro-Tip: Write down regular interior angles first: Equilateral Triangle $= 60^\circ$, Square $= 90^\circ$, Regular Pentagon $= 108^\circ$, Regular Hexagon $= 120^\circ$. Then draw parallel lines through polygon vertices.

🌿

Yul's Insight: Those Who Draw Auxiliary Lines Master Geometry

These 16 essential types form the absolute bedrock of middle school geometry, directly connecting to high school coordinate proofs, vectors, and 3D spatial geometry.

When students struggle with geometry, it is rarely due to a lack of formulas. It is because they stare only at the printed lines. Visualizing a virtual bounding box to filter out skew lines, or sketching a single parallel auxiliary line through a zig-zag vertex, instantly converts chaotic diagrams into elementary linear equations.

Do not memorize proofs mechanically. When you understand why paper folding births isosceles triangles and why skew lines are best counted through complements, geometry ceases to be tedious memorization and becomes the most satisfying visual puzzle.

SELF-TEST WORKSHEET

[Practice Test] Basic Geometric Figures & Angle Mastery (16 Types)

💡 Instructions: Clicking [Print Practice Test] formats this page into a clean, 2-column A4 exam paper with all solutions hidden. Work through the problems at your own pace to evaluate your mastery!

Subject: Grade 7–8 Geometry Core

Name: ____________________

Score: _________ / 100

01. [Polyhedron Components]

In a heptagonal prism, find $a - b + c$, where $a$ is vertices, $b$ is edges, and $c$ is faces.

02. [Collinear Rays & Lines]

Given 6 points where 4 are collinear, find the sum of distinct lines and distinct segments.

03. [Segment Midpoint Chaining]

$M, N$ are midpoints of $AB, CD$. $\overline{AD} = 28\text{ cm}, \overline{BC} = 8\text{ cm}$. Find $\overline{MN}$.

04. [Trisection Coordinate Mapping]

$\overline{AC} : \overline{CB} = 3 : 2$. $M$ is midpoint of $AC$, $N$ is midpoint of $AB$. If $\overline{MN} = 4\text{ cm}$, find $\overline{AB}$.

05. [Complement & Supplement]

The supplement of $\angle A$ is $15^\circ$ less than $4$ times its complement. Find $\angle A$.

06. [Clock Separation Angle]

At what exact time between 4:00 and 5:00 do the hands form a straight line ($180^\circ$)?

07. [Vertical Angle Multi-Intersection]

Find the total number of vertical angle pairs formed by 4 lines with no two parallel and no three concurrent.

08. [Distance to Line via Area]

In right $\triangle ABC$ ($9, 12, 15$), find the shortest distance from vertex $C$ to hypotenuse $AB$.

09. [Plane Region Maximum Bounds]

Find $M - m$, where $M$ is max and $m$ is min regions divided by 4 coplanar lines.

10. [Skew Edges in Regular Octahedron]

Find the number of edges skew to a given edge $e$ in a regular octahedron.

11. [Line & Plane Perpendicularity]

In a rectangular prism, find $x + y$ where $x$ is edges $\parallel$ diagonal plane and $y$ is edges $\perp$ base.

12. [Spatial Proposition Deductions]

If $l \perp P, m \parallel Q, P \parallel Q$, determine all positional relationships between $l$ and $m$.

13. [Alternate Angles & Triangle Exterior]

$l \parallel m$. Alternate angle $= 68^\circ$, $\angle BAC = 52^\circ$. Find obtuse angle to $l$.

14. [Zig-Zag Left/Right Sums]

$l \parallel m$. Left angles: $35^\circ, 45^\circ$. Right angles: $50^\circ, x^\circ$. Find $x$.

15. [Paper Folding Crease Angle]

Rectangular strip folded with $\angle C'FE = 65^\circ$. Find $\angle GEF$.

16. [Regular Polygon on Parallel Lines]

Regular pentagon between $l \parallel m$. Acute angle to $l$ is $22^\circ$. Find angle to $m$.

📁 [Quick Answer Key] Solution Matrix (Click to Expand)

#	Answer	#	Answer	#	Answer
01	$2$	07	$12$ pairs	13	$120^\circ$
02	$25$	08	$7.2\text{ cm}$	14	$30^\circ$
03	$18\text{ cm}$	09	$6$	15	$65^\circ$
04	$40\text{ cm}$	10	$4$ edges	16	$50^\circ$
05	$55^\circ$	11	$6$	-
06	$4:54\frac{6}{11}$	12	$\perp$ Intersect or Skew	-