1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area

Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area

Geometry on the ACET emphasizes fast shape identification, spatial reasoning, area/volume ratios, and circle properties. Calculators are unavailable, so memorizing key geometric ratios saves time.

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1. Angle Relationships & Parallel Lines

When two parallel lines $L_1 \parallel L_2$ are cut by a transversal $T$:

• Corresponding Angles are equal ($1 = 5$).
• Alternate Interior Angles are equal ($3 = 6$).
• Alternate Exterior Angles are equal ($2 = 7$).
• Consecutive Interior Angles are supplementary ($3 + 5 = 180^\circ$).

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2. Triangles: Properties, Special Triangles & Area Formulas

Fundamental Triangle Inequalities & Angle Theorems

• Angle Sum: $\theta_1 + \theta_2 + \theta_3 = 180^\circ$.
• Exterior Angle Theorem: The measure of an exterior angle equals the sum of its two remote interior angles.
• Triangle Inequality Theorem: For sides $a, b, c$: $|a - b| < c < a + b$.

Special Right Triangles & Pythagorean Triples

$$\begin{aligned} \mathbf{45^\circ-45^\circ-90^\circ \text{ Triangle:}} & \quad \text{Legs } x, x \implies \text{Hypotenuse } x\sqrt{2} \\ \mathbf{30^\circ-60^\circ-90^\circ \text{ Triangle:}} & \quad \text{Short leg } x \implies \text{Long leg } x\sqrt{3}, \text{ Hypotenuse } 2x \end{aligned}$$

High-Frequency ACET Pythagorean Triples:

• $3 : 4 : 5 \quad (6:8:10, \; 9:12:15, \; 12:16:20)$
• $5 : 12 : 13 \quad (10:24:26)$
• $8 : 15 : 17$
• $7 : 24 : 25$

Comprehensive Area Formulas for Triangles

1. Standard Formula: $A = \frac{1}{2} b h$
2. SAS Trigonometric Formula: $A = \frac{1}{2} a b \sin C$
3. Equilateral Triangle Area: $A = \frac{s^2 \sqrt{3}}{4}$
4. Heron's Formula (Given 3 sides $a, b, c$): $$A = \sqrt{s(s - a)(s - b)(s - c)}, \quad s = \frac{a + b + c}{2}$$

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3. Polygon Properties & Interior/Exterior Angles

For a convex polygon with $n$ sides:

• Sum of Interior Angles: $S_{\text{int}} = (n - 2) \times 180^\circ$
• Interior Angle of Regular Polygon: $I = \frac{(n - 2) \times 180^\circ}{n}$
• Sum of Exterior Angles: $S_{\text{ext}} = 360^\circ$
• Exterior Angle of Regular Polygon: $E = \frac{360^\circ}{n}$
• Number of Diagonals: $D = \frac{n(n - 3)}{2}$

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4. Circle Theorems & Sector Measures

Primary Circle Formulas

• Circumference: $C = 2\pi r = \pi d$
• Area: $A = \pi r^2$
• Arc Length ($L$ for central angle $\theta^\circ$): $L = \frac{\theta}{360^\circ} \times 2\pi r$
• Sector Area ($A_{\text{sector}}$): $A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$

Core Circle Theorems

1. Inscribed Angle Theorem: An inscribed angle is half the central angle intercepting the same arc: $\theta_{\text{inscribed}} = \frac{1}{2} \theta_{\text{central}}$.
2. Thales' Theorem: An angle inscribed in a semicircle is always a right angle ($90^\circ$).
3. Tangent Line Theorem: A line tangent to a circle is perpendicular to the radius at the point of tangency.
4. Intersecting Chords Theorem: $PA \cdot PB = PC \cdot PD$.
5. Secant-Tangent Theorem: $PT^2 = PA \cdot PB$.

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Worked Step-by-Step ACET Exam Problems

Problem 1: Sector Area and Shaded Segment Calculation

A circle has a radius of $6\text{ cm}$. A central angle of $60^\circ$ intercepts an arc. What is the area of the shaded segment formed between the chord and the arc?

Step 1: Calculate total sector area.

$$A_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \pi (6)^2 = \frac{1}{6} \times 36 \pi = 6\pi \text{ cm}^2$$

Step 2: Calculate area of the subtended triangle. Since the central angle is $60^\circ$ and two sides are radii ($r=6$), the triangle is equilateral with side length $s = 6$.

$$A_{\text{triangle}} = \frac{s^2 \sqrt{3}}{4} = \frac{6^2 \sqrt{3}}{4} = \frac{36 \sqrt{3}}{4} = 9\sqrt{3} \text{ cm}^2$$

Step 3: Subtract triangle area from sector area.

$$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = (6\pi - 9\sqrt{3}) \text{ cm}^2$$

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Problem 2: Similar Triangles and Area Scale Factors

In $\triangle ABC$, a line parallel to base $BC$ intersects side $AB$ at $D$ and side $AC$ at $E$. If $AD = 3\text{ cm}$ and $DB = 5\text{ cm}$, find the ratio of the area of $\triangle ADE$ to the area of trapezoid $DBCE$.

Step 1: Determine side ratio between similar triangles. $\triangle ADE \sim \triangle ABC$. Side length $AB = AD + DB = 3 + 5 = 8\text{ cm}$. Scale factor $k = \frac{AD}{AB} = \frac{3}{8}$.

Step 2: Determine area ratio.

$$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = k^2 = \left(\frac{3}{8}\right)^2 = \frac{9}{64}$$

Step 3: Calculate area ratio relative to trapezoid. Let $\text{Area}(\triangle ABC) = 64x$, so $\text{Area}(\triangle ADE) = 9x$.

$$\text{Area}(\text{Trapezoid } DBCE) = 64x - 9x = 55x$$ $$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\text{Trapezoid } DBCE)} = \frac{9x}{55x} = \frac{9}{55}$$

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ACET Speed Tactics & Traps Summary