3.3 Informal Geometry, Perimeter, Area, and Volume

Geometric Fundamentals, Lines, and Angle Relationships

Geometry on the LET evaluates spatial reasoning, geometric proofs, perimeter, area, and volume calculations. Understanding core angle relationships and polygon properties is prerequisite to solving measurement problems.

Core Angle Definitions

• Complementary Angles: Two angles whose measures sum to $90^\circ$.
• Supplementary Angles: Two angles whose measures sum to $180^\circ$.
• Vertical Angles: Opposite angles formed by intersecting lines; vertical angles are congruent (equal in measure).
• Linear Pair: Adjacent angles formed when two lines intersect; they are supplementary ($a + b = 180^\circ$).

Parallel Lines Cut by a Transversal

When two parallel lines ($l_1 \parallel l_2$) are intersected by a transversal line, specific angle congruences and equivalences are established:

• Corresponding Angles: Equal in measure (e.g., Position top-left = Position top-left).
• Alternate Interior Angles: Equal in measure (e.g., inside parallel lines on opposite sides of transversal).
• Alternate Exterior Angles: Equal in measure (e.g., outside parallel lines on opposite sides of transversal).
• Consecutive Interior Angles: Supplementary (sum to $180^\circ$).

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Triangles and the Pythagorean Theorem

Triangles are three-sided polygons whose interior angles always sum to $180^\circ$.

Triangle Classification

• By Sides: Equilateral (3 equal sides), Isosceles (2 equal sides), Scalene (no equal sides).
• By Angles: Acute (all angles $< 90^\circ$), Right (one angle $= 90^\circ$), Obtuse (one angle $> 90^\circ$).

The Pythagorean Theorem and Triples

For any right-angled triangle with legs $a$ and $b$ and hypotenuse $c$:

$$a^2 + b^2 = c^2 \implies c = \sqrt{a^2 + b^2}$$

Common Pythagorean Triples tested on the LET:

• $3 - 4 - 5$ (and multiples: $6-8-10, 9-12-15, 12-16-20$)
• $5 - 12 - 13$
• $8 - 15 - 17$
• $7 - 24 - 25$

Special Right Triangles

Worked Example 1: Special Right Triangle Calculation

Problem: A square garden has a diagonal measuring $14\text{ meters}$. What is the exact perimeter of the garden?

Step 1: Relate diagonal to side length using the $45^\circ-45^\circ-90^\circ$ ratio.
In a square, the diagonal cuts the square into two $45^\circ-45^\circ-90^\circ$ triangles.

$$\text{Diagonal} = s\sqrt{2} = 14$$

Step 2: Solve for side length $s$.

$$s = \frac{14}{\sqrt{2}} = \frac{14\sqrt{2}}{2} = 7\sqrt{2}\text{ meters}$$

Step 3: Calculate Perimeter ($P = 4s$).

$$P = 4 \times (7\sqrt{2}) = 28\sqrt{2}\text{ meters}$$

Final Answer: $28\sqrt{2}\text{ meters} \approx 39.60\text{ meters}$

Two-Dimensional Areas and Perimeter Formulas

Polygon Angle Formulas

For any convex polygon with $n$ sides:

• Sum of Interior Angles: $S = (n - 2) \times 180^\circ$
• Each Interior Angle of Regular Polygon: $I = \frac{(n - 2) \times 180^\circ}{n}$
• Sum of Exterior Angles: Always equals $360^\circ$ for all convex polygons.

Worked Example 2: Shaded Region Area

Problem: A circular swimming pool with a radius of $6\text{ meters}$ is surrounded by a concrete walkway of uniform width $2\text{ meters}$. What is the area of the concrete walkway? (Use $\pi \approx 3.1416$).

Step 1: Calculate the area of the inner circle (pool).

$$r_{\text{inner}} = 6\text{ m} \implies A_{\text{inner}} = \pi (6)^2 = 36\pi$$

Step 2: Calculate the radius and area of the outer circle (pool + walkway).

$$r_{\text{outer}} = 6 + 2 = 8\text{ m} \implies A_{\text{outer}} = \pi (8)^2 = 64\pi$$

Step 3: Subtract inner area from outer area.

$$A_{\text{walkway}} = A_{\text{outer}} - A_{\text{inner}} = 64\pi - 36\pi = 28\pi$$ $$A_{\text{walkway}} = 28 \times 3.1416 = 87.9644\text{ m}^2$$

Final Answer: $28\pi\text{ m}^2 \approx 87.96\text{ m}^2$

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Three-Dimensional Volume and Surface Area

Worked Example 3: Volume of a Right Cylinder

Problem: A cylindrical water storage tank has a base diameter of $4\text{ meters}$ and a height of $7\text{ meters}$. How many cubic meters of water can it hold when completely full?

Step 1: Calculate the radius from diameter.

$$r = \frac{d}{2} = \frac{4}{2} = 2\text{ meters}$$

Step 2: Apply cylinder volume formula $V = \pi r^2 h$.

$$V = \pi (2)^2 (7) = \pi (4) (7) = 28\pi\text{ m}^3$$ $$V \approx 28 \times 3.1416 = 87.9644\text{ m}^3$$

Final Answer: $28\pi\text{ m}^3 \approx 87.96\text{ m}^3$