Parallel Lines and Transversals

When a transversal crosses parallel lines:

• Corresponding angles: Equal
• Alternate interior angles: Equal
• Alternate exterior angles: Equal
• Co-interior (same-side interior) angles: Supplementary (sum to 180°)

Triangle Properties

• Angle sum: 180°
• Exterior angle = sum of two remote interior angles
• Triangle Inequality: Sum of any two sides > third side

Special Triangles

• Equilateral: All sides equal, all angles 60°
• Isosceles: Two sides equal, base angles equal
• 30-60-90: Sides in ratio $1 : \sqrt{3} : 2$
• 45-45-90: Sides in ratio $1 : 1 : \sqrt{2}$

Area Formulas

Volume Formulas

SAT Tips

1. These formulas are given on the SAT reference sheet — know how to use them quickly
2. Draw diagrams for word problems
3. Mark equal angles and sides on your figure

$5^2 + b^2 = 13^2$ $25 + b^2 = 169$ $b^2 = 144$ $b = 12$

Answer: $12$

Recognition: This is the 5-12-13 Pythagorean triple!

SAT Tip: Knowing common triples (3-4-5, 5-12-13, 8-15-17) saves time!

• Opposite $30°$: $x$ (shortest)
• Opposite $60°$: $x\sqrt{3}$
• Opposite $90°$ (hypotenuse): $2x$

Given: Side opposite $30°$ is 6 $x = 6$

Hypotenuse: $2x = 2(6) = 12$

Answer: $12$

SAT Tip: Memorize 30-60-90 ratios - they appear frequently! Side opposite 30° is half the hypotenuse!

Area: $A = lw = 12 \times 5 = 60$

Answer: Perimeter = 34, Area = 60

Area: $A = lw = 12 \times 5 = 60$

Answer: Perimeter = 34, Area = 60

• Four angles of 115° (the angle, its vertical angle, and corresponding angles)
• Four angles of 65° (supplementary to the 115° angles)

Answer: Four angles are 115° and four are 65°.

Key relationships used: vertical angles, corresponding angles, supplementary angles.

• Four angles of 115° (the angle, its vertical angle, and corresponding angles)
• Four angles of 65° (supplementary to the 115° angles)

Answer: Four angles are 115° and four are 65°.

Key relationships used: vertical angles, corresponding angles, supplementary angles.

$d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$

Answer: The space diagonal is 13.

Alternatively: First find the diagonal of the base: $\sqrt{3^2 + 4^2} = 5$ (3-4-5 triple), then use that with the height: $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$ (5-12-13 triple).

$d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$

Answer: The space diagonal is 13.

Alternatively: First find the diagonal of the base: $\sqrt{3^2 + 4^2} = 5$ (3-4-5 triple), then use that with the height: $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$ (5-12-13 triple).

Step 2: Calculate total surface area (two circles + lateral surface): $SA = 2\pi r^2 + 2\pi rh$ $= 2\pi(25) + 2\pi(5)(4)$ $= 50\pi + 40\pi$ $= 90\pi \approx 282.74 \text{ cm}^2$

Answer: $90\pi$ cm² (approximately 282.74 cm²)

Step 2: Calculate total surface area (two circles + lateral surface): $SA = 2\pi r^2 + 2\pi rh$ $= 2\pi(25) + 2\pi(5)(4)$ $= 50\pi + 40\pi$ $= 90\pi \approx 282.74 \text{ cm}^2$

Answer: $90\pi$ cm² (approximately 282.74 cm²)

Side ratio: $\frac{3}{5}$

Area ratio: $\left(\frac{3}{5}\right)^2 = \frac{9}{25}$

$\frac{27}{A_{\text{large}}} = \frac{9}{25}$ $A_{\text{large}} = \frac{27 \times 25}{9} = 75 \text{ cm}^2$

Answer: 75 cm²

Remember: Side ratio = $k$, Area ratio = $k^2$, Volume ratio = $k^3$.

Side ratio: $\frac{3}{5}$

Area ratio: $\left(\frac{3}{5}\right)^2 = \frac{9}{25}$

$\frac{27}{A_{\text{large}}} = \frac{9}{25}$ $A_{\text{large}} = \frac{27 \times 25}{9} = 75 \text{ cm}^2$

Answer: 75 cm²

Remember: Side ratio = $k$, Area ratio = $k^2$, Volume ratio = $k^3$.