Geometry Basics

Master area, perimeter, and volume formulas for common shapes, understand angle relationships, and solve problems involving geometric properties.

Written and reviewed by the Study Mondo Education TeamLast updated April 18, 2026

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Geometry Basics

Angle Relationships

Supplementary angles: Sum to 180°
Complementary angles: Sum to 90°
Vertical angles: Equal (formed by intersecting lines)
Linear pair: Supplementary and adjacent

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

Shape	Formula
Triangle	$A = \frac{1}{2}bh$
Rectangle	$A = lw$
Parallelogram	$A = bh$
Trapezoid	$A = \frac{1}{2}(b_1 + b_2)h$
Circle	$A = \pi r^2$

Volume Formulas

Shape	Formula
Rectangular prism	$V = lwh$
Cylinder	$V = \pi r^2 h$
Cone	$V = \frac{1}{3}\pi r^2 h$
Sphere	$V = \frac{4}{3}\pi r^3$

SAT Tips

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

📚 Practice Problems

1Problem 1easy

❓ Question:

In a triangle, two angles measure $45°$ and $70°$ . What is the measure of the third angle?

💡 Show Solution

Solution:

Triangle angle sum: All angles add to $180°$

$45° + 70° + x = 180°$ $115° + x = 180°$ $x = 65°$

Answer: $65°$

SAT Tip: Triangle angles ALWAYS sum to $180°$ - use this constantly!

2Problem 2medium

❓ Question:

In a right triangle, one leg is 5 and the hypotenuse is 13. What is the length of the other leg?

💡 Show Solution

Solution:

Use Pythagorean theorem: $a^2 + b^2 = c^2$

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

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!

3Problem 3hard

❓ Question:

In a 30-60-90 triangle, the side opposite the $30°$ angle is 6. What is the length of the hypotenuse?

💡 Show Solution

Solution:

30-60-90 ratio: $x : x\sqrt{3} : 2x$

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!

4Problem 4easy

❓ Question:

A rectangle has a length of 12 and a width of 5. What is its perimeter and area?

💡 Show Solution

Perimeter: $P = 2l + 2w = 2(12) + 2(5) = 24 + 10 = 34$

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

Answer: Perimeter = 34, Area = 60

5Problem 5easy

❓ Question:

A rectangle has a length of 12 and a width of 5. What is its perimeter and area?

💡 Show Solution

Perimeter: $P = 2l + 2w = 2(12) + 2(5) = 24 + 10 = 34$

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

Answer: Perimeter = 34, Area = 60

6Problem 6medium

❓ Question:

Two parallel lines are cut by a transversal. One of the angles formed is 115°. What are the measures of all eight angles?

💡 Show Solution

When parallel lines are cut by a transversal, we get two types of angles:

The angle of 115° and its vertical angle are both 115°. The supplementary angles are $180° - 115° = 65°$ .

All eight angles are either 115° or 65°:

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.

7Problem 7medium

❓ Question:

Two parallel lines are cut by a transversal. One of the angles formed is 115°. What are the measures of all eight angles?

💡 Show Solution

When parallel lines are cut by a transversal, we get two types of angles:

The angle of 115° and its vertical angle are both 115°. The supplementary angles are $180° - 115° = 65°$ .

All eight angles are either 115° or 65°:

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.

8Problem 8medium

❓ Question:

A rectangular box has dimensions 3 × 4 × 12. What is the length of the longest diagonal inside the box?

💡 Show Solution

3D diagonal formula: $d = \sqrt{l^2 + w^2 + h^2}$

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

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).

9Problem 9medium

❓ Question:

A rectangular box has dimensions 3 × 4 × 12. What is the length of the longest diagonal inside the box?

💡 Show Solution

3D diagonal formula: $d = \sqrt{l^2 + w^2 + h^2}$

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

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).

10Problem 10hard

❓ Question:

The volume of a cylinder is $100\pi$ cubic cm and its height is 4 cm. What is the total surface area?

💡 Show Solution

Step 1: Find the radius using the volume formula: $V = \pi r^2 h$ $100\pi = \pi r^2 (4)$ $r^2 = 25$ $r = 5 \text{ cm}$

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)

11Problem 11hard

❓ Question:

The volume of a cylinder is $100\pi$ cubic cm and its height is 4 cm. What is the total surface area?

💡 Show Solution

Step 1: Find the radius using the volume formula: $V = \pi r^2 h$ $100\pi = \pi r^2 (4)$ $r^2 = 25$ $r = 5 \text{ cm}$

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)

12Problem 12expert

❓ Question:

Two similar triangles have corresponding sides in the ratio $3:5$ . If the area of the smaller triangle is 27 cm², what is the area of the larger triangle?

💡 Show Solution

Key property of similar figures: If corresponding sides have ratio $k$ , then areas have ratio $k^2$ .

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 = $k^3$ .

13Problem 13expert

❓ Question:

Two similar triangles have corresponding sides in the ratio $3:5$ . If the area of the smaller triangle is 27 cm², what is the area of the larger triangle?

💡 Show Solution

Key property of similar figures: If corresponding sides have ratio $k$ , then areas have ratio $k^2$ .

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 = $k^3$ .

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Master area, perimeter, and volume formulas for common shapes, understand angle relationships, and solve problems involving geometric properties.

How can I study Geometry Basics effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 13 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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Geometry Basics is part of the SAT Prep course on Study Mondo, specifically in the Geometry and Trigonometry section. You can explore the full course for more related topics and practice resources.

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Yes, this page includes 13 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Geometry Basics

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Geometry Basics

Angle Relationships

Parallel Lines and Transversals

Triangle Properties

Special Triangles

Area Formulas

Volume Formulas

SAT Tips

📚 Practice Problems

1Problem 1easy

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2Problem 2medium

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3Problem 3hard

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4Problem 4easy

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5Problem 5easy

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6Problem 6medium

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7Problem 7medium

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8Problem 8medium

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9Problem 9medium

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10Problem 10hard

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11Problem 11hard

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12Problem 12expert

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