Geometry Basics

Study Sheet

Geometry Basics

Points, angles, shapes, and the formulas that measure them

Basic Vocabulary: Points, Lines, and Angles

Concept
The Building Blocks
55°
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  • A point is an exact location. It has no size, just position.
  • A line goes straight forever in both directions (two arrowheads).
  • A segment is part of a line with two endpoints.
  • A ray starts at one endpoint and goes forever in one direction.
  • An angle is formed by two rays that share an endpoint, called the vertex. We measure angles in degrees (^\circ).

An angle measures a turn.

Concept
Types of Angles
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  • Acute: greater than 00^\circ and less than 9090^\circ.
  • Right: exactly 9090^\circ (marked with a small square).
  • Obtuse: greater than 9090^\circ and less than 180180^\circ.
  • Straight: exactly 180180^\circ (a straight line).
Concept
Complementary and Supplementary Angles

Two angles are complementary if their measures add to 9090^\circ. Two angles are supplementary if their measures add to 180180^\circ.

Example
Finding a Complement and a Supplement

The complement of a 3535^\circ angle is 9035=5590^\circ - 35^\circ = 55^\circ. The supplement of a 3535^\circ angle is 18035=145180^\circ - 35^\circ = 145^\circ.

Example
Classifying by Measure

An angle of 118118^\circ is obtuse (between 9090^\circ and 180180^\circ). An angle of 7272^\circ is acute (less than 9090^\circ).

Tip

Tip: Complementary comes before Supplementary in the alphabet, just as 9090 comes before 180180. So C goes with 9090^\circ and S goes with 180180^\circ.

Triangles

Concept
Classifying Triangles
55°

By sides:

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  • Equilateral: all three sides equal.
  • Isosceles: exactly two sides equal.
  • Scalene: no sides equal.

By angles:

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  • Acute: all three angles less than 9090^\circ.
  • Right: one angle exactly 9090^\circ.
  • Obtuse: one angle greater than 9090^\circ.

An angle measures a turn.

Concept
The Angle Sum Rule

The three angles of any triangle always add up to 180180^\circ:

A+B+C=180.\angle A + \angle B + \angle C = 180^\circ.
Example
Finding a Missing Angle

Two angles of a triangle are 5555^\circ and 6565^\circ. Find the third. Add the known angles: 55+65=12055^\circ + 65^\circ = 120^\circ. Subtract from 180180^\circ: 180120=60180^\circ - 120^\circ = 60^\circ. The third angle is 60\mathbf{60^\circ}.

Example
Classifying a Triangle

A triangle has sides 66, 66, and 99, and angles 5050^\circ, 5050^\circ, 8080^\circ. Two sides are equal, so it is isosceles. All angles are below 9090^\circ, so it is also acute.

Tip

Tip: A triangle can have at most one right or obtuse angle. If it had two, they would already use up 180180^\circ or more with nothing left for the third angle.

Perimeter of Polygons

Concept
What Perimeter Means
55°

The perimeter is the total distance around a shape: just add up the lengths of all the sides. Perimeter is a length, so it uses plain units (cm, m, in).

Prectangle=2l+2wPsquare=4sP_{\text{rectangle}} = 2l + 2w \qquad P_{\text{square}} = 4s

An angle measures a turn.

Example
Perimeter of a Rectangle

A rectangle is 99 cm long and 44 cm wide.

P=2(9)+2(4)=18+8=26 cm.P = 2(9) + 2(4) = 18 + 8 = 26 \text{ cm}.
Example
Perimeter of a Triangle

A triangle has sides 55 in, 88 in, and 1111 in.

P=5+8+11=24 in.P = 5 + 8 + 11 = 24 \text{ in}.
Tip

Tip: Perimeter is measured in single units (like cm). Area is measured in square units (like cm2^2). Don't mix them up!

Area of Rectangles, Squares, Parallelograms, Triangles, and Trapezoids

Concept
The Area Formulas
55°

Area is the amount of surface a flat shape covers, measured in square units.

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  • Rectangle: A=l×wA = l \times w
  • Square: A=s2A = s^2
  • Parallelogram: A=b×hA = b \times h
  • Triangle: A=12bhA = \dfrac{1}{2} b h
  • Trapezoid: A=12(b1+b2)hA = \dfrac{1}{2}(b_1 + b_2)\,h

Here hh is always the perpendicular height (straight up from the base), not a slanted side.

An angle measures a turn.

Example
Triangle and Parallelogram

Triangle, base 1010 cm, height 66 cm:

A=12(10)(6)=12(60)=30 cm2.A = \tfrac{1}{2}(10)(6) = \tfrac{1}{2}(60) = 30 \text{ cm}^2.

Parallelogram, base 1212 m, height 55 m:

A=(12)(5)=60 m2.A = (12)(5) = 60 \text{ m}^2.
Example
Trapezoid

A trapezoid has parallel sides b1=7b_1 = 7 cm and b2=11b_2 = 11 cm and height h=4h = 4 cm.

A=12(7+11)(4)=12(18)(4)=12(72)=36 cm2.A = \tfrac{1}{2}(7 + 11)(4) = \tfrac{1}{2}(18)(4) = \tfrac{1}{2}(72) = 36 \text{ cm}^2.
Tip

Tip: For a triangle, remember the 12\tfrac{1}{2}! A triangle is exactly half of a rectangle (or parallelogram) with the same base and height.

Circles

Concept
Parts of a Circle
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  • The radius rr is the distance from the center to the edge.
  • The diameter dd goes all the way across through the center: d=2rd = 2r, so r=d2r = \dfrac{d}{2}.
  • Circumference (distance around): C=πdC = \pi d, which is the same as C=2πrC = 2\pi r.
  • Area: A=πr2A = \pi r^2.

We use π3.14\pi \approx 3.14.

Example
Circumference

A circle has diameter d=10d = 10 cm.

C=πd3.14×10=31.4 cm.C = \pi d \approx 3.14 \times 10 = 31.4 \text{ cm}.
Example
Area of a Circle

A circle has radius r=6r = 6 m.

A=πr23.14×62=3.14×36=113.04 m2.A = \pi r^2 \approx 3.14 \times 6^2 = 3.14 \times 36 = 113.04 \text{ m}^2.
Tip

Tip: For area you square the radius first, then multiply by π\pi. Compute r2r^2 before multiplying by 3.143.14. And remember: area answers get square units.

The Pythagorean Theorem

Concept
The Rule for Right Triangles
55°

In a right triangle, the two shorter sides (the legs, aa and bb) and the longest side (the hypotenuse, cc, opposite the right angle) are related by

a2+b2=c2.a^2 + b^2 = c^2.

The hypotenuse cc is always the longest side and sits across from the 9090^\circ angle.

An angle measures a turn.

Example
Finding the Hypotenuse

A right triangle has legs a=6a = 6 and b=8b = 8.

c2=62+82=36+64=100,c=100=10.c^2 = 6^2 + 8^2 = 36 + 64 = 100, \qquad c = \sqrt{100} = 10.
Example
Finding a Leg

A right triangle has hypotenuse c=13c = 13 and one leg a=5a = 5. Find leg bb.

52+b2=132    25+b2=169    b2=144    b=12.5^2 + b^2 = 13^2 \;\Rightarrow\; 25 + b^2 = 169 \;\Rightarrow\; b^2 = 144 \;\Rightarrow\; b = 12.
Tip

Tip: Only the hypotenuse sits alone on one side of the equation. If you are missing a leg, subtract; if you are missing the hypotenuse, add.

Surface Area and Volume of Prisms and Cylinders

Concept
Rectangular Prisms

For a box with length ll, width ww, and height hh:

V=l×w×hSA=2(lw+lh+wh).V = l \times w \times h \qquad SA = 2(lw + lh + wh).

Volume is measured in cubic units (cm3^3); surface area in square units (cm2^2).

Concept
Cylinders

For a cylinder with radius rr and height hh:

V=πr2h.V = \pi r^2 h.

(The base is a circle of area πr2\pi r^2; multiply by the height.)

Example
Volume and Surface Area of a Box

A box is l=5l = 5 cm, w=4w = 4 cm, h=3h = 3 cm.

V=5×4×3=60 cm3.V = 5 \times 4 \times 3 = 60 \text{ cm}^3.
SA=2(54+53+43)=2(20+15+12)=2(47)=94 cm2.SA = 2(5\cdot 4 + 5\cdot 3 + 4\cdot 3) = 2(20 + 15 + 12) = 2(47) = 94 \text{ cm}^2.
Example
Volume of a Cylinder

A cylinder has radius r=3r = 3 cm and height h=10h = 10 cm.

V=πr2h3.14×32×10=3.14×9×10=282.6 cm3.V = \pi r^2 h \approx 3.14 \times 3^2 \times 10 = 3.14 \times 9 \times 10 = 282.6 \text{ cm}^3.
Tip

Tip: Surface area is the total of all the outside faces (like the paper needed to wrap a gift). Volume is how much fills the inside. Square units for surface, cubic units for volume.

Real-World Geometry Problems

Concept
A Plan of Attack
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  • Decide what the question is really asking: distance around (perimeter/circumference), surface covered (area), or space filled (volume).
  • Pick the matching formula and substitute the numbers.
  • Keep the units, and label the answer with the correct kind of unit.
Example
Fencing and Sod

A rectangular garden is 1212 ft by 88 ft. Fence around it == perimeter =2(12)+2(8)=24+16=40= 2(12) + 2(8) = 24 + 16 = 40 ft. Grass to cover it == area =12×8=96= 12 \times 8 = 96 ft2^2.

Example
Cost of Carpet

A floor is 1010 ft by 99 ft, and carpet costs $44 per square foot. Area =10×9=90= 10 \times 9 = 90 ft2^2. Cost =90×$4=$360= 90 \times \$4 = \$360.

Tip

Tip: “Around” problems (fencing, trim, edging) use perimeter. “Cover” problems (paint, carpet, sod) use area. “Fill” problems (water, sand) use volume.

Going Deeper: Advanced Geometry

Concept
Why the Pythagorean Theorem Is True (A Proof)
55°

Take four copies of a right triangle with legs aa, bb and hypotenuse cc, and arrange them inside a big square whose side is a+ba + b. The four hypotenuses form a tilted square of side cc in the middle.

Compute the big square's area two ways:

(a+b)2outer square  =  412abfour triangles  +  c2tilted square.\underbrace{(a+b)^2}_{\text{outer square}} \;=\; \underbrace{4\cdot\tfrac{1}{2}ab}_{\text{four triangles}} \;+\; \underbrace{c^2}_{\text{tilted square}}.

Expand the left side and simplify:

a2+2ab+b2=2ab+c2    a2+b2=c2.a^2 + 2ab + b^2 = 2ab + c^2 \;\Longrightarrow\; a^2 + b^2 = c^2.

The 2ab2ab cancels from both sides, leaving the theorem. This shows the rule is not a coincidence: it follows from just adding up areas.

An angle measures a turn.

Reminder — The Pythagorean theorem:a2+b2=c2a^2+b^2=c^2
Concept
Heron's Formula: Area From Three Sides

You can find a triangle's area from its side lengths alone---no height needed. First compute the semiperimeter (half the perimeter):

s=a+b+c2.s = \frac{a+b+c}{2}.

Then

A=s(sa)(sb)(sc).A = \sqrt{\,s(s-a)(s-b)(s-c)\,}.

This is the tool to reach for when you know all three sides but cannot easily measure a perpendicular height.

Example
Heron's Formula in Action

Find the area of a triangle with sides a=13a = 13, b=14b = 14, c=15c = 15. Semiperimeter: s=13+14+152=422=21s = \dfrac{13+14+15}{2} = \dfrac{42}{2} = 21. Now the differences: sa=8s-a = 8,   sb=7\; s-b = 7,   sc=6\; s-c = 6.

A=21876=7056=84 square units.A = \sqrt{21\cdot 8\cdot 7\cdot 6} = \sqrt{7056} = \mathbf{84}\ \text{square units}.

(Check the product step by step: 218=16821\cdot 8 = 168, then 1687=1176168\cdot 7 = 1176, then 11766=70561176\cdot 6 = 7056.)

Concept
Similar Triangles and Proportional Reasoning

Two triangles are similar if they have the same shape but not necessarily the same size: their corresponding angles are equal, and their corresponding sides are in the same ratio (scale factor). We write ABCDEF\triangle ABC \sim \triangle DEF.

The key fact is that matching sides form equal fractions:

ABDE=BCEF=ACDF.\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}.

This lets you find a distance you cannot measure directly (like the height of a tree) by comparing it to one you can.

Example
Measuring a Tree With Its Shadow

At the same time of day, a 66-ft person casts a 44-ft shadow, and a tree casts a 2424-ft shadow. The sun's rays make similar triangles, so heights and shadows are proportional:

tree heighttree shadow=person heightperson shadow    h24=64.\frac{\text{tree height}}{\text{tree shadow}} = \frac{\text{person height}}{\text{person shadow}} \;\Longrightarrow\; \frac{h}{24} = \frac{6}{4}.

Cross-multiply: 4h=24×6=1444h = 24 \times 6 = 144, so h=1444=36h = \dfrac{144}{4} = \mathbf{36} ft. The tree is 3636 feet tall.

Concept
Angle Chasing: The Exterior Angle Rule

Because a triangle's angles sum to 180180^\circ and a straight line also measures 180180^\circ, an exterior angle (formed by extending one side) equals the sum of the two remote interior angles:

Here the exterior angle at BB equals x+yx + y. Why: the interior angle at BB is 180xy180^\circ - x - y (angle sum), and the exterior angle is its supplement, 180(180xy)=x+y180^\circ - (180^\circ - x - y) = x + y.

Example
Chasing a Missing Angle

A triangle has interior angles x=50x = 50^\circ and y=70y = 70^\circ at two vertices. The exterior angle at the third vertex is

50+70=120,50^\circ + 70^\circ = 120^\circ,

and a quick check: the third interior angle is 1805070=60180^\circ - 50^\circ - 70^\circ = 60^\circ, whose supplement is 18060=120180^\circ - 60^\circ = 120^\circ. The two methods agree.

Concept
A Circle Fact: The Inscribed Angle (Preview)

An inscribed angle has its vertex on the circle; a central angle has its vertex at the center. If both open onto the same arc, then

inscribed angle=12(central angle).\text{inscribed angle} = \tfrac{1}{2}\,(\text{central angle}).

A famous special case (Thales' theorem): any angle inscribed in a semicircle is a right angle, because it opens onto a 180180^\circ arc, and 12(180)=90\tfrac{1}{2}(180^\circ) = 90^\circ. So if one side of an inscribed triangle is a diameter, the triangle is a right triangle.

Concept
Deriving the Surface Area of a Cylinder

Imagine unwrapping a can. You get two circles (the top and bottom) plus one rectangle (the curved side, or “label”). Each circle has area πr2\pi r^2. When you unroll the side, its width is the circumference 2πr2\pi r and its height is hh, so the rectangle's area is 2πrh2\pi r \cdot h. Add the pieces:

SA=2πr2two circles+2πrhunrolled side.SA = \underbrace{2\pi r^2}_{\text{two circles}} + \underbrace{2\pi r h}_{\text{unrolled side}}.

This is why the formula looks the way it does---it is just areas of shapes you already know, glued together.

Example
Composite Figure: Rectangle Plus a Semicircle

A window is a 1010-cm-wide by 66-cm-tall rectangle topped by a semicircle whose diameter is the rectangle's width (1010 cm, so radius r=5r = 5 cm). Find its total area by decomposing it into familiar pieces.

Arectangle=10×6=60 cm2.A_{\text{rectangle}} = 10 \times 6 = 60 \text{ cm}^2.
Asemicircle=12πr212(3.14)(52)=12(3.14)(25)=39.25 cm2.A_{\text{semicircle}} = \tfrac{1}{2}\pi r^2 \approx \tfrac{1}{2}(3.14)(5^2) = \tfrac{1}{2}(3.14)(25) = 39.25 \text{ cm}^2.

Total: 60+39.25=99.25 cm260 + 39.25 = \mathbf{99.25}\ \text{cm}^2. Break any odd shape into rectangles, triangles, and circle-parts, then add (or subtract) the areas.

Tip

Tip: Most “advanced” geometry is really old rules used together. Angle chasing leans on the 180180^\circ angle sum and the straight-line 180180^\circ; the Pythagorean proof and composite figures are just areas added up; similar triangles are proportions in disguise. When a problem looks new, ask: which basic facts can I combine?

Formulas, Proofs & Tips

Tip
Area formulas
Arect=bh,A=12bh,Apar=bh,Atrap=12(b1+b2)h,A=πr2A_{\text{rect}}=bh,\quad A_{\triangle}=\tfrac12 bh,\quad A_{\text{par}}=bh,\quad A_{\text{trap}}=\tfrac12(b_1+b_2)h,\quad A_{\odot}=\pi r^{2}

What it means. Every one of these is really "base times height", adjusted.

Example. A triangle with base 66 and height 44 has area 1264=12\tfrac12\cdot6\cdot4=12.

Why it works. A parallelogram becomes a rectangle when you cut a triangle off one end and slide it to the other, so its area is bhbh. A triangle is half a parallelogram (two copies make one), giving 12bh\tfrac12 bh. Two copies of a trapezoid form a parallelogram of base b1+b2b_1+b_2, giving 12(b1+b2)h\tfrac12(b_1+b_2)h.

Tip. The height must be perpendicular to the base, not a slanted side. In an obtuse triangle the height can fall outside the triangle.