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
PpointABsegment ABCDray CDEFline EFvertex50°
A point, a segment, a ray, a line, and an angle with its vertex.
  • 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).

Perimeter is the distance around; area is the space inside. Angle facts do most of the work in geometry: angles on a straight line add to 180∘180^\circ, angles around a point to 360∘360^\circ, and the three angles of any triangle add to 180∘180^\circ. Formulas for rectangles, triangles, and circles follow from cutting and rearranging.

An angle is measured by how far one ray must turn to lie on the other, and a full turn is 360∘360^\circ. That single idea explains the special numbers: a quarter turn is 360÷4=90∘360 \div 4 = 90^\circ (a right angle) and a half turn is 180∘180^\circ (a straight line). Complementary angles fill a quarter turn together; supplementary angles fill a half turn.

Concept
Types of Angles
  • Acute: greater than 0∘0^\circ and less than 90∘90^\circ.
  • Right: exactly 90∘90^\circ (marked with a small square).
  • Obtuse: greater than 90∘90^\circ and less than 180∘180^\circ.
  • Straight: exactly 180∘180^\circ (a straight line).
Concept
Complementary and Supplementary Angles

Two angles are complementary if their measures add to 90∘90^\circ. Two angles are supplementary if their measures add to 180∘180^\circ.

Example
Finding a Complement and a Supplement
35°55°35°145°
Left: 35∘+55∘35^\circ + 55^\circ fill a right angle. Right: 35∘+145∘35^\circ + 145^\circ fill a straight angle.

The complement of a 35∘35^\circ angle is 90∘−35∘=55∘90^\circ - 35^\circ = 55^\circ. The supplement of a 35∘35^\circ angle is 180∘−35∘=145∘180^\circ - 35^\circ = 145^\circ.

Side note
Where it goes wrong

Mixing up complement and supplement: "the complement of 35∘35^\circ is 145∘145^\circ." Complementary angles fill a right angle, so they add to 90∘90^\circ; 35∘+145∘=180∘35^\circ + 145^\circ = 180^\circ is the supplementary pair.

Example
Classifying by Measure

An angle of 118∘118^\circ is obtuse (between 90∘90^\circ and 180∘180^\circ). An angle of 72∘72^\circ is acute (less than 90∘90^\circ).

Tip

Tip: Complementary comes before Supplementary in the alphabet, just as 9090 comes before 180180. So C goes with 90∘90^\circ and S goes with 180∘180^\circ.

Try it
Try it: complement and supplement

An angle measures 62∘62^\circ. Find its complement and its supplement, and classify all three angles.

Show the answer

Complement: 90∘−62∘=28∘90^\circ - 62^\circ = 28^\circ. Supplement: 180∘−62∘=118∘180^\circ - 62^\circ = 118^\circ. The 62∘62^\circ and 28∘28^\circ angles are acute (less than 90∘90^\circ), and 118∘118^\circ is obtuse (between 90∘90^\circ and 180∘180^\circ).

Triangles

Concept
Classifying Triangles

By sides:

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

By angles:

  • Acute: all three angles less than 90∘90^\circ.
  • Right: one angle exactly 90∘90^\circ.
  • Obtuse: one angle greater than 90∘90^\circ.
Concept
The Angle Sum Rule
55°65°60°
Drawn to scale: 55∘+65∘+60∘=180∘55^\circ + 65^\circ + 60^\circ = 180^\circ.

The three angles of any triangle always add up to 180∘180^\circ:

∠A+∠B+∠C=180∘.\angle A + \angle B + \angle C = 180^\circ.

Why 180∘180^\circ? Draw the line through the top vertex parallel to the base. The two angles it makes with the sides are copies of the two base angles (alternate interior angles), and together with the top angle they fill that straight line, which is 180∘180^\circ. So the three angles of any triangle, however it is shaped, add to a straight angle.

Example
Finding a Missing Angle

Two angles of a triangle are 55∘55^\circ and 65∘65^\circ. Find the third. Add the known angles: 55∘+65∘=120∘55^\circ + 65^\circ = 120^\circ. Subtract from 180∘180^\circ: 180∘−120∘=60∘180^\circ - 120^\circ = 60^\circ. The third angle is 60∘\mathbf{60^\circ}.

Side note
Where it goes wrong

Believing a triangle can have two obtuse angles, say 100∘100^\circ and 95∘95^\circ. Those two alone already total 195∘195^\circ, more than the 180∘180^\circ available for all three.

Example
Classifying a Triangle
977
Two equal sides (tick marks): isosceles.

A triangle has sides 77, 77, and 99, and (to the nearest degree) angles 50∘50^\circ, 50∘50^\circ, 80∘80^\circ. Two sides are equal, so it is isosceles. All angles are below 90∘90^\circ, so it is also acute. The two equal angles sit opposite the two equal sides, as they always do in an isosceles triangle.

Tip

Tip: A triangle can have at most one right or obtuse angle. If it had two, they would already use up 180∘180^\circ or more with nothing left for the third angle.

Try it
Try it: two unknown angles

An isosceles triangle has a top angle of 40∘40^\circ and two equal base angles. Find each base angle.

Show the answer

The base angles together get what is left of 180∘180^\circ: 180∘−40∘=140∘180^\circ - 40^\circ = 140^\circ. They are equal, so each is 140∘÷2=70∘140^\circ \div 2 = 70^\circ. Check: 40+70+70=18040 + 70 + 70 = 180.

Perimeter of Polygons

Concept
What Perimeter Means
9 cm4 cm
Walk around: 9+4+9+4=269 + 4 + 9 + 4 = 26 cm.

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

Perimeter is a walk around the edge, so it is measured the same way any walk is: add up the lengths. The rectangle formula 2l+2w2l + 2w is just that sum organized, since a rectangle has two sides of each length. There is no need to memorize a formula for an odd shape; list its sides and add.

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}.
Side note
Where it goes wrong

Adding only the two labelled sides of a rectangle: 9+4=139 + 4 = 13 cm. That is half the walk; the opposite sides are unlabelled but still there. 2(9)+2(4)=262(9) + 2(4) = 26 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!

Try it
Try it: a missing side

A rectangle has perimeter 3030 m and length 1111 m. Find its width.

Show the answer

The two lengths use 2×11=222 \times 11 = 22 m of the perimeter, leaving 30−22=830 - 22 = 8 m for the two widths. So each width is 8÷2=48 \div 2 = 4 m. Check: 2(11)+2(4)=302(11) + 2(4) = 30.

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

Concept
The Area Formulas

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

  • 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.

All five formulas grow from the rectangle, A=lwA = lw, which counts unit squares in rows. A parallelogram can be cut along its height and the piece slid across to make a rectangle of the same base and height, so A=bhA = bh. A triangle is exactly half of such a parallelogram, so A=12bhA = \tfrac12 bh. A trapezoid plus a rotated copy of itself forms a parallelogram with base b1+b2b_1 + b_2, and taking half gives 12(b1+b2)h\tfrac12(b_1 + b_2)h.

Example
Triangle and Parallelogram
106
The triangle is half of the 10×610 \times 6 rectangle around it: 3030 cm2^2.

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.
Side note
Where it goes wrong

Using a slanted side as the height. For a parallelogram with base 1212, slanted side 66 and height 55, "12×6=7212 \times 6 = 72" overcounts; the area is 12×5=6012 \times 5 = 60. The height is always measured at a right angle to the base.

Example
Trapezoid
1174
Parallel sides 77 and 1111, height 44.

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.

Try it
Try it: work backward

A triangle has area 4242 cm2^2 and base 1212 cm. How tall is it?

Show the answer

Use A=12bhA = \tfrac12 bh: 42=12(12)h=6h42 = \tfrac12(12)h = 6h. Divide by 66: h=7h = 7 cm. Check: 12⋅12⋅7=42\tfrac12 \cdot 12 \cdot 7 = 42.

Circles

Concept
Parts of a Circle
  • 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 π≈3.14\pi \approx 3.14.

Why π\pi in both formulas? For every circle, big or small, the distance around divided by the distance across is the same number, about 3.143.14; that constant ratio is named π\pi, and C=πdC = \pi d is its definition rewritten. The area formula follows by slicing the circle into thin wedges and laying them head to tail: they form a near-rectangle of height rr and width half the circumference, πr\pi r, so the area is πr⋅r=πr2\pi r \cdot r = \pi r^2.

Example
Circumference
d = 10
The diameter goes all the way across: d=10d = 10 cm.

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

C=πd≈3.14×10=31.4 cm.C = \pi d \approx 3.14 \times 10 = 31.4 \text{ cm}.
Side note
Where it goes wrong

Using the diameter in A=πr2A = \pi r^2. For d=10d = 10: "3.14×102=3143.14 \times 10^2 = 314" is four times too big. Halve first, r=5r = 5, then A≈3.14×25=78.5A \approx 3.14 \times 25 = 78.5.

Example
Area of a Circle
r = 6
The radius goes from center to edge: r=6r = 6 m.

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

A=πr2≈3.14×62=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.

Try it
Try it: from the diameter

A round table has diameter 44 ft. Find its circumference and its area (use π≈3.14\pi \approx 3.14).

Show the answer

Circumference: C=πd≈3.14×4=12.56C = \pi d \approx 3.14 \times 4 = 12.56 ft. For area, first halve the diameter: r=2r = 2 ft. Then A=πr2≈3.14×4=12.56A = \pi r^2 \approx 3.14 \times 4 = 12.56 ft2^2. (The two numbers match only because r=2r = 2; the units, ft and ft2^2, do not.)

The Pythagorean Theorem

Concept
The Rule for Right Triangles
16925
Squares on the legs: 9+169 + 16. Square on the hypotenuse: 2525.

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 90∘90^\circ angle.

The theorem is a statement about squares built on the sides. On a 33-44-55 right triangle the squares on the legs hold 99 and 1616 unit squares, and the square on the hypotenuse holds exactly 9+16=259 + 16 = 25. That is not a coincidence of these numbers; it happens for every right triangle, and the proof in Going Deeper shows why. It fails for other triangles, which is why the right angle must be there before the formula is used.

Example
Finding the Hypotenuse
86c = 10
Legs 66 and 88; the hypotenuse is across from the right angle.

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.
Side note
Where it goes wrong

Adding when a leg is missing: legs are unknown, hypotenuse 1313, other leg 55, and "b2=132+52=194b^2 = 13^2 + 5^2 = 194." The hypotenuse is the longest side, so bb must be less than 1313; subtract instead: b2=169−25=144b^2 = 169 - 25 = 144, b=12b = 12.

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.

Try it
Try it: a ladder

A 1010-ft ladder leans against a wall with its foot 66 ft from the wall. How high up the wall does it reach?

Show the answer

The wall and ground form the right angle, so the ladder is the hypotenuse: 62+h2=1026^2 + h^2 = 10^2. Then h2=100−36=64h^2 = 100 - 36 = 64 and h=8h = 8 ft.

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).

Volume counts unit cubes. A 5×45 \times 4 base holds 2020 cubes in one layer, and a height of 33 stacks 33 layers, so V=20×3=60V = 20 \times 3 = 60. That "base area times height" reasoning never used the shape of the base, which is why a cylinder works the same way: its base is a circle of area πr2\pi r^2, stacked hh high. Surface area is different: it is the wrapping paper, the sum of the areas of all the faces.

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
543
A 5×4×35 \times 4 \times 3 box: three pairs of matching faces.

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(5⋅4+5⋅3+4⋅3)=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.
Side note
Where it goes wrong

Adding only three faces for surface area: 5⋅4+5⋅3+4⋅3=475\cdot4 + 5\cdot3 + 4\cdot3 = 47 cm2^2. A box has six faces in three matching pairs (top and bottom, front and back, two sides), so the total is 2×47=942 \times 47 = 94 cm2^2.

Example
Volume of a Cylinder
r = 3h = 10
A circular base of radius 33, stacked 1010 high.

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

V=πr2h≈3.14×32×10=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.

Try it
Try it: fill a tank

A fish tank is 6060 cm long, 3030 cm wide and 4040 cm tall. How many liters does it hold? (11 liter =1000= 1000 cm3^3.)

Show the answer

Volume =60×30×40=72,000= 60 \times 30 \times 40 = 72{,}000 cm3^3. Each liter is 10001000 cm3^3, so the tank holds 72,000÷1000=7272{,}000 \div 1000 = 72 liters.

Real-World Geometry Problems

Concept
A Plan of Attack
  • 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.

The key decision is which kind of measurement the question wants, and the units give it away. Fencing is laid along a line, so it is a length (feet): perimeter. Grass covers a surface, so it is counted in square feet: area. Water fills a space: cubic units, volume. Choosing by units first prevents the most common real-world error, using the right numbers in the wrong formula.

Example
Fencing and Sod
12 ft8 ft
Fence goes around the edge; sod covers the inside.

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.

Side note
Where it goes wrong

Pricing fence by area: "9696 ft2^2 of fence." Fence is sold by length and runs around the edge, 4040 ft; 9696 ft2^2 is how much ground it encloses.

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.

Try it
Try it: paint a wall

A wall is 1212 ft wide and 88 ft tall and has a 33 ft by 44 ft window. One can of paint covers 4040 ft2^2. How many cans are needed?

Show the answer

Paint covers a surface, so find area. Wall: 12×8=9612 \times 8 = 96 ft2^2. Subtract the window: 3×4=123 \times 4 = 12 ft2^2, leaving 8484 ft2^2. Two cans cover only 8080 ft2^2, so 33 cans are needed.

Going Deeper: Advanced Geometry

Concept
Why the Pythagorean Theorem Is True (A Proof)
ababc²c
Drawn with a=3a = 3, b=4b = 4: the tilted square has side c=5c = 5.

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)2⏟outer square  =  4⋅12ab⏟four triangles  +  c2⏟tilted 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.

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(s−a)(s−b)(s−c) .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
14131512
The altitude to the side 1414 is 1212, so the area is 12⋅14⋅12=84\tfrac12 \cdot 14 \cdot 12 = 84 too.

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: s−a=8s-a = 8,   s−b=7\; s-b = 7,   s−c=6\; s-c = 6.

A=21⋅8⋅7⋅6=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: 21⋅8=16821\cdot 8 = 168, then 168⋅7=1176168\cdot 7 = 1176, then 1176⋅6=70561176\cdot 6 = 7056.)

A check without Heron: the altitude from the top vertex to the side of length 1414 splits it into 55 and 99, making right triangles 55-1212-1313 and 99-1212-1515. So the height is 1212 and the area is 12⋅14⋅12=84\tfrac12 \cdot 14 \cdot 12 = 84, the same answer.

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 △ABC∼△DEF\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
4624h
Same sun angle, same shape: heights and shadows are proportional.

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 180∘180^\circ and a straight line also measures 180∘180^\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 180∘−x−y180^\circ - x - y (angle sum), and the exterior angle is its supplement, 180∘−(180∘−x−y)=x+y180^\circ - (180^\circ - x - y) = x + y.

Example
Chasing a Missing Angle
50°70°120°
The exterior angle equals the two far interior angles combined.

A triangle has interior angles x=50∘x = 50^\circ and y=70∘y = 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 180∘−50∘−70∘=60∘180^\circ - 50^\circ - 70^\circ = 60^\circ, whose supplement is 180∘−60∘=120∘180^\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 180∘180^\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 2πr⋅h2\pi r \cdot h. Add the pieces:

SA=2πr2⏟two circles+2πrh⏟unrolled 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
106r = 5
A 10×610 \times 6 rectangle topped by a semicircle of radius 55.

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=12πr2≈12(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 180∘180^\circ angle sum and the straight-line 180∘180^\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?

Why It Works

The angles of a triangle add to 180∘180^\circ: draw the line through one vertex parallel to the opposite side. The two angles it makes with the triangle's sides equal the two far angles of the triangle (alternate interior angles), and together with the vertex angle they fill a straight line. The area of a circle is πr2\pi r^2 because cutting it into many thin wedges and laying them alternately point-up and point-down makes a near-rectangle of height rr and width half the circumference, πr\pi r.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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 12⋅6⋅4=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.