MATH 130 - Polygon Area, Triangles, and Similarity

Exterior Angles: One Complete Turn

Predict first

Why is the exterior-angle sum always 360° for a convex polygon?

Commit to a response before discussion.

Five exterior turns, measured between an extended incoming side and the outgoing side, accompany one clockwise circuit of a convex pentagon. — Measure each turn in the same direction.

Activate · 1.0 minSource: Authored

Polygon Angle Sums and Area

1

Interior-angle sum: $(n-2)180^\circ$; convex-polygon exterior-angle sum: $360^\circ$.

2

Parallelogram area: $A=bh$. Trapezoid area: $A=\tfrac12(b_1+b_2)h$; use the perpendicular height.

3

For a regular hexagon, $e=360/6=60^\circ$ and $i=180-60=120^\circ$.

Five diagonals from vertex A partition a convex octagon into six nonoverlapping triangles. The eight vertices are A through H. — Diagonals from one vertex give n − 2 triangles.

Explain · 2.0 minSource: 20, 21

Octagon, Trapezoid, and Parallelogram

Find an octagon's interior-angle sum, a trapezoid's area (bases 12 and 20, height 7), and a parallelogram's area (base 9, height 8).

1

$(8-2)180^\circ=1080^\circ$.

2

$A_{trap}=\tfrac12(12+20)(7)=112$ square units.

3

$A_{para}=bh=9(8)=72$ square units.

4

Use perpendicular height; angle measure uses degrees and area uses square units.

An octagon has eight sides. A trapezoid has parallel bases 12 and 20 and perpendicular height 7. A parallelogram has base 9 and perpendicular height 8. Heights have right-angle marks. — Dashed heights meet the bases at right angles.

Model · 4.0 minSource: 22, 23

Area Practice: Parallelogram and Trapezoid

Student practice

Find the parallelogram’s area. Explain which displayed measurement is unnecessary.
Find the trapezoid’s area. Identify its two parallel bases and perpendicular height.

Feedback and solution

$A_{para}=11(4)=44$ square meters. The 5-meter slanted side is unnecessary.
$A_{trap}=\tfrac12(6+12)(4)=36$ square meters.

The first figure has base 11, perpendicular height 4, and slanted side 5. The second has parallel bases 6 and 12 with perpendicular height 4. All lengths are in meters. Neither area is displayed. — Use the perpendicular height in both formulas.

Practice · 2.0 minSource: Authored

A Regular Polygon with a 24° Exterior Angle

Student practice

A regular polygon has exterior angle 24°. Find the number of sides and each interior angle.

Feedback and solution

$n=360/24=15$ sides.
Each interior angle is $180-24=156^\circ$.

The incoming side extends horizontally through the vertex. The outgoing side turns 24 degrees. The adjacent interior angle is labeled i equals question mark. The complete polygon and its number of sides are intentionally not shown. — Only one vertex is shown. The polygon is regular.

Practice · 1.5 minSource: 24, 25

Compound Area: Partition and Subtract

Find the shaded area. All dimensions are in meters.

1

The outside rectangle has area $12(8)=96$ square meters.

2

The corner cutout has area $4(3)=12$. The interior hole has area $2(1)=2$.

3

The shaded area is $96-12-2=82$ square meters.

4

A partition gives the same result: $12(5)+8(3)-2=82$.

The outside bounding rectangle is 12 by 8 meters. A bottom-right cutout is 4 by 3 meters. An interior rectangular hole is 2 by 1 meter. The cutout and hole are separate regions, each subtracted once. — Subtract each missing region exactly once.

Model · 2.0 minSource: Authored

Compound Area Practice

Student practice

Find the shaded area using the diagram. Show which regions you add or subtract. All dimensions are in meters.

Feedback and solution

$A=15(10)-5(4)-3(2)=124$ square meters.
An alternative partition gives $15(6)+10(4)-3(2)=124$. Subtract the hole exactly once.

An L-shaped region fits inside a 15 by 10 meter rectangle. A bottom-right cutout is 5 by 4 meters and an interior rectangular hole is 3 by 2 meters. A dashed line shows one possible partition. The shaded area is not provided. — The cutout and the interior hole are separate regions.

Feedback · 2.0 minSource: 26, 27

Pacing checkpoint

Polygon Exit Practice

Student practice

A regular 12-gon has how large an interior-angle sum and how large each interior angle?
A pentagon has interior angles 95°, 120°, 110°, 100°, and x°. Find x.

Feedback and solution

For $n=12$, the sum is $(12-2)180^\circ=1800^\circ$. Each angle is $1800^\circ/12=150^\circ$.
The pentagon sum is $540^\circ$. Thus $x=540-(95+120+110+100)=115$, so the missing angle is $115^\circ$.

Synthesize · 1.5 minSource: 28, 29

Choosing a Triangle Area Formula

Predict first

What information determines whether to use $A=bh/2$, Heron’s formula, or the equilateral formula?

Commit to a response before discussion.

Activate · 0.5 minSource: Authored

Triangle Area: Match the Given Information

1

Known base and height: $A=\tfrac12 bh$. For $b=10$ and $h=6$, $A=30$ square units.

2

Right triangle with a missing leg: use $a^2+b^2=c^2$, then $A=\tfrac12 ab$. Here $c$ is the hypotenuse.

3

Three sides: use Heron’s formula, $A=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=(a+b+c)/2$.

4

Equilateral triangle with side $s$: $A=s^2\sqrt3/4$.

The first triangle labels a base b and a perpendicular height h. The second labels perpendicular legs a and b and hypotenuse c with a right-angle mark. The third labels three side lengths a, b, c. The fourth has three equal sides s with matching tick marks. — Use the given measurements to choose a method.

Explain · 2.0 minSource: 31, 32

Equilateral Triangle: Exact Area

Apply the equilateral formula for side 4 units, then for side 1600 km.

1

For $s=4$, $A=4^2\sqrt3/4=4\sqrt3$ square units.

2

For $s=1600$ km, $A=1600^2\sqrt3/4=640000\sqrt3$ square kilometers.

3

$A\approx1108512.52$ square kilometers.

4

Keep the exact radical form unless an approximation is requested.

A geometrically equilateral triangle has three sides of 1600 kilometers. A dashed perpendicular altitude bisects the base into two 800 kilometer segments. The altitude is labeled h, not precomputed. — Equal sides give two congruent right triangles.

Model · 3.0 minSource: 33, 34

Heron’s Formula: A Worked Example

Find the area of a triangle with sides 5, 6, and 7 units. Give exact form, then round to two decimal places.

1

Check $5+6>7$. The semiperimeter is $s=(5+6+7)/2=9$.

2

Then $s-a=4$, $s-b=3$, and $s-c=2$.

3

$A=\sqrt{9(4)(3)(2)}=\sqrt{216}=6\sqrt6$ square units.

4

$A\approx14.70$ square units. Round after finding the exact value.

A non-right triangle has side lengths a equals 5, b equals 6, and c equals 7, drawn with a common scale. No altitude or computed area is shown. — The semiperimeter is half the perimeter.

Model · 2.0 minSource: Authored

Right Triangle: Height Before Area

A right triangle has base 8 and hypotenuse 17 units. Find its perpendicular height, then its area.

1

The 17-unit side is opposite the right angle, so it is the hypotenuse.

2

$h^2+8^2=17^2$, so $h^2=289-64=225$.

3

$h=\sqrt{225}=15$ units. A length is positive.

4

$A=\tfrac12(8)(15)=60$ square units.

A right triangle has base 8, hypotenuse 17, and unknown perpendicular height h. Its sides are drawn proportionally. No height or area answer is printed. — Find the missing leg before using the area formula.

Model · 1.5 minSource: Authored

Triangle Area Practice

Student practice

Triangle 1 has sides 7, 8, and 9. Find its area using Heron’s formula.
Triangle 2 has base 9 and hypotenuse 15. Find its perpendicular height with $a^2+b^2=c^2$, then its area.

Feedback and solution

$s=12$, so $A=\sqrt{12(5)(4)(3)}=12\sqrt5\approx26.8$ square units.
$h=\sqrt{15^2-9^2}=12$ units, so $A=\tfrac12(9)(12)=54.0$ square units.

The first triangle has side lengths 7, 8, and 9 drawn proportionally, with no right-angle mark. The second is a right triangle with base 9, hypotenuse 15, and unknown perpendicular height h. Answers are absent. — Lengths use the same units within each triangle.

Practice · 4.0 minSource: 35, 36

Error Analysis: Slanted Side or Height?

Student practice

A student uses $A=\tfrac12(10)(13)=65$. Identify the error and correct the area.
Can $a^2+b^2=c^2$ be used for this whole triangle? Explain.

Feedback and solution

The 13-unit side is slanted. Use $h=12$: $A=\tfrac12(10)(12)=60$ square units.
The whole triangle is obtuse. Apply the Pythagorean theorem only to the small right triangle formed by the altitude.

An obtuse triangle has base 10 and perpendicular height 12. Its 13-unit slanted side joins the right end of the base to the opposite vertex. The altitude meets the extended base outside the triangle. No area answer is printed. — The perpendicular height can lie outside an obtuse triangle.

Feedback · 1.0 minSource: 37, 38

Pacing checkpoint

Which Triangle Area Method?

Student practice

Choose an appropriate method for each of the three triangles shown. Find the exact areas in square units.

Feedback and solution

Triangle 1: $A=\tfrac12(14)(5)=35$.
Triangle 2: $A=6^2\sqrt3/4=9\sqrt3$.
Triangle 3: $s=8$ and $A=\sqrt{8(3)(3)(2)}=12$. Alternatively, find height 4 and use $A=bh/2$.

Triangle 1 has base 14 and perpendicular height 5. Triangle 2 has all three sides equal to 6. Triangle 3 has sides 5, 5, and 6, and no height is given. Figures have uniform scales within each triangle and no computed areas. — The problem gives measurements, so you choose the method.

Synthesize · 2.5 minSource: 39, 40

Corresponding Vertices and Sides

Predict first

How does vertex order prevent a similarity proportion from using mismatched sides?

Commit to a response before discussion.

A right triangle ABC and a larger rotated right triangle DEF are similar. A corresponds to D, B to E, C to F. Blue sides AB and DE correspond, teal sides AC and DF correspond, and orange sides BC and EF correspond. Angle marks confirm the mapping. — Rotation preserves the matching angles and sides.

Activate · 0.5 minSource: Authored

Similarity and Navigation Angles

Concept

Corresponding vertices must remain in the same order.
DMS conversion uses 60 minutes per degree and 60 seconds per minute.
One minute of latitude is approximately one nautical mile. One nautical mile is exactly 1852 m.
Example: $5$ nmi $=5(1852)=9260$ m $=9.26$ km.

One degree contains 60 angular minutes, one minute contains 60 angular seconds, and one second is 1/3600 degree. A schematic Earth meridian arc is magnified to illustrate that one minute of latitude is approximately one nautical mile. A nautical mile is exactly 1852 meters. — Primes denote angular minutes and seconds.

Explain · 2.0 minSource: 42, 43

Similarity Proportions and DMS

Match the sides to solve $6/9=10/x$. Convert $45^\circ22'18''$ to decimal degrees, rounded to six decimal places.

$6x=90$, so $x=15$.
Add $45+22/60+18/3600$ to obtain approximately $45.371667^\circ$.
Corresponding sides stay in the same order and DMS uses factors of 60.

The similar triangles label AB as 6, DE as 9, BC as 10, and EF as x. The second visual separates 45 degrees, 22 minutes, and 18 seconds and shows divisors 60 for minutes and 3600 for seconds when converting to degrees. The answer for x is not printed. — Match sides by vertex order. Convert each angular part.

Model · 2.0 minSource: 44, 45

Similar Triangles: Angles and Shrinking

For $\triangle ABC\sim\triangle DEF$, find $\angle D$, $\angle F$, $DF$, and $EF$. Round angles to 0.1° and lengths to 0.01 unit.

$\angle D\approx59.5^\circ$. The third angle is $180-59.5-53.1=67.4^\circ$, so $\angle F\approx67.4^\circ$.
The scale from ABC to DEF is $k=DE/AB=10/15=2/3<1$.
$DF=13(2/3)=26/3\approx8.67$ units.
$EF=14(2/3)=28/3\approx9.33$ units. Apply the same factor to both sides.

Triangle ABC has exact sides AB equals 15, AC equals 13, and BC equals 14. Angle A is approximately 59.5 degrees and angle B approximately 53.1 degrees, rounded to a tenth. The rotated similar triangle DEF has DE equals 10, while DF and EF are unknown. The target is a uniform two-thirds scale of the source. — The matching target sides are shorter by the same factor.

Model · 2.5 minSource: Authored

Decimal Degrees to DMS: Successive Remainders

Convert $135.721^\circ$ to DMS. Keep seconds to one decimal place.

1

The whole degrees are $135$. The fractional degree part is $0.721$.

2

$0.721(60)=43.26$: keep $43$ minutes and the remainder $0.26$.

3

$0.26(60)=15.6$ seconds.

4

$135.721^\circ=135^\circ43'15.6''$.

The angle 135.721 degrees separates into 135 whole degrees and a 0.721 degree remainder. Multiply fractional degrees by 60 for minutes, retain the whole minutes, and multiply only the fractional minutes by 60 for seconds. Computed minutes and seconds are not printed. — Multiply only the fractional remainder at each stage.

Model · 1.5 minSource: Authored

Scale Factor and Decimal Degrees

Student practice

Convert $73.625^\circ$ to DMS. A larger similar triangle has scale factor $3/2$ from a smaller triangle whose corresponding side is 14; find the larger side.

Feedback and solution

$0.625(60)=37.5$: keep $37'$ and convert the remainder, $0.5(60)=30''$.
$73.625^\circ=73^\circ37'30''$.
The larger side is $14(3/2)=21$ units.

The smaller similar triangle has blue side AB equal to 14 and the rotated larger triangle has corresponding blue side DE labeled question mark. The scale factor is 3/2. The angle 73.625 degrees is separated into 73 whole degrees and 0.625 of a degree. The answers are not displayed. — Apply the scale factor to corresponding lengths.

Practice · 2.0 minSource: 46, 47

Error Analysis: DMS and Correspondence

Error analysis

A student reports $0.5^\circ$ as $0^\circ50'$. Correct and explain the conversion.
For $\triangle ABC\sim\triangle DEF$, correct $AB/DE=EF/BC$.

Feedback and solution

$0.5(60)=30$, giving $0^\circ30'00''$.
$AB/DE=BC/EF$. Keep the same triangle in each numerator.

If rounding produces 12 degrees 59 minutes 60 seconds, carry the 60 seconds into one minute. Then carry 60 minutes into one degree, obtaining 13 degrees 0 minutes 0 seconds. In normalized DMS, minutes and seconds each lie from zero inclusive to 60 exclusive. — Carry seconds into minutes, then minutes into degrees.

Feedback · 1.5 minSource: 49, 50

Pacing checkpoint

Quiz 7 Practice: Similarity and DMS

Student practice

Use $k$ for the scale from ABC to DEF. Find $k$, $\angle E$, $\angle F$, $DF$, and $EF$ (angles to 0.1°).
Convert $120^\circ48'50''$ to decimal degrees (six decimal places).

Feedback and solution

Angles: $\angle E\approx22.6^\circ$ and $\angle F\approx180-90-22.6=67.4^\circ$.
Scale: $k=21/12=7/4$.
$DF=5k=8.75$, $EF=13k=22.75$ units.
Degrees: $120+48/60+50/3600$ gives approximately $120.813889^\circ$.

A right triangle ABC has AB equals 12, AC equals 5, BC equals 13, and angle B approximately 22.6 degrees. A rotated similar triangle DEF has DE equals 21. Students must find corresponding angles E and F and sides DF and EF. No requested values are displayed. — Determine the scale from the corresponding given sides.

Synthesize · 3.0 minSource: 51, 52

Final Exit: Area and Navigation

Student practice

Find the area of a trapezoid with bases 8 m and 14 m and perpendicular height 5 m. State which measurement must be perpendicular.
A ship moves north by $6'$ of latitude. Estimate the distance in nautical miles and kilometers.
Record one skill to review before Quiz 7.

Synthesize · 2.0 minSource: 53, 54, 55, 57

Polygon Area, Triangles, and Similarity

Today’s Learning Roadmap

Session 2 · Polygons, Triangles, and Applications

Exterior Angles: One Complete Turn

Polygon Angle Sums and Area

Octagon, Trapezoid, and Parallelogram

Area Practice: Parallelogram and Trapezoid

A Regular Polygon with a 24° Exterior Angle

Compound Area: Partition and Subtract

Compound Area Practice

Polygon Exit Practice

Choosing a Triangle Area Formula

Triangle Area: Match the Given Information

Equilateral Triangle: Exact Area

Heron’s Formula: A Worked Example

Right Triangle: Height Before Area

Triangle Area Practice

Error Analysis: Slanted Side or Height?

Which Triangle Area Method?

Corresponding Vertices and Sides

Similarity and Navigation Angles

Similarity Proportions and DMS

Similar Triangles: Angles and Shrinking

Decimal Degrees to DMS: Successive Remainders

Scale Factor and Decimal Degrees

Error Analysis: DMS and Correspondence

Quiz 7 Practice: Similarity and DMS

Final Exit: Area and Navigation