What is a triangle in geometry?

In Geometry, a triangle is a three-sided polygon that consists of three edges and three vertices . The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees. This property is called angle sum property of triangle.

Why is geometry difficult? Geometry is creative rather than analytical, and students often have trouble making the leap between Algebra and Geometry . They are required to use their spatial and logical skills instead of the analytical skills they were accustomed to using in Algebra.

Is geometry math easy?

It requires logical and deductive reasoning, which can be challenging for students who need help with abstract thinking or have difficulty following formal proof structures . Complex Language: Geometry has specific terminology and vocabulary, which can overwhelm some students.

What does △ mean in geometry?

The triangle symbol (△) is used in math to reference a triangle in a diagram . Typically, the symbol is used in an expression like this: △ABC. In plain language, the expression △ABC can be read as the triangle formed by the three points A, B and C.

What are the 3 dots in a triangle maths?

In logical argument and mathematical proof, the therefore sign, ∴ , is generally used before a logical consequence, such as the conclusion of a syllogism. The symbol consists of three dots placed in an upright triangle and is read therefore.

What is triangle class 9?

A figure formed by the intersection of three lines is said to be a triangle. A triangle has three vertices, three sides, and three angles. The above figure shows ABC, here AB, BC, AC are the sides of the triangle. A, B, C are the vertex and ∠ A, ∠B, ∠ C are the three angles.

What is the easiest way to learn geometry?

To understand geometry, it is easier to visualize the problem and then draw a diagram . If you're asked about some angles, draw them. Relationships like vertical angles are much easier to see in a diagram; if one isn't provided, draw it yourself.

(PDF) Geometry Unit 8 Practice Test QandA - MathGuy.US · Geometry Practice Test – Unit 8 Page | 2 A C B The first thing to notice about this right triangle is that the short leg is half

Geometry Name_________________________________Period:___

Practice Test – Unit 8

Note:thispagewillnotbeavailabletoyouforthetest.Memorizeit!

TrigonometricFunctions(p.53oftheGeometryHandbook,version2.1)

SpecialTriangles(p.52oftheGeometryHandbook,version2.1)

45⁰‐45⁰‐90⁰Triangle

√

30⁰‐60⁰‐90⁰Triangle

√ 2

SOH‐CAH‐TOA

sin sin sin

cos cos cos

tan tan tan

Ina45⁰‐45⁰‐90⁰triangle,thecongruenceoftwoanglesguaranteesthecongruenceofthetwolegsofthetriangle.Theproportionsofthethree

sidesare: ∶ ∶ √ .Thatis,thetwolegshave

thesamelengthandthehypotenuseis√ timesaslongaseitherleg.

Ina30⁰‐60⁰‐90⁰triangle,theproportionsofthe

threesidesare: ∶ √ ∶ .Thatis,thelongleg

is√ timesaslongastheshortleg,andthehypotenuseis timesaslongastheshortleg.

GeometryPracticeTest–Unit8

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Thefirstthingtonoticeaboutthisrighttriangleisthattheshortlegishalfthelengthofthehypotenuse.Thatmakesthisa30° 60° 90°triangle,whichhassideproportions:1 ∶ √3 ∶ 2.

So,wehave:

6 ∙ √3 √

∠ °

Thefirstthingtonoticeaboutthisrighttriangleisthatithasa45°angle.Thatmakesthisa45° 45° 90°triangle,whichhassideproportions:1 ∶ 1 ∶ √2.

So,wehave:

√ √

∙ √√

√ √

∠ °

√

C A

Thisisnotaspecialtriangle,sowemustuseTrigfunctionstosolveit.

First,wehave:

Then,

∠ ° ° °

tan 15°9

⇒

9tan 15°

sin 15°9

⇒9

sin 15°.

Solve each triangle below. Remember, each triangle has three answers.

Fortheseproblems,wehaveaddednamesfortheanglesandthemissingsides.Wesuggestyoudothesame.

1) 2)

612 45°

15°

GeometryPracticeTest–Unit8

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Thisisnotaspecialtriangle,sowemustuseTrigfunctionstosolveit.

First,wehave:

Then,

2 7 ⇒ 4 49

⇒ 45

√45 √ ~ .

sin27

⇒ ∠ sin

. °

∠ ° . ° . °

HopeyoulikeTrigfunctions!

First,wehave:

Then,

5 12 ⇒ 25 144

⇒ 169

√169

tan512

⇒ ∠ tan

512

. °

∠ ° . ° . °

Heretheyareagain!

First,wehave:

Then,

∠ ° ° °

tan 22°10

⇒ 10 ∙ tan 22°~ .

cos 22°10

⇒

10cos 22°

~ .

Tip:Whencalculatinganglesinproblemswheretwosidelengthsaregiven,baseyourtrigfunctionsonthegivenlengths,evenifyouhavealreadycalculatedthelengthoftheremainingside.Thiswillproducemoreaccurateanswers. 4) 5)

22°

512

GeometryPracticeTest–Unit8

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7) A regular octagon has a perimeter of 80 inches and an apothem of 12.07 inches. Find the area of the regular octagon, rounded to one decimal place.

Theformulafortheareaofaregularpolygonis ,where isthe

lengthoftheapothemand istheperimeterofthepolygon.

Wearegiven: 80, 12.07

So, 12.07 80 . in2

8) Find the area of the regular hexagon shown below. Leave your answer as a radical.

Step1:Howmanysides?6

Step2:Findtheperimeter: 6 ∙ 18mm 108mm

Step3:Findtheapothem:

Createthe“littleguy”triangle:

Ourgoalistofind .

Thelengthofthebaseofthe“littleguy”triangleis: 2

Thesumoftheanglesinthefigure(upperright)is: 6 2 ∙ 180˚ 720˚

Eachangleofthefiguremeasures:720˚ 6 120˚

∠ 120˚ 2 60˚

Then,the“littleguy”triangleisa30˚‐60˚‐90˚triangle.So, √ .

Step4:Calculatethearea:

∙ √ ∙ √

Step5:(Optional)Compareresulttotheareaofasquarewithside2 .

TheareainStep4is486 ∙ √3~ .

Theareaofasquarewithside2 18√3shouldbealittlemorethanthis.

Squareareais: 18√3 ∙ 18√3 324 ∙ 3 vs .

GeometryPracticeTest–Unit8

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9) A regular pentagon has each side = 8 cm. Find the area of the regular pentagon, rounded to one decimal place.

Step1:Howmanysides?5

Step2:Findtheperimeter: 5 ∙ 8cm 40cm

Step3:Findtheapothem:

Createthe“littleguy”triangle:

Ourgoalistofind .

Thelengthofthebaseofthe“littleguy”triangleis: 2

Thesumoftheanglesinthefigure(upperright)is: 5 2 ∙ 180˚ 540˚

Eachangleofthefiguremeasures:540˚ 5 108˚

∠ 108˚ 2 54˚

Then,tan 54° ⇒ 4 tan 54° ~ . .

(keeplotsofdecimalsinyouranswersuntilthefinalcalculation)

Step4:Calculatethearea:

∙ . ∙ .

Step5:(Optional)Compareresulttotheareaofasquarewithside2 .

TheareainStep4is .

Theareaofasquarewithside2 ~11.01shouldbealittlemorethanthis.

Squareareais: 11.01 ∙ 11.01 . vs .

GeometryPracticeTest–Unit8

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10) An equilateral triangle has a side of 7√3inches. Find the area of the equilateral triangle. Leave your answer as a radical.

Step1:Howmanysides?3

Step2:Findtheperimeter: 3 ∙ √ mm 21√3in

Step3:Findtheapothem:

Createthe“littleguy”triangle:

Ourgoalistofind .

Thelengthofthebaseofthe“littleguy”triangleis: √ 2 √

Thesumoftheanglesinatriangleis180˚

Eachangleofthefiguremeasures:180˚ 3 60˚

∠ 60˚ 2 30˚

Then,the“littleguy”triangleisa30˚‐60˚‐90˚triangle.So,√ √ √

√.

Step4:Calculatethearea:

∙ ∙ √

Alternative:Findtheheightandthenusetheformula

Thelengthofthebaseofthelefttriangleis: √ 2 √

Thesumoftheanglesinatriangleis180˚

Eachangleofthefiguremeasures:180˚ 3 60˚

Then,thelefttriangleisa30˚‐60˚‐90˚triangle.

So, ∙ √37√3 ∙ √3

7√3∙√3 7∙3.

Calculatetheareaoftheentiretriangle:

∙ √ ∙

√

√ in

° √ in

GeometryPracticeTest–Unit8

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11) A regular hexagon has an apothem of15inches. Each side is 10√3 . Find the area.

Wearegiven: 15 10√3

Perimeter 6 6 ∙ 10√3 60√3

So, 15 60√3 √ in2

12) A regular octagon has a perimeter of 8 cm. Find the area of the regular octagon. Round your answer to the nearest tenth.

Step1:Howmanysides?8(notealsothatwearegiven )

Step2:Findthelengthofaside: 8 1

Step3:Findtheapothem:

Createthe“littleguy”triangle:

Ourgoalistofind .

Thelengthofthebaseofthe“littleguy”triangleis:1 2 .

Thesumoftheanglesinthefigure upperright is: 8 2 ∙ 180˚ 1,080˚

Eachangleofthefiguremeasures:1,080˚ 8 135˚

∠ 135˚ 2 67.5˚

Then,tan 67.5˚ .

So, . ∙ tan 67.5˚ .

Step4:Calculatethearea:

∙ . ∙ .

Step5:(Optional)Compareresulttotheareaofasquarewithside2 .

TheareainStep4is .

Theareaofasquarewithside2 ~2.4142shouldbealittlemorethanthis.

Squareareais: 2.4142 ∙ 2.4142 . vs .

√ in

GeometryPracticeTest–Unit8

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7cm

Find the surface area for each solid.

13)

TotalSurfaceArea 2 ∙ 36 2 ∙ 12 2 ∙ 27 ft2 14)

TotalSurfaceArea 2 ∙ 56 640 698.57 280 , . cm2

9ft

4ft.

3ft

Thetwomainoptionstodealwiththisproblemaretouseaformula,whichworksifyouhaveitavailable,ortodeconstructtheshape.WewillillustratethedeconstructionmethodinProblems13and14.

FrontandBack

9 ∙ 4 36ft2

9ft

4ft

TwoSides

3 ∙ 4 12ft2

TopandBottom

9 ∙ 3 27ft2

3ft

4ft

9ft

3ft

LeftRectangularFace

40 ∙ 16 640 cm2

RightRectangularFace

40 ∙ 17.46425 698.57cm2

BottomRectangle

40 ∙ 7 280cm2

12∙ 16 ∙ 7

FrontandBackTriangularFaces

56cm2

7cm

40cm40cm

16cm

7ft

40cm

17.46425 cm

16ft7 16

305

~17.46425

Findlengthofhypotenuse:

GeometryPracticeTest–Unit8

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Wewilluseformulastocalculatethesurfaceareainthebalanceoftheseexercises.

Find the surface area in terms of x. Leave pi in the answer. 15) 16) 17) Find the surface area given the square base of side equal to 16m, and height of 10m.

10in

13in

2 2 2 7 9 2 7

Theformulaforthesurfaceareaofacylinderis 2 2 ,where istheradiusofthecircularfacesand istheheightofthecylinder.

Forthisproblem, 7and 9 .So,

7in

9xin

2 2 2 6.5 10 2 6.5

Forthisproblem, 6.5and 10.So,

1264 12.80625 256 .

Theformulaforthesurfaceareaofapyramidis ,where

istheperimeterofthebase, istheslant‐heightofaface,and istheareaofthebase.

Forthisproblem,

4 ∙ 16 64,

12.80625(seeboxatleft),and

16 256

8 10

164

~ .

FindtheSlant‐Height

Notethatthelengthofthebaseofthetriangularsemi‐cross‐sectionifthepyramidis16 2 8.

Then,

10m

GeometryPracticeTest–Unit8

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Find the surface area for the following solids. Leave pi in the answer.

18) 19) 20) Find the surface area of the cone in terms of x.

10cm

13cm

l =26x

r=10x

5 16.763 5

Theformulaforthesurfaceareaofaconeis ,where istheradiusofthecircularbaseand istheheightofthecone.

Forthisproblem, 5and √5 16 16.763.So,

5 13 5

Forthisproblem, 5and 13.So,

10 26 10

Forthisproblem, 10 and 26 .So,

16mm

5mm

GeometryPracticeTest–Unit8

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Find the surface area of the solids. Leave pi in the answer. 21) radius = 14 in 22) diameter = 14 cm Find the Surface Area of the solid. 23) 24) radius = 6 mm

2.7m

3.1m

2.6m

10mm

4 4 14

Theformulaforthesurfaceareaofasphereis 4 ,where istheradiusofthesphere.

Forthisproblem, 14.So,

4 4 7

Forthisproblem, 7.So,

2 2 2 2 ∙ 2.7 3.1 2.7 2.6 3.1 2.6

Theformulaforthesurfaceareaofarectangularprismis 2 ,where , and arethedimensionsoftheprism.

Forthisproblem, 2.7, 3.1, 2.6.So,

2 2 2 6 10 2 6

Theformulaforthesurfaceareaofacylinderis 2 2 ,where istheradiusofthecircularfacesand istheheightofthecylinder.

Forthisproblem, 6and 10.So,

GeometryPracticeTest–Unit8

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25) Find the surface area of both the right hexagonal prism and the cylinder whose bases are drawn below. Notice that both figures have the same radius. Assume the hexagon is regular and that the height of both solids is 10 m.

SurfaceAreaoftheHexagonalPrism

First,findtheareaoftheHexagonalBase

Step1:Howmanysides?6

Step2:Findtheperimeter:

Notethatthelengthofasideofahexagonisthesameasitsradius.So, 6 ∙ 16mm 96m

Step3:Findtheapothem:

Createthe“littleguy”triangle:

Ourgoalistofind .

Thelengthofthebaseofthe“littleguy”triangleis: 2 hypotenuse

Therefore,the“littleguy”triangleisa30˚‐60˚‐90˚triangle.

So, ∙ √3 √ .

Step4:Calculatetheareaofeachhexagonalbase:

∙ √ ∙ √ ~ .

Next,findtheareaofeachrectangularface:

∙ 16 ∙ 10

Then,findthesurfaceareaoftheregularhexagonalprism:

Theprismhas2basesand6faces,so 2 6 ,where

istheareaofahexagonalbaseand istheareaofarectangularface.

2 . 6 , .

16m

2 2 2 2 16 10 2 16 2

~ , .

SurfaceAreaoftheCylinder: 2 2

Forthisproblem, 16and 10.So,

(PDF) Geometry Unit 8 Practice Test QandA - MathGuy.US · Geometry Practice Test – Unit 8 Page | 2 A C B The first thing to notice about this right triangle is that the short leg is half - DOKUMEN.TIPS (2024)

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