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Wiskunde

Grade 8

Verhoudings en eweredigheid

Meting en vormleer

Konstruksies

Module 17

Konstruksie van verskillende soorte hoeke

AKTIWITEIT 1

Om verskillende soorte hoeke en driehoeke te konstrueer

[LU 3.4, 3.5, 4.7]

1. Hoe om ‘n hoek te teken:Benodigdhede: potlood, liniaal, gradeboog.

1.1 Begin altyd met ‘n basislyn.

1.2 Maak ‘n merkie vir ‘n begin bv. links en plaas jou gradeboog op jou merkie.

1.3 Lees óf van die buitekant óf die binnekant van jou gradeboog vanaf 0°.

1.4 By hoeke groter as 180° moet jy eers die bepaalde hoek van 360° aftrek, en dan daardie betrokke hoek teken. Die hoek buitekant om (die inspringende hoek) sal dan die betrokke hoek wees wat jy moet teken.Bv. 320°: (360° – 320° = 40°). Teken nou ‘n hoek van 40°. Die inspringende hoek verteenwoordig nou die 320°.

2. Konstrueer nou die volgende hoeke en benoem elke hoek:

  • A B ˆ size 12{ { hat {B}}} {} C = 75°

Soort hoek: ______

2.2 C D ˆ size 12{ { hat {D}}} {} E = 135°

Soort hoek: ______

2.3 F G ˆ size 12{ { hat {G}}} {} H = 215°

Soort hoek: ______

3. Hoe om ‘n driehoek te konstrueer:

Benodigdhede: potlood, liniaal, gradeboog en passer.

  • Begin altyd eers deur ‘n rowwe skets te maak.
  • Gebruik dan een van die sye waarvan die lengte gegee is, as basis.
  • Bv. konstrueer Δ size 12{Δ} {} ABC met BC = 40 mm, B ˆ size 12{ { hat {B}}} {} = 70° en C ˆ size 12{ { hat {C}}} {} = 50°.

Rowwe skets:

  • Om ‘n sylengte akkuraat te meet moet jy die lengte met jou passer op jou liniaal meet en dan jou passer se punt op B sit en met die potlood ‘n “kapmerk” maak waar C moet wees.
  • Konstruksie:

4. Konstrueer nou elk van die volgende driehoeke:

4.2 Δ size 12{Δ} {} PQR met QR = 58 mm, P Q ˆ size 12{ { hat {Q}}} {} R = 62° en Q P ˆ size 12{ { hat {P}}} {} R = 69°.

Meet:

  1. PQ = mm
  2. R ˆ size 12{ { hat {R}}} {} =

4.2 Gelykbenige Δ size 12{Δ} {} ABC met BC = 42 mm, AB = AC en B ˆ size 12{ { hat {B}}} {} = 63°.

Meet:

a) PQ = mm

AKTIWITEIT 2

Om enige gegewe lyn of hoek te halveer [LU 3.4, 3.5, 4.7]

  1. Halvering van ‘n gegewe lyn AB :
  • Meet lynstuk AB (bv. 40 mm).
  • Neem jou passer en meet bietjie meer as die helfte van jou lyn (d.w.s. ± 22-25 mm).
  • Plaas jou passer se skerppunt op A en maak ‘n “kapmerk” onder en bo die lyn.
  • Plaas dan jou passer op B en maak ook ‘n “kapmerk” bo en onder die lyn.
  • Verbind die kruispunte van die twee “kapmerke” met mekaar.
  • Benoem die punt op lyn AB , P. P is nou die middelpunt van lyn AB .

2. Probeer nou self die volgende:

  • Teken ‘n lynstuk PQ = 70 mm.
  • Halveer nou lynstuk PQ , soos in nr. 1 verduidelik.

3. Halvering van π ABC :

  • Plaas jou passer se skerppunt op B .
  • Trek enige grootte boog soos aangedui.
  • Plaas jou passer se punt op die punt waar die twee lyne mekaar kruis en maak ‘n “kapmerk” binne die hoek.
  • Plaas nou jou passer se punt op die ander punt waar die twee lyne mekaar kruis en maak ‘n “kapmerk” binne die hoek, sodat jou twee “kapmerke” mekaar kruis.
  • Verbind B ˆ size 12{ { hat {B}}} {} (hoek B ) met die plek waar jou “kapmerke” mekaar kruis.
  • B ˆ size 12{ { hat {B}}} {} 1 sal nou net so groot wees soos B ˆ size 12{ { hat {B}}} {} 2 . Meet beide hoeke. Is hulle ewe groot?

4. Probeer nou self die volgende doen:

  • Teken D E ˆ size 12{ { hat {E}}} {} F . = 125°.
  • Halveer nou D E ˆ size 12{ { hat {E}}} {} F .

AKTIWITEIT 3

Om ‘n loodlyn vanuit ‘n punt op ‘n lyn te konstrueer [LU 3.4, 3.5, 4.7]

1. Konstrueer AD size 12{ ortho } {} BC .

  • Plaas jou passer se skerppunt op A (want jy wil uit A ‘n lyn loodreg op BC trek.)
  • Maak nou ‘n boog op BC .
  • Plaas jou passer se punt eers op die een punt waar die boog en BC mekaar kruis en maak ‘n “kapmerk” onder BC en dan op die ander kruispunt en maak weer ‘n “kapmerk” onder BC , sodat jou twee “kapmerke” mekaar kruis.
  • Verbind nou A met die kruispunt van die twee “kapmerke”.
  • Merk die plek waar die twee lyne mekaar sny, D .
  • AD is nou loodreg op BC . ( AD size 12{ ortho } {} BC .)

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Wiskunde graad 8. OpenStax CNX. Sep 11, 2009 Download for free at http://cnx.org/content/col11033/1.1
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