9.1 In die diagram is DEFG 'n koordevierhoek met DE || GF - NSC Mathematics - Question 9 - 2022 - Paper 2
Question 9
9.1 In die diagram is DEFG 'n koordevierhoek met DE || GF. DE word na R verleng. T is nog 'n punt op die sirkel. EG, FT en ET word getrek. Eã = 72° en Gã = 16°.
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Worked Solution & Example Answer:9.1 In die diagram is DEFG 'n koordevierhoek met DE || GF - NSC Mathematics - Question 9 - 2022 - Paper 2
Step 1
Bepaal, met redes, die grootte van die hoek DãGF.
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Answer
Omdat DE || GF is en D is 'n hoek van die koordevierhoek, is DãGF = Eã = 72° (omdat die hoeke aan die buitekant van die koordevierhoek gelyk is).
Step 2
Bepaal, met redes, die grootte van die hoek Tã.
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Answer
Aangesien die hoeke aan die sirkel aanling, en Gã = 16°, beteken dit dat Tã = 72° - 16° = 56°.
Step 3
Bepaal, met redes, die grootte van die hoek GãEF.
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Answer
Omdat Gã = 16° en DE || GF, is GãEF = 180° - 16° = 164°, aangesien hierdie 'n alternatiewe binnehoek is.
Step 4
Bepaal, met redes, die waarde van die verhouding NP : PT in eenvoudigste vorm.
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Answer
Die verhouding NP : PT is 4 : 1, aangesien NM : MS = 4 : 1. Gegewe dat PT = 4, kan ons NP vind as NP = 16 : 4 = 4 : 1.
Step 5
Bewys, met redes, dat KM || RS.
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Hierdie bewys volg uit die druk van parallelogramme. Aangesien NP en PT 'n regte lyn vorm, is die hoeke aan die teenoorgestelde kant gelyk, dus is KM || RS.
Step 6
Bepaal, met redes, die lengte van RL.
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Answer
Om RL te bereken, gebruik ons die proporetionale eienskappe: RL = \frac{12 * 21}{16} = 15,75 cm.
