PR is die middellyn van die sirkel PRSU - NSC Mathematics - Question 8 - 2017 - Paper 2
Question 8
PR is die middellyn van die sirkel PRSU. QU is ewewydig aan RS geteken en sny SP by T.
8.1.1 Skryf, met 'n rede, die grootte van S neer.
8.1.2 As die middellyn 20 ... show full transcript
Worked Solution & Example Answer:PR is die middellyn van die sirkel PRSU - NSC Mathematics - Question 8 - 2017 - Paper 2
Step 1
8.1.1 Skryf, met 'n rede, die grootte van S neer.
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Answer
Volgens die eienskappe van 'n halwe sirkel, is die hoek ildeS wat teen die diameter gestel is, altyd 90ext°. Dus, ildeS=90ext°.
Step 2
8.1.2 As die middellyn 20 cm en SP = 16 cm, bereken die lengte van TU.
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Answer
Kom ons gebruik die Piramiedteoreem. Omdat T die middelpunt van SP is, is ST = 8 cm. Van riangleSQT kan ons aflei:
QT2=SQ2+ST2
QT2=(10)2−(8)2=100−64=36
Dus, QT=ext6cm. Aangesien TU = QP = 10 cm, dan TU = 4 cm.
Step 3
8.2.1 $ ilde{B}_1$
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Answer
Volgens die Chord Theorem, kan ons afsluit ildeB1=30ext°.
Step 4
8.2.2 $ ilde{P}_2$
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Answer
Aangesien PQ || BS, volg dat ildeP2=70ext°.
Step 5
8.2.3 $ ilde{R}$
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Aangesien riangleABC 'n siklies vierhoek is, weet ons dat die teenoorgestelde hoeke saam 180° is. Dus, ildeR=180ext°−ildeB−ildeP=180ext°−30ext°−70ext°=15ext°.
Step 6
8.2.4 $ ilde{Q}_2$
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In riangleABC, gebruik die som van die hoeke: ildeQ2=180ext°−(30ext°+70ext°)=80ext°.