Fun Geometry Problem with Solution #120
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٨ x = 30°
٨(Angel Lazo HK)



(1)∠ACD = 90°- x Р∠BDC = 130°- x

(2) ˹ش O 繨شٹҧͧǧՠ∆ACD Ṻ AO = CO = DO =
͡ҡ ѧҠ∠COD = 2(∠CAD) ∠COD = 80°
∵ CO = DO ∆CDO 繠∆˹Ҩ ՠ∠O (= 80°) ʹ ∠DCO = 50°(∠ACO = 40°- x) Р∠CDO = 50°
∵ AO = CO ∆ACO 繠∆˹Ҩ ՠ∠O ʹ ∠CAO =∠ACO ∠CAO = 40°- x

(3) ˹ش P BC CP =
Ҡ∆CDP∆CDO ¤ѹẺ -- (CD = CD,∠DCP =∠DCO, CP = CO) DP = DO DP =
͡ҡ ѧҠ∠CDP =∠CDO ∠CDP = 50° ∠BDP = 80°- x Р∠BPD = 100°

(4) ˹ش Q ˹ BD BQ = DQ =
Ҡ∆BDQ∆ACO ¤ѹẺ -- (BD = AC, BQ = AO, DQ = CO) ∠DBQ =∠CAO Р∠BDQ =∠ACO ∠DBQ = 40°- x Р∠BDQ = 40°- x ∠PBQ = 2x - 40°Р∠PDQ = 40°

(5) α= 20°
ѧࡵ ☐BPDQ ՠBQ = DP = DQ,∠PDQ = 2αР∠BPD = 120°-α ∠PBQ =α (Click ʹԸվ٨⨷ 4) 2x - 40°= 20° x = 30° Q.E.D.




Create Date : 16 ¹ 2558
Last Update : 29 áҤ 2558 23:56:00 .
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Fun Geometry Problem with Solution #119
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======================================




˹∆ABC 繠∆ҹ
٨ BP = AP + CP
٨



(1)∵∆ABC 繠∆ҹ AC = BC Р∠BAC =∠ACB = 60°
ѧࡵҠ∠BAC =∠BPC ☐ABCP öṺǧ APB =ACB APB = 60°
͡ҡ ѧ ∠CAP =∠CBP

(2) ˹ش Q BP PQ = CP
∵ CP = PQ Р∠CPQ = 60° ∆CPQ 繠∆ҹ ∠CQP = 60° ∠BQC = 120°

(3) ѧࡵҠ∆BCQ∆ACP ¤ѹẺ -- (∠BQC =∠APC,∠CBQ =∠CAP, BC = AC) BQ = AP
∵BP = BQ + PQ BP = AP + CP Q.E.D.




Create Date : 13 ¹ 2558
Last Update : 21 ¹ 2558 19:22:00 .
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Fun Geometry Problem with Solution #118
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٨ x = 10°
٨ 1



(1)∠ABC = 40°
∵ ∠CAP =∠ACP ∆ACP 繠∆˹Ҩ ՠ∠P ʹ AP = CP

(2) ˹ش Q AB PQ = AP (= CP) ∆APQ 繠∆˹Ҩ ՠ∠P ʹ ∠AQP =∠PAQ ∠AQP = 20°

(3) ԨóҠ☐ACPQ Ҡ∠CPQ (˭) = 360°- 60° ∠CPQ () = 60°
∵ CP = PQ Р∠CPQ = 60° ∆CPQ 繠∆ҹ CQ = PQ
͡ҡ ѧҠ∠PCQ = 60° ∠BCQ = 40° ∠BCQ =∠CBQ ∆BCQ 繠∆˹Ҩ ՠ∠Q ʹ BQ = CQ BQ = PQ ∆BPQ 繠∆˹Ҩ ՠ∠Q ʹ ∠BPQ =∠PBQ ∠BPQ = x

(4) ԨóҠ∆BPQ Ҡ∠PBQ +∠BPQ =∠AQP x + x = 20° x = 10° Q.E.D.

٨ 2



(1)∠ABC = 40°
∵ ∠CAP =∠ACP ∆ACP 繠∆˹Ҩ ՠ∠P ʹ AP = CP

(2) ˹ش Q Ҿз͹ͧش C ҹ AB ∆ABQ∆ABC AQ = AC,∠BAQ =∠BAC = 30°Р∠AQB =∠ACB = 110°
∵ AC = AQ Р∠CAQ = 60° ∆ACQ 繠∆ҹ AQ = CQ Р∠AQC = 60°(∠BQC = 50°)

(3) ѧࡵҠ∆CPQ∆APQ ¤ѹẺ -- (CP = AP, CQ = AQ, PQ = PQ) ∠CQP (=∠AQP) = (∠AQC)/2 = 30°

(4) ԨóҠ☐BCPQ Ҡ∠BCP +∠BQP = 180° ☐BCPQ öṺǧ ∠CBP =∠CQP ∠CBP = 30° ∠ABP = x = 10° Q.E.D.




Create Date : 10 ¹ 2558
Last Update : 10 ¹ 2558 0:00:01 .
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Fun Geometry Problem with Solution #117
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٨ x = 6°
٨



(1) ˹ش O 繨شٹҧͧǧՠ∆ABP Ṻ ...
AO = BO
∠AOP = 2(∠ABP) ∠AOP = 12°
∠BOP = 2(∠BAP) ∠BOP = 48°
∵ AO = BO Р∠AOB = 60° ∆ABO 繠∆ҹ AB = BO Р∠ABO = 60°

(2) ˹ش Q BC BQ = AB (= BO) ∆ABQ 繠∆˹Ҩ ՠ∠B (= 72°) ʹ ∠AQB (=∠BAQ) = 54°
Ҡ∆BPQ∆BOP ¤ѹẺ -- (BP = BP,∠PBQ =∠OBP, BQ = BO) ∠BQP =∠BOP ∠BQP = 48° ∠AQP = 6°

(3) ԨóҠ☐ACQP Ҡ∠CAP =∠BQP ☐ACQP öṺǧ ∠ACP =∠AQP x = 6° Q.E.D.




Create Date : 07 ¹ 2558
Last Update : 7 ¹ 2558 0:00:00 .
Counter : 557 Pageviews.

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Fun Geometry Problem with Solution #116
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======================================




٨ x = 30°
٨



(1)∠ADC = 80°

(2) ˹ش P CD AP = AD ∆ADP 繠∆˹Ҩ ՠ∠A ʹ ∠APD =∠ADP ∠APD = 80° ∠CAP = 40°

(3) ˹ش Q Ҿз͹ͧش A ҹ BC ∆BCQ∆ABC CQ = AC,∠BCQ =∠ACB = 30°Р∠BQC =∠BAC = 20°
∵ AC = CQ Р∠ACQ = 60° ∆ACQ 繠∆ҹ AC = AQ Р∠CAQ =∠AQC = 60°(∠BAQ =∠AQB = 40°)

(4) ѧࡵҠ∆ABQ∆ACP ¤ѹẺ -- (∠BAQ =∠CAP, AQ = AC,∠AQB =∠ACP) AB = AP AB = AD ∆ABD 繠∆˹Ҩ ՠ∠A (= 80°) ʹ ∠ADB (=∠ABD) = 50° ∠BDC = x = 30° Q.E.D.




Create Date : 04 ¹ 2558
Last Update : 4 ¹ 2558 0:03:00 .
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