Fun Geometry Problem with Solution #145



٨Ҡα+β+ γ = 90°
٨



(1)∵AB = BC ∆ABC 繠∆˹Ҩ ՠ∠B (= 90°) ʹ ∠ACB (=∠BAC) = 45° α= 45°

(2) ҡٻ Ҡ∆APQ∆EQR∆ABD ¤ѹẺ -- (AP = QR = AB,∠APQ =∠ERQ =∠ABD, PQ = ER = BD) ∠AQP =∠QER =∠ADB = β

(3) ԨóҠ∆EQR Ҡ∠EQR = 90°- β
∵ ∠AQP +∠AQE +∠EQR = 180° β+∠AQE + (90°-β) = 180° ∠AQE = 90°

(4)∵∆APQ∆EQR AQ = EQ ∆AEQ 繠∆˹Ҩ ՠ∠Q (= 90°) ʹ ∠AEQ (=∠EAQ) = 45° ∠AEQ =α
∵ ∠AEQ +∠QER +∠AEB = 90° α+β+ γ = 90°Q.E.D.




Create Date : 30 Զع¹ 2558
Last Update : 30 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #144



˹∠BCDҹ
٨ x = 15°
٨



(1) ԨóҠ∆ABD ∠ADB = 90°- x
ԨóҠ∆ACD ∠CAD = 60°- x

(2) BC ͡ѧش P ·∠BPD = 90° ∠DCP = 3x
Ҡ∆BDP∆ABD ¤ѹẺ -- (∠BPD =∠BAD,∠DBP =∠ABD, BD = BD) BP = AB ∆ABP ∆˹Ҩ ՠ∠B (= 2x) ʹ ∠BAP =∠APB = 90°- x ∠DAP =∠APD = x

(3) ˹ش Q Ҿз͹ͧش A ҹ CD ∆CDQ∆ACD CQ = AC,∠DCQ =∠ACD = 30°Р∠CQD = ∠CAD = 60°- x
∵ AC = CQ Р∠ACQ = 60° ∆ACQ 繠∆ҹ AC = AQ
͡ҡ ѧҠ∠CAQ =∠AQC = 60° ∠DAQ =∠AQD = x

(4) ѧࡵҠ∆ADP∆ADQ ¤ѹẺ -- (∠APD =∠AQD,∠DAP =∠DAQ, AD = AD) AP = AQ AP = AC ∆ACP 繠∆˹Ҩ ՠ∠A ʹ ∠ACP =∠APC ∠ACP = 90°- x ∠DCP = 60°- x 3x = 60°- x x = 15° Q.E.D.




Create Date : 27 Զع¹ 2558
Last Update : 27 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #143



٨ x = 40° ɮպǡѺǧ
٨



(1) AP ͡仾 BC ش Q ∠BPQ = 40°,∠AQB = 100°Р∠AQC = 80°
∵ ∠BPQ =∠PBQ ∆BPQ 繠∆˹Ҩ ՠ∠Q ʹ BQ = PQ

(2) ˹ش R Ҿз͹ͧش Q ҹ AC ∆ACR∆ACQ AR = AQ,∠CAR =∠CAQ = 30° ∠ARC = ∠AQC = 80°
∵ AQ = AR Р∠QAR = 60° ∆AQR 繠∆ҹ AQ = QR
͡ҡ ѧҠ∠AQR =∠ARQ = 60° ∠CQR =∠CRQ = 20°

(3) ˹ش S AB ∠AQS = 20° ∠BQS = 80°Р∠BSQ = 40°
Ҡ∆AQS∆CQR ¤ѹẺ -- (∠QAS =∠CRQ, AQ = QR,∠AQS =∠CQR) QS = CQ

(4) ѧࡵҠ∆CPQ∆BQS ¤ѹẺ -- (CQ = QS,∠CQP =∠BQS, PQ = BQ) ∠PCQ =∠BSQ x = 40° Q.E.D.




Create Date : 24 Զع¹ 2558
Last Update : 24 Զع¹ 2558 0:00:00 .
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Fun Geometry Problem with Solution #142



٨ x = 30°
٨



AP = L

(1) AB ͡ѧش Q · PQ = L ∠CBQ = 140°
∵ AP = PQ ∆APQ 繠∆˹Ҩ ՠ∠P ʹ ∠AQP =∠PAQ ∠AQP = 10°

(2) AC ͡ѧش R · PR = L ∠BCR = 70°
∵ AP = PR ∆APR 繠∆˹Ҩ ՠ∠P ʹ ∠ARP =∠PAR ∠ARP = 20°
ԨóҠ∆CPR Ҡ∠CPR = 80° ∠CPR =∠PCR ∆CPR 繠∆˹Ҩ ՠ∠R ʹ CR = PR CR = L

(3) ԨóҠ☐AQPR Ҡ∠QPR (˭) = 360°- 60° ∠QPR () = 60°
∵ PQ = PR Р∠QPR = 60° ∆PQR 繠∆ҹ QR = L Р∠PQR =∠PRQ = 60°
∵CR = QR ∆CQR 繠∆˹Ҩ ՠ∠R (= 80°) ʹ ∠QCR =∠CQR = 50° ∠BCQ =∠BQC = 20° ∆BCQ 繠∆˹Ҩ ՠ∠B (= 140°) ʹ BC = BQ

(4) ѧࡵҠ∆BQR∆BCR ¤ѹẺ -- (BQ = BC, BR = BR, QR = CR) ∠BRQ (=∠BRC) = (∠CRQ)/2 = 40° ∠BRP = 20°
ԨóҠ∆BQR Ҡ∠QBR = 70° (∠ABR = 110°) ∠QBR =∠BQR ∆BQR 繠∆˹Ҩ ՠ∠R ʹ BR = QR BR = L
∵ BR = PR ∆BPR 繠∆˹Ҩ ՠ∠R (= 20°) ʹ ∠PBR (=∠BPR) = 80° ∠ABP = x = 30° Q.E.D.




Create Date : 21 Զع¹ 2558
Last Update : 21 Զع¹ 2558 0:00:00 .
Counter : 549 Pageviews.

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Fun Geometry Problem with Solution #141



٨ x = 30°
٨



(1)∠ACB = 120°- 2αР∠ADB = 60°-α

(2) DA ͡ѧش P · AP = AC ∠BAP = 60°
Ҡ∆ABP∆ABC ¤ѹẺ -- (AB = AB,∠BAP =∠BAC, AP = AC) BP = BC
͡ҡ ѧҠ∠APB =∠ACB ∠APB = 120°- 2α

(3) DP ͡ѧش Q ∠BQD = 60°-α ∠BQD =∠BDQ ∆BDQ 繠∆˹Ҩ ՠ∠B ʹ BD = BQ
ԨóҠ∆BPQ Ҡ∠PBQ = 60°-αР∠BPQ = 60°+ 2α
∵ ∠PBQ =∠BQP ∆BPQ 繠∆˹Ҩ ՠ∠P ʹ PQ = BP PQ = BC

(4) ˹ش R ҧҹҢͧ PQ PR = QR = PQ (= BC) ∆PQR 繠∆ҹ ∠QPR = 60°Р∠PQR = 60° ∠BPR = 2αР∠BQR =α
∵ BP = PR ∆BPR 繠∆˹Ҩ ՠ∠P (= 2α) ʹ ∠PBR (=∠BRP) = 90°-α ∠QBR = 30°

(5) ѧࡵҠ∆BCD∆BQR ¤ѹẺ -- (BC = QR,∠CBD =∠BQR, BD = BQ) ∠BDC =∠QBR x = 30° Q.E.D.




Create Date : 18 Զع¹ 2558
Last Update : 18 Զع¹ 2558 0:00:00 .
Counter : 460 Pageviews.

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TIYHz
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