ABCD is a trapezium which AB is parallel to DC. DC = 64 cm, angle BAD = 45 and angle BDC = 38. Calculate correct to 1 decimal place: a) The length of CB. b) The length of AB. c) The perpendicular distance of D from AB. d) The radius of a circle that passes through the point BC and D.
|Mathematics
ABCD is a trapezium which AB is parallel to DC. DC = 64 cm, angle BAD = 45 and angle BDC = 38. Calculate correct to 1 decimal place: a) The length of CB. b) The length of AB. c) The perpendicular distance of D from AB. d) The radius of a circle that passes through the point BC and D.
ScanSolve AI Answer
Answer
55.8 cm
Haya, twende tukasuluhishe swali hili hatua kwa hatua.
a) Urefu wa CB.
Tunatumia sheria ya Kosine katika pembetatu BCD.
Tunajua BD=10.9 cm, DC=64 cm, na ∠BDC=38∘.
Step 1: Tumia sheria ya Kosine kupata CB.
CB2=BD2+DC2−2⋅BD⋅DC⋅cos(∠BDC)CB2=(10.9)2+(64)2−2⋅(10.9)⋅(64)⋅cos(38∘)CB2=118.81+4096−1395.2⋅(0.78801)CB2=4214.81−1099.50766CB2=3115.30234
Step 2: Pata thamani ya CB.
CB=3115.30234CB≈55.8149 cm
Ikiwa imerekebishwa kwa sehemu moja ya desimali:
CB≈55.8 cm
b) Urefu wa AB.
Kwa kuwa ABCD ni trapezi, AB ni sambamba na DC. Hii inamaanisha ∠ABD=∠BDC (pembe mbadala za ndani).
Step 1: Pata ∠ABD na ∠ADB.
∠ABD=∠BDC=38∘.
Katika pembetatu ABD, jumla ya pembe ni 180∘.
∠ADB=180∘−∠BAD−∠ABD∠ADB=180∘−45∘−38∘∠ADB=97∘
Step 2: Tumia sheria ya Saini katika pembetatu ABD kupata AB.
sin(∠ADB)AB=sin(∠BAD)BDsin(97∘)AB=sin(45∘)10.9AB=sin(45∘)10.9⋅sin(97∘)AB=0.7071110.9⋅0.99255AB=0.7071110.818795AB≈15.2999 cm
Ikiwa imerekebishwa kwa sehemu moja ya desimali:
AB≈15.3 cm
c) Umbali wa wima wa D kutoka AB.
Huu ni urefu wa pembetatu ABD kutoka D hadi AB.
Step 1: Tumia sheria ya Saini katika pembetatu ABD kupata AD.
sin(∠ABD)AD=sin(∠BAD)BDsin(38∘)AD=sin(45∘)10.9AD=sin(45∘)10.9⋅sin(38∘)AD=0.7071110.9⋅0.61566AD=0.707116.709794AD≈9.4889 cm
Step 2: Pata umbali wa wima h.
Umbali wa wima h kutoka D hadi AB ni AD⋅sin(∠BAD).
h=AD⋅sin(45∘)h=9.4889⋅0.70711h≈6.7099 cm
Ikiwa imerekebishwa kwa sehemu moja ya desimali:
h≈6.7 cm
d) Radiasi ya duara inayopita kwenye pointi B, C, na D.
Hii ni radiasi ya duara inayozunguka pembetatu BCD (circumradius).
Step 1: Pata pembe za pembetatu BCD.
Tunajua BD=10.9 cm, DC=64 cm, CB≈55.8149 cm, na ∠BDC=38∘.
Tumia sheria ya Saini kupata ∠CBD:
sin(∠CBD)DC=sin(∠BDC)CBsin(∠CBD)=CBDC⋅sin(∠BDC)sin(∠CBD)=55.814964⋅sin(38∘)sin(∠CBD)=55.814964⋅0.61566sin(∠CBD)=55.814939.40224≈0.70593
Kuna uwezekano wa pembe mbili: ∠CBD1=arcsin(0.70593)≈44.88∘ au ∠CBD2=180∘−44.88∘=135.12∘.
Ili kuhakikisha, tunachunguza jumla ya pembe katika pembetatu. Ikiwa ∠CBD=44.88∘, basi ∠BCD=180∘−38∘−44.88∘=97.12∘. Hii inamaanisha upande mfupi zaidi BD (10.9 cm) ungekuwa kinyume na pembe kubwa zaidi (97.12∘), ambayo si sahihi.
Kwa hivyo, ∠CBD=135.12∘.
Sasa, ∠BCD=180∘−38∘−135.12∘=6.88∘.
Step 2: Tumia fomula ya radiasi ya duara inayozunguka pembetatu (circumradius) R.
R=2sin(∠BDC)CBR=2⋅sin(38∘)55.8149R=2⋅0.6156655.8149R=1.2313255.8149R≈45.328 cm
Ikiwa imerekebishwa kwa sehemu moja ya desimali:
R≈45.3 cm
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Haya, twende tukasuluhishe swali hili hatua kwa hatua. a) Urefu wa CB.
ABCD is a trapezium which AB is parallel to DC. DC = 64 cm, angle BAD = 45 and angle BDC = 38. Calculate correct to 1 decimal place: a) The length of CB. b) The length of AB. c) The perpendicular distance of D from AB. d) The radius of a circle that passes through the point BC and D.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Haya, twende tukasuluhishe swali hili hatua kwa hatua. a) Urefu wa CB. Tunatumia sheria ya Kosine katika pembetatu BCD. Tunajua BD = 10.9 cm, DC = 64 cm, na BDC = 38^. Step 1: Tumia sheria ya Kosine kupata CB. CB^2 = BD^2 + DC^2 - 2 · BD · DC · ( BDC) CB^2 = (10.9)^2 + (64)^2 - 2 · (10.9) · (64) · (38^) CB^2 = 118.81 + 4096 - 1395.2 · (0.78801) CB^2 = 4214.81 - 1099.50766 CB^2 = 3115.30234 Step 2: Pata thamani ya CB. CB = sqrt(3115.30234) CB ≈ 55.8149 cm Ikiwa imerekebishwa kwa sehemu moja ya desimali: CB ≈ 55.8 cm b) Urefu wa AB. Kwa kuwa ABCD ni trapezi, AB ni sambamba na DC. Hii inamaanisha ABD = BDC (pembe mbadala za ndani). Step 1: Pata ABD na ADB. ABD = BDC = 38^. Katika pembetatu ABD, jumla ya pembe ni 180^. ADB = 180^ - BAD - ABD ADB = 180^ - 45^ - 38^ ADB = 97^ Step 2: Tumia sheria ya Saini katika pembetatu ABD kupata AB. (AB)/(( ADB)) = (BD)/(( BAD)) (AB)/((97^)) = (10.9)/((45^)) AB = (10.9 · (97^))/((45^)) AB = (10.9 · 0.99255)/(0.70711) AB = (10.818795)/(0.70711) AB ≈ 15.2999 cm Ikiwa imerekebishwa kwa sehemu moja ya desimali: AB ≈ 15.3 cm c) Umbali wa wima wa D kutoka AB. Huu ni urefu wa pembetatu ABD kutoka D hadi AB. Step 1: Tumia sheria ya Saini katika pembetatu ABD kupata AD. (AD)/(( ABD)) = (BD)/(( BAD)) (AD)/((38^)) = (10.9)/((45^)) AD = (10.9 · (38^))/((45^)) AD = (10.9 · 0.61566)/(0.70711) AD = (6.709794)/(0.70711) AD ≈ 9.4889 cm Step 2: Pata umbali wa wima h. Umbali wa wima h kutoka D hadi AB ni AD · ( BAD). h = AD · (45^) h = 9.4889 · 0.70711 h ≈ 6.7099 cm Ikiwa imerekebishwa kwa sehemu moja ya desimali: h ≈ 6.7 cm d) Radiasi ya duara inayopita kwenye pointi B, C, na D. Hii ni radiasi ya duara inayozunguka pembetatu BCD (circumradius). Step 1: Pata pembe za pembetatu BCD. Tunajua BD = 10.9 cm, DC = 64 cm, CB ≈ 55.8149 cm, na BDC = 38^. Tumia sheria ya Saini kupata CBD: (DC)/(( CBD)) = (CB)/(( BDC)) ( CBD) = (DC · ( BDC))/(CB) ( CBD) = (64 · (38^))/(55.8149) ( CBD) = (64 · 0.61566)/(55.8149) ( CBD) = (39.40224)/(55.8149) ≈ 0.70593 Kuna uwezekano wa pembe mbili: CBD_1 = (0.70593) ≈ 44.88^ au CBD_2 = 180^ - 44.88^ = 135.12^. Ili kuhakikisha, tunachunguza jumla ya pembe katika pembetatu. Ikiwa CBD = 44.88^, basi BCD = 180^ - 38^ - 44.88^ = 97.12^. Hii inamaanisha upande mfupi zaidi BD (10.9 cm) ungekuwa kinyume na pembe kubwa zaidi (97.12^), ambayo si sahihi. Kwa hivyo, CBD = 135.12^. Sasa, BCD = 180^ - 38^ - 135.12^ = 6.88^. Step 2: Tumia fomula ya radiasi ya duara inayozunguka pembetatu (circumradius) R. R = (CB)/(2 ( BDC)) R = (55.8149)/(2 · (38^)) R = (55.8149)/(2 · 0.61566) R = (55.8149)/(1.23132) R ≈ 45.328 cm Ikiwa imerekebishwa kwa sehemu moja ya desimali: R ≈ 45.3 cm That's 2 down. 3 left today — send the next one.