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Question-61048




Question Number 61048 by Gulay last updated on 28/May/19
Commented by Gulay last updated on 28/May/19
AB=30    BD=24   Find DC   <B=90  sir plz help me
$$\mathrm{AB}=\mathrm{30}\:\:\:\:\mathrm{BD}=\mathrm{24}\:\:\:\mathrm{Find}\:\mathrm{DC}\:\:\:<\mathrm{B}=\mathrm{90} \\ $$$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$
Answered by mr W last updated on 28/May/19
AD=(√(AB^2 −BD^2 ))=(√(30^2 −24^2 ))=18  ((DC)/(BD))=((BD)/(AD))  ⇒DC=((BD^2 )/(AD))=((24^2 )/(18))=32
$${AD}=\sqrt{{AB}^{\mathrm{2}} −{BD}^{\mathrm{2}} }=\sqrt{\mathrm{30}^{\mathrm{2}} −\mathrm{24}^{\mathrm{2}} }=\mathrm{18} \\ $$$$\frac{{DC}}{{BD}}=\frac{{BD}}{{AD}} \\ $$$$\Rightarrow{DC}=\frac{{BD}^{\mathrm{2}} }{{AD}}=\frac{\mathrm{24}^{\mathrm{2}} }{\mathrm{18}}=\mathrm{32} \\ $$
Commented by kaivan.ahmadi last updated on 28/May/19
if AD^� B=90 this is true
$${if}\:{A}\hat {{D}B}=\mathrm{90}\:{this}\:{is}\:{true}\: \\ $$

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