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Question-67728




Question Number 67728 by TawaTawa last updated on 30/Aug/19
Commented by mr W last updated on 30/Aug/19
question is not clear.  is ∠BAD=90° as marked or is  ∠BAC=90° as shown?
$${question}\:{is}\:{not}\:{clear}. \\ $$$${is}\:\angle{BAD}=\mathrm{90}°\:{as}\:{marked}\:{or}\:{is} \\ $$$$\angle{BAC}=\mathrm{90}°\:{as}\:{shown}? \\ $$
Commented by TawaTawa last updated on 30/Aug/19
yes sir
$$\mathrm{yes}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 30/Aug/19
∠ BAD = 90
$$\angle\:\mathrm{BAD}\:=\:\mathrm{90} \\ $$
Commented by TawaTawa last updated on 30/Aug/19
That is how the question is. Sir do it correctly sir.  I will study  your work sir
$$\mathrm{That}\:\mathrm{is}\:\mathrm{how}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}.\:\mathrm{Sir}\:\mathrm{do}\:\mathrm{it}\:\mathrm{correctly}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{will}\:\mathrm{study} \\ $$$$\mathrm{your}\:\mathrm{work}\:\mathrm{sir} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 31/Aug/19
Sir,  ∠ BAD = 90
$$\mathrm{Sir},\:\:\angle\:\mathrm{BAD}\:=\:\mathrm{90} \\ $$
Answered by mr W last updated on 31/Aug/19
Commented by mr W last updated on 31/Aug/19
∠BAD=90°  draw AF//ED  ∠ABD=90°−α  ∠BAF=90°−α=∠ABD  BF=AF=FD=(a/2)  ((DC)/(FC))=((DE)/(AF))  ⇒(a/((a/2)+a))=(3/(a/2))  ⇒(2/3)=(6/a)  ⇒a=9  BC=2a=18
$$\angle{BAD}=\mathrm{90}° \\ $$$${draw}\:{AF}//{ED} \\ $$$$\angle{ABD}=\mathrm{90}°−\alpha \\ $$$$\angle{BAF}=\mathrm{90}°−\alpha=\angle{ABD} \\ $$$${BF}={AF}={FD}=\frac{{a}}{\mathrm{2}} \\ $$$$\frac{{DC}}{{FC}}=\frac{{DE}}{{AF}} \\ $$$$\Rightarrow\frac{{a}}{\frac{{a}}{\mathrm{2}}+{a}}=\frac{\mathrm{3}}{\frac{{a}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{6}}{{a}} \\ $$$$\Rightarrow{a}=\mathrm{9} \\ $$$${BC}=\mathrm{2}{a}=\mathrm{18} \\ $$
Commented by TawaTawa last updated on 31/Aug/19
Great.  Really appreciate you sir. God bless you sir.
$$\mathrm{Great}.\:\:\mathrm{Really}\:\mathrm{appreciate}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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