Question-67072

Question Number 67072 by TawaTawa last updated on 22/Aug/19

Answered by mr W last updated on 22/Aug/19

Commented by mr W last updated on 22/Aug/19

side length =25+15=40 ((SQ)/(AP))=((QB)/(AB))=(2/(2+3))=(2/5) ⇒SQ=(2/5)×25=10 ⇒QB=(2/(2+3))×40=16 ((RT)/(QB−RT))=((BC)/(SQ))=((40)/(10))=4 ((RT)/(16−RT))=4 5×RT=64 ⇒RT=((64)/5) A_(green) =(1/2)×40×((64)/5)=256 cm^2

$${side}\:{length}\:=\mathrm{25}+\mathrm{15}=\mathrm{40} \\ $$$$\frac{{SQ}}{{AP}}=\frac{{QB}}{{AB}}=\frac{\mathrm{2}}{\mathrm{2}+\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Rightarrow{SQ}=\frac{\mathrm{2}}{\mathrm{5}}×\mathrm{25}=\mathrm{10} \\ $$$$\Rightarrow{QB}=\frac{\mathrm{2}}{\mathrm{2}+\mathrm{3}}×\mathrm{40}=\mathrm{16} \\ $$$$\frac{{RT}}{{QB}−{RT}}=\frac{{BC}}{{SQ}}=\frac{\mathrm{40}}{\mathrm{10}}=\mathrm{4} \\ $$$$\frac{{RT}}{\mathrm{16}−{RT}}=\mathrm{4} \\ $$$$\mathrm{5}×{RT}=\mathrm{64} \\ $$$$\Rightarrow{RT}=\frac{\mathrm{64}}{\mathrm{5}} \\ $$$${A}_{{green}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{40}×\frac{\mathrm{64}}{\mathrm{5}}=\mathrm{256}\:{cm}^{\mathrm{2}} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}.\: \\ $$

Commented by TawaTawa last updated on 22/Aug/19

Sir, please i don′t understand this rule: ((RT)/(QB − RT)) = ((BC)/(SQ)) i understand the rest as similar triangle. Thanks for your time sir

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{this}\:\mathrm{rule}:\:\:\:\frac{\mathrm{RT}}{\mathrm{QB}\:−\:\mathrm{RT}}\:\:=\:\:\frac{\mathrm{BC}}{\mathrm{SQ}} \\ $$$$\mathrm{i}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{as}\:\mathrm{similar}\:\mathrm{triangle}.\:\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 22/Aug/19

Commented by mr W last updated on 22/Aug/19

((RT)/(UR))=((CB)/(SQ)) UR=QB−RT ((RT)/(QB−RT))=((CB)/(SQ)) ((RT)/(16−RT))=((40)/(10))=4 5×RT=64 ⇒RT=((64)/5)

$$\frac{{RT}}{{UR}}=\frac{{CB}}{{SQ}} \\ $$$${UR}={QB}−{RT} \\ $$$$\frac{{RT}}{{QB}−{RT}}=\frac{{CB}}{{SQ}} \\ $$$$\frac{{RT}}{\mathrm{16}−{RT}}=\frac{\mathrm{40}}{\mathrm{10}}=\mathrm{4} \\ $$$$\mathrm{5}×{RT}=\mathrm{64} \\ $$$$\Rightarrow{RT}=\frac{\mathrm{64}}{\mathrm{5}} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

I understand this now sir. very clear, God bless you sir. But what rule is: ((RT)/(UR)) = ((CB)/(SQ))

$$\mathrm{I}\:\mathrm{understand}\:\mathrm{this}\:\mathrm{now}\:\mathrm{sir}.\:\mathrm{very}\:\mathrm{clear},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{what}\:\mathrm{rule}\:\mathrm{is}:\:\:\:\:\:\:\:\frac{\mathrm{RT}}{\mathrm{UR}}\:=\:\frac{\mathrm{CB}}{\mathrm{SQ}} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

I know similar triangle but i dont know when to use the above sir

$$\mathrm{I}\:\mathrm{know}\:\mathrm{similar}\:\mathrm{triangle}\:\mathrm{but}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{when}\:\mathrm{to}\:\mathrm{use}\:\mathrm{the}\:\mathrm{above}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 22/Aug/19

$$\frac{{RT}}{{UR}}=\frac{{RC}}{{RQ}}=\frac{{CB}}{{SQ}} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

I appreciate your time sir. I have new question on the group sir.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{have}\:\mathrm{new}\:\mathrm{question}\:\mathrm{on}\:\mathrm{the}\:\mathrm{group}\:\mathrm{sir}. \\ $$

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