Question-211079

Question Number 211079 by Ari last updated on 27/Aug/24

Answered by A5T last updated on 27/Aug/24

Suppose DB⊥AC A_2 =((DB×AB)/2)=63;A_1 =((EA×AB)/2)=144 (A_2 /A_1 )=((63)/(144))=((DB)/(EA))=(7/(16))⇒(A_3 /(A_1 +A_2 +A_3 ))=((7/(16)))^2 ⇒A_3 =49u^2

$${Suppose}\:{DB}\bot{AC} \\ $$$${A}_{\mathrm{2}} =\frac{{DB}×{AB}}{\mathrm{2}}=\mathrm{63};{A}_{\mathrm{1}} =\frac{{EA}×{AB}}{\mathrm{2}}=\mathrm{144} \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }=\frac{\mathrm{63}}{\mathrm{144}}=\frac{{DB}}{{EA}}=\frac{\mathrm{7}}{\mathrm{16}}\Rightarrow\frac{{A}_{\mathrm{3}} }{{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} }=\left(\frac{\mathrm{7}}{\mathrm{16}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{A}_{\mathrm{3}} =\mathrm{49}{u}^{\mathrm{2}} \\ $$

Commented by Ari last updated on 27/Aug/24

$${but}\:{DB}\:{is}\:{not}\:{given}\:{perpendicular}\:{to}\:{AC} \\ $$$$ \\ $$

Commented by A5T last updated on 27/Aug/24

$${Then}\:{it}\:{cannot}\:{be}\:{unique}\:{or}\:{it}\:{won}'{t}\:{be}\: \\ $$$${dependent}\:{on}\:{it}\:{being}\:{perpendicular} \\ $$

Commented by Ari last updated on 27/Aug/24

$${Ok},{you}\:{are}\:{right}\:! \\ $$

Commented by mm1342 last updated on 27/Aug/24

if △_1 ≈ △_2 ⇒ (s_1 /s_2 )=k^2 ⇒(A_3 /(A_1 +A_2 +A_3 ))=(((DB)/(EA)))^2 =((49)/(256)) ⇒A_3 =49 ✓

$${if}\:\:\:\bigtriangleup_{\mathrm{1}} \approx\:\bigtriangleup_{\mathrm{2}} \:\Rightarrow\:\frac{{s}_{\mathrm{1}} }{{s}_{\mathrm{2}} }={k}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{A}_{\mathrm{3}} }{{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} }=\left(\frac{{DB}}{{EA}}\right)^{\mathrm{2}} =\frac{\mathrm{49}}{\mathrm{256}} \\ $$$$\Rightarrow{A}_{\mathrm{3}} =\mathrm{49}\:\checkmark \\ $$$$ \\ $$

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