Question-222646

Question Number 222646 by Tawa11 last updated on 03/Jul/25

Answered by som(math1967) last updated on 03/Jul/25

AB=BC−d, AC=BC+d AC^2 =AB^2 +BC^2 (BC+d)^2 =(BC−d)^2 +BC^2 ⇒(BC+d)^2 −(BC−d)^2 =BC^2 ⇒4BC×d=BC^2 ∴BC=4d ar △ABC=(1/2)×AB×BC =(1/2)×(BC−d)(BC) =(1/2)×(3d)(4d)=6d^2 squnit

$$\:{AB}={BC}−{d},\:{AC}={BC}+{d} \\ $$$$\:\:{AC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} \\ $$$$\left({BC}+{d}\right)^{\mathrm{2}} =\left({BC}−{d}\right)^{\mathrm{2}} +{BC}^{\mathrm{2}} \\ $$$$\Rightarrow\left({BC}+{d}\right)^{\mathrm{2}} −\left({BC}−{d}\right)^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{BC}×{d}={BC}^{\mathrm{2}} \\ $$$$\:\therefore{BC}=\mathrm{4}{d} \\ $$$$\: \\ $$$${ar}\:\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{BC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\left({BC}−{d}\right)\left({BC}\right)\: \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{3}{d}\right)\left(\mathrm{4}{d}\right)=\mathrm{6}{d}^{\mathrm{2}} {squnit} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa11 last updated on 03/Jul/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{part}. \\ $$

Answered by mr W last updated on 04/Jul/25

(a−d)^2 +a^2 =(a+d)^2 ⇒a=4d ⇒sides: 3d, 4d, 5d (C) (i) BD×AC=DA×BC+DC×AB BD×5d=DA×4d+DC×3d ⇒BD=(4/5)×DA+(3/5)×DC (ii) DA=p−r DC=p DB=p+r p+r=(4/5)(p−r)+(3/5)p ⇒r=((2p)/9) (p−r)^2 +p^2 =(5d)^2 [((7/9))^2 +1]p^2 =25d^2 p^2 =((25d^2 )/( 1+((7/9))^2 ))=((405d^2 )/(26)) ΔADC=(((p−r)p)/2)=((7p^2 )/(18))=((315d^2 )/(52))=((7875)/(52))

$$\left({a}−{d}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} =\left({a}+{d}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\mathrm{4}{d} \\ $$$$\Rightarrow{sides}:\:\mathrm{3}{d},\:\mathrm{4}{d},\:\mathrm{5}{d} \\ $$$$\left({C}\right) \\ $$$$\left({i}\right) \\ $$$${BD}×{AC}={DA}×{BC}+{DC}×{AB} \\ $$$${BD}×\mathrm{5}{d}={DA}×\mathrm{4}{d}+{DC}×\mathrm{3}{d} \\ $$$$\Rightarrow{BD}=\frac{\mathrm{4}}{\mathrm{5}}×{DA}+\frac{\mathrm{3}}{\mathrm{5}}×{DC} \\ $$$$\left({ii}\right) \\ $$$${DA}={p}−{r} \\ $$$${DC}={p} \\ $$$${DB}={p}+{r} \\ $$$${p}+{r}=\frac{\mathrm{4}}{\mathrm{5}}\left({p}−{r}\right)+\frac{\mathrm{3}}{\mathrm{5}}{p} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{p}}{\mathrm{9}} \\ $$$$\left({p}−{r}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} =\left(\mathrm{5}{d}\right)^{\mathrm{2}} \\ $$$$\left[\left(\frac{\mathrm{7}}{\mathrm{9}}\right)^{\mathrm{2}} +\mathrm{1}\right]{p}^{\mathrm{2}} =\mathrm{25}{d}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} =\frac{\mathrm{25}{d}^{\mathrm{2}} }{\:\mathrm{1}+\left(\frac{\mathrm{7}}{\mathrm{9}}\right)^{\mathrm{2}} }=\frac{\mathrm{405}{d}^{\mathrm{2}} }{\mathrm{26}} \\ $$$$\Delta{ADC}=\frac{\left({p}−{r}\right){p}}{\mathrm{2}}=\frac{\mathrm{7}{p}^{\mathrm{2}} }{\mathrm{18}}=\frac{\mathrm{315}{d}^{\mathrm{2}} }{\mathrm{52}}=\frac{\mathrm{7875}}{\mathrm{52}} \\ $$

Commented by Tawa11 last updated on 04/Jul/25