Question-182542

Question Number 182542 by cortano1 last updated on 11/Dec/22

Commented by cortano1 last updated on 11/Dec/22

Radius semi circle PQR = 24 cm radius semi circle BML = 21cm find the area ABCD .

$$\mathrm{Radius}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{PQR}\:=\:\mathrm{24}\:\mathrm{cm} \\ $$$$\mathrm{radius}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{BML}\:=\:\mathrm{21cm} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{ABCD}\:. \\ $$

Answered by mr W last updated on 11/Dec/22

Commented by mr W last updated on 11/Dec/22

((x−r_1 )/r_1 )=(r_1 /(y−r_1 )) ⇒x=((r_1 y)/(y−r_1 )) ...(i) AC=y+(√((x−r_2 )^2 −r_2 ^2 ))=(√(x^2 +y^2 )) y^2 +x(x−2r_2 )+2(√(x(x−2r_2 )))=x^2 +y^2 y(√(x(x−2r_2 )))=r_2 x y^2 (x−2r_2 )=r_2 ^2 x ⇒x=((2r_2 y^2 )/(y^2 −r_2 ^2 )) ...(ii) ((2r_2 y^2 )/(y^2 −r_2 ^2 ))=((r_1 y)/(y−r_1 )) (2r_2 −r_1 )y^2 −2r_1 r_2 y+r_1 r_2 ^2 =0 ⇒y=((r_1 r_2 +r_2 (√(2r_1 (r_1 −r_2 ))))/(2r_2 −r_1 )) =((24×21+21(√(2×24(24−21))))/(2×21−24)) =42 ✓ ⇒x=((24×42)/(42−24))=56 ✓ area of rectangle xy=2352

$$\frac{{x}−{r}_{\mathrm{1}} }{{r}_{\mathrm{1}} }=\frac{{r}_{\mathrm{1}} }{{y}−{r}_{\mathrm{1}} } \\ $$$$\Rightarrow{x}=\frac{{r}_{\mathrm{1}} {y}}{{y}−{r}_{\mathrm{1}} }\:\:\:…\left({i}\right) \\ $$$${AC}={y}+\sqrt{\left({x}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} +{x}\left({x}−\mathrm{2}{r}_{\mathrm{2}} \right)+\mathrm{2}\sqrt{{x}\left({x}−\mathrm{2}{r}_{\mathrm{2}} \right)}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${y}\sqrt{{x}\left({x}−\mathrm{2}{r}_{\mathrm{2}} \right)}={r}_{\mathrm{2}} {x} \\ $$$${y}^{\mathrm{2}} \left({x}−\mathrm{2}{r}_{\mathrm{2}} \right)={r}_{\mathrm{2}} ^{\mathrm{2}} {x} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{r}_{\mathrm{2}} {y}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} }\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{2}{r}_{\mathrm{2}} {y}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{{r}_{\mathrm{1}} {y}}{{y}−{r}_{\mathrm{1}} }\: \\ $$$$\left(\mathrm{2}{r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right){y}^{\mathrm{2}} −\mathrm{2}{r}_{\mathrm{1}} {r}_{\mathrm{2}} {y}+{r}_{\mathrm{1}} {r}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{{r}_{\mathrm{1}} {r}_{\mathrm{2}} +{r}_{\mathrm{2}} \sqrt{\mathrm{2}{r}_{\mathrm{1}} \left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)}}{\mathrm{2}{r}_{\mathrm{2}} −{r}_{\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{24}×\mathrm{21}+\mathrm{21}\sqrt{\mathrm{2}×\mathrm{24}\left(\mathrm{24}−\mathrm{21}\right)}}{\mathrm{2}×\mathrm{21}−\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{42}\:\checkmark \\ $$$$\Rightarrow{x}=\frac{\mathrm{24}×\mathrm{42}}{\mathrm{42}−\mathrm{24}}=\mathrm{56}\:\checkmark \\ $$$${area}\:{of}\:{rectangle}\:{xy}=\mathrm{2352} \\ $$

Commented by cortano1 last updated on 11/Dec/22