Question-67386

Question Number 67386 by mr W last updated on 26/Aug/19

Commented by mr W last updated on 26/Aug/19

$${Find}\:{AB}=? \\ $$$${Find}\:{shaded}\:{area}=? \\ $$

Commented by TawaTawa last updated on 26/Aug/19

Hahaha. Am just looking at this. I only know that the area of each circle from lowest to highest are: 1.5cm, 2cm, and 3cm And the length of the rectangle = 6 cm

$$\mathrm{Hahaha}.\:\mathrm{Am}\:\mathrm{just}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{this}. \\ $$$$\mathrm{I}\:\mathrm{only}\:\mathrm{know}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{each}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{lowest}\:\mathrm{to}\:\mathrm{highest}\: \\ $$$$\mathrm{are}:\:\:\:\:\:\mathrm{1}.\mathrm{5cm},\:\:\:\:\:\:\mathrm{2cm},\:\:\:\:\mathrm{and}\:\:\mathrm{3cm} \\ $$$$\mathrm{And}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:=\:\mathrm{6}\:\mathrm{cm} \\ $$

Commented by Cmr 237 last updated on 27/Aug/19

$$\mathrm{6}.\mathrm{5cm} \\ $$

Commented by mr W last updated on 27/Aug/19

$$\mathrm{6}.\mathrm{5}\:{is}\:{wrong}\:{sir}. \\ $$

Commented by Prithwish sen last updated on 27/Aug/19

$$\mathrm{Sir}\:\mathrm{is}\:\mathrm{AB}\backsim\:\mathrm{9}.\mathrm{2}\:\mathrm{cm}\:? \\ $$

Commented by mr W last updated on 27/Aug/19

$${AB}=\mathrm{3}\sqrt{\mathrm{6}},\:{see}\:{solution}\:{from}\:{MJS}\:{sir}. \\ $$

Commented by TawaTawa last updated on 27/Aug/19

$$\mathrm{Please}\:\mathrm{solution}\: \\ $$

Commented by Prithwish sen last updated on 27/Aug/19

$$\mathrm{ok}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

Answered by MJS last updated on 27/Aug/19

(1) (x−(3/2))^2 +(y−(9/2))^2 =(9/4) x^2 +y^2 −3x−9y+((81)/4)=0 (2) (x−p)^2 +(y−2)^2 =4 x^2 +y^2 −2px−4y+p^2 =0 (3) (x−q)^2 +(y−3)^2 =9 x^2 +y^2 −2qx−6y+q^2 =0 (1)∩(2) is unique ⇒ p=(3/2)+(√6) (2)∩(3) is unique ⇒ q=(3/2)+3(√6) ∣AB∣=q−(3/2)=3(√6) ...going to bed now, the area has to wait...

$$\left(\mathrm{1}\right)\:\:\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{9}{y}+\frac{\mathrm{81}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{px}−\mathrm{4}{y}+{p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\left({x}−{q}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{qx}−\mathrm{6}{y}+{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\:\mathrm{is}\:\mathrm{unique}\:\Rightarrow \\ $$$${p}=\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{6}} \\ $$$$\left(\mathrm{2}\right)\cap\left(\mathrm{3}\right)\:\mathrm{is}\:\mathrm{unique}\:\Rightarrow \\ $$$${q}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{6}} \\ $$$$\mid{AB}\mid={q}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{3}\sqrt{\mathrm{6}} \\ $$$$…\mathrm{going}\:\mathrm{to}\:\mathrm{bed}\:\mathrm{now},\:\mathrm{the}\:\mathrm{area}\:\mathrm{has}\:\mathrm{to}\:\mathrm{wait}… \\ $$

Commented by mr W last updated on 27/Aug/19

$${thanks}\:{sir}! \\ $$

Answered by mr W last updated on 27/Aug/19

Commented by mr W last updated on 27/Aug/19

FK=2+2−3=1 FG=2+3=5 ⇒KG=(√(5^2 −1^2 ))=2(√6) FH=2+2−(3/2)=(5/2) FE=2+(3/2)=(7/2) ⇒EH=((√(7^2 −5^2 ))/2)=(√6) AB=EH+KG=(√6)+2(√6)=3(√6) cos α=(5/7)⇒α=cos^(−1) (5/7) cos β=(1/5)⇒β=cos^(−1) (1/5) A_(shade) =((((3/2)+4)(√6))/2)+(((4+3)2(√6))/2)−(1/2)((3/2))^2 (π−α)−((2^2 (α+β))/2)−((3^2 (π−β))/2) =((78(√6)−45π−7α+20β)/8) ≈8.956

$${FK}=\mathrm{2}+\mathrm{2}−\mathrm{3}=\mathrm{1} \\ $$$${FG}=\mathrm{2}+\mathrm{3}=\mathrm{5} \\ $$$$\Rightarrow{KG}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{6}} \\ $$$${FH}=\mathrm{2}+\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${FE}=\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow{EH}=\frac{\sqrt{\mathrm{7}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }}{\mathrm{2}}=\sqrt{\mathrm{6}} \\ $$$${AB}={EH}+{KG}=\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{6}}=\mathrm{3}\sqrt{\mathrm{6}} \\ $$$$ \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{5}}{\mathrm{7}}\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{1}}{\mathrm{5}}\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}} \\ $$$${A}_{{shade}} =\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4}\right)\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\left(\mathrm{4}+\mathrm{3}\right)\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\pi−\alpha\right)−\frac{\mathrm{2}^{\mathrm{2}} \left(\alpha+\beta\right)}{\mathrm{2}}−\frac{\mathrm{3}^{\mathrm{2}} \left(\pi−\beta\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{78}\sqrt{\mathrm{6}}−\mathrm{45}\pi−\mathrm{7}\alpha+\mathrm{20}\beta}{\mathrm{8}} \\ $$$$\approx\mathrm{8}.\mathrm{956} \\ $$

Commented by TawaTawa last updated on 27/Aug/19

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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