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Question-188095

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Question Number 188095 by normans last updated on 25/Feb/23
Commented by infinityaction last updated on 25/Feb/23
20
$$\mathrm{20} \\ $$
Commented by Rasheed.Sindhi last updated on 26/Feb/23
Welcome after long absence infinityaction sir!
$${Welcome}\:{after}\:{long}\:{absence}\: \\ $$$${infinityaction}\:{sir}! \\ $$
Commented by infinityaction last updated on 26/Feb/23
no sir my second id is universe
$${no}\:{sir}\:{my}\:{second}\:{id}\:{is}\:{universe} \\ $$
Commented by Rasheed.Sindhi last updated on 26/Feb/23
�� thank you!
Answered by mr W last updated on 25/Feb/23
Commented by mr W last updated on 25/Feb/23
tan α=(1/3) 2β+(π/2)−2α=π ⇒β=(π/4)+α γ+2((π/2)−α)=π ⇒γ=2α A_(pink) =((12x×2×12x cos β×cos β)/2)=30 (12x)^2 cos^2 β=30 (12x)^2 (cos α−sin α)^2 =60 (12x)^2 ((3/( (√(10))))−(1/( (√(10)))))^2 =60 (12x)^2 =150 ⇒x^2 =((150)/(144))=((25)/(24)) A_(blue) =((4x×8x×sin γ)/2)=16x^2 sin (2α) =16×((25)/(24))×2×(1/( (√(10))))×(3/( (√(10)))) =10 ✓
$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{2}\beta+\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha=\pi \\ $$$$\Rightarrow\beta=\frac{\pi}{\mathrm{4}}+\alpha \\ $$$$\gamma+\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\alpha\right)=\pi \\ $$$$\Rightarrow\gamma=\mathrm{2}\alpha \\ $$$${A}_{{pink}} =\frac{\mathrm{12}{x}×\mathrm{2}×\mathrm{12}{x}\:\mathrm{cos}\:\beta×\mathrm{cos}\:\beta}{\mathrm{2}}=\mathrm{30} \\ $$$$\left(\mathrm{12}{x}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\beta=\mathrm{30} \\ $$$$\left(\mathrm{12}{x}\right)^{\mathrm{2}} \left(\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha\right)^{\mathrm{2}} =\mathrm{60} \\ $$$$\left(\mathrm{12}{x}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}\right)^{\mathrm{2}} =\mathrm{60} \\ $$$$\left(\mathrm{12}{x}\right)^{\mathrm{2}} =\mathrm{150} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\mathrm{150}}{\mathrm{144}}=\frac{\mathrm{25}}{\mathrm{24}} \\ $$$${A}_{{blue}} =\frac{\mathrm{4}{x}×\mathrm{8}{x}×\mathrm{sin}\:\gamma}{\mathrm{2}}=\mathrm{16}{x}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{2}\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{16}×\frac{\mathrm{25}}{\mathrm{24}}×\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{10}\:\checkmark \\ $$
Commented by manxsol last updated on 26/Feb/23
2α=45 tgα=(1/3) α=18.43?
$$\mathrm{2}\alpha=\mathrm{45}\:\:\:\:\:{tg}\alpha=\frac{\mathrm{1}}{\mathrm{3}}\:\alpha=\mathrm{18}.\mathrm{43}? \\ $$
Commented by mr W last updated on 26/Feb/23
look at my comment below!
$${look}\:{at}\:{my}\:{comment}\:{below}! \\ $$
Answered by manxsol last updated on 26/Feb/23
((12xh)/2)=30⇒xh=5 ((8(xh))/2)=((8×5)/2)=20
$$\frac{\mathrm{12}{xh}}{\mathrm{2}}=\mathrm{30}\Rightarrow{xh}=\mathrm{5} \\ $$$$\frac{\mathrm{8}\left({xh}\right)}{\mathrm{2}}=\frac{\mathrm{8}×\mathrm{5}}{\mathrm{2}}=\mathrm{20} \\ $$
Commented by manxsol last updated on 26/Feb/23
$$ \\ $$$$ \\ $$
Commented by mr W last updated on 26/Feb/23
be careful! the diagonal which looks like diagonal and you took as diagonal is in fact not the diagonal! from the proportion of 4x and 8x in the diagram, you can see 8x is not 2 times as long as 4x, that means the diagram is not drawn to scale. when you draw the diagram to scale, it should look like this:
$${be}\:{careful}! \\ $$$${the}\:{diagonal}\:{which}\:{looks}\:{like}\:{diagonal} \\ $$$${and}\:{you}\:{took}\:{as}\:{diagonal}\:{is}\:{in}\:{fact}\: \\ $$$${not}\:{the}\:{diagonal}! \\ $$$${from}\:{the}\:{proportion}\:{of}\:\mathrm{4}{x}\:{and}\:\mathrm{8}{x}\:{in} \\ $$$${the}\:{diagram},\:{you}\:{can}\:{see}\:\mathrm{8}{x}\:{is}\:{not} \\ $$$$\mathrm{2}\:{times}\:{as}\:{long}\:{as}\:\mathrm{4}{x},\:{that}\:{means} \\ $$$${the}\:{diagram}\:{is}\:{not}\:{drawn}\:{to}\:{scale}. \\ $$$${when}\:{you}\:{draw}\:{the}\:{diagram}\:{to}\:{scale}, \\ $$$${it}\:{should}\:{look}\:{like}\:{this}: \\ $$
Commented by mr W last updated on 26/Feb/23
Commented by mr W last updated on 26/Feb/23
therefore the solution is not as easy as ?=((8x)/(12x))×30=20 i was not misguided by the “wrong” diagram, so i selected a “complicated”, but correct way to solve.
$${therefore}\:{the}\:{solution}\:{is}\:{not}\:{as}\:{easy} \\ $$$${as} \\ $$$$?=\frac{\mathrm{8}{x}}{\mathrm{12}{x}}×\mathrm{30}=\mathrm{20} \\ $$$${i}\:{was}\:{not}\:{misguided}\:{by}\:{the}\:“{wrong}'' \\ $$$${diagram},\:{so}\:{i}\:{selected}\:{a}\:“{complicated}'',\: \\ $$$${but}\:{correct}\:{way}\:{to}\:{solve}. \\ $$
Commented by manxsol last updated on 26/Feb/23
Commented by manxsol last updated on 26/Feb/23
12a(2cos^2 −1,−2sinαcosα) (9.6a,−7.2a) v=(12,0)−(9.6,7.2) v=(2.4,7.2) P=C−v P=(12a,12a)−(2.4,7.2) P=(9.6a ;4.8a) h_(blue) =12a−9.6a=2.4a h_(pink) =4.8a (4.8a)(12a)=60 a^2 =((25)/(24)) A_(blue) =(8a)(2.4a)(1/2) A_(blue) =(((48)/5))(((25)/(24)))=10 You are great and generous. Muchas gracias
$$\mathrm{12}{a}\left(\mathrm{2}{cos}^{\mathrm{2}} −\mathrm{1},−\mathrm{2}{sin}\alpha{cos}\alpha\right) \\ $$$$\left(\mathrm{9}.\mathrm{6}{a},−\mathrm{7}.\mathrm{2}{a}\right) \\ $$$${v}=\left(\mathrm{12},\mathrm{0}\right)−\left(\mathrm{9}.\mathrm{6},\mathrm{7}.\mathrm{2}\right) \\ $$$${v}=\left(\mathrm{2}.\mathrm{4},\mathrm{7}.\mathrm{2}\right) \\ $$$${P}={C}−{v} \\ $$$${P}=\left(\mathrm{12}{a},\mathrm{12}{a}\right)−\left(\mathrm{2}.\mathrm{4},\mathrm{7}.\mathrm{2}\right) \\ $$$${P}=\left(\mathrm{9}.\mathrm{6}{a}\:;\mathrm{4}.\mathrm{8}{a}\right) \\ $$$${h}_{{blue}} =\mathrm{12}{a}−\mathrm{9}.\mathrm{6}{a}=\mathrm{2}.\mathrm{4}{a} \\ $$$${h}_{{pink}} =\mathrm{4}.\mathrm{8}{a} \\ $$$$\left(\mathrm{4}.\mathrm{8}{a}\right)\left(\mathrm{12}{a}\right)=\mathrm{60} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{24}} \\ $$$${A}_{{blue}} =\left(\mathrm{8}{a}\right)\left(\mathrm{2}.\mathrm{4}{a}\right)\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}_{{blue}} =\left(\frac{\mathrm{48}}{\mathrm{5}}\right)\left(\frac{\mathrm{25}}{\mathrm{24}}\right)=\mathrm{10} \\ $$$${You}\:{are}\:{great}\:{and}\:{generous}. \\ $$$${Muchas}\:{gracias} \\ $$
Commented by manxsol last updated on 26/Feb/23
all understood. is a clear case of origami
$${all}\:{understood}.\:{is}\:{a}\:{clear} \\ $$$$\:{case}\:{of}\:{origami} \\ $$
Commented by manxsol last updated on 26/Feb/23
$$ \\ $$

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