Question-199234

Question Number 199234 by mr W last updated on 30/Oct/23

Commented by mr W last updated on 30/Oct/23

the side length of the squares is 1. find the area of the isosceles right angled triangle.

$${the}\:{side}\:{length}\:{of}\:{the}\:{squares}\:{is}\:\mathrm{1}. \\ $$$${find}\:{the}\:{area}\:{of}\:{the}\:{isosceles}\:{right} \\ $$$${angled}\:{triangle}. \\ $$

Answered by AST last updated on 30/Oct/23

GF=(√(1^2 +2^2 ))=(√5) ((sin45)/1)=((sinAFE)/(AE))=(((2(√5))/( 5))/(AE))⇒AE=(((2(√5))/5)/((√2)/2))=((2(√(10)))/5) sinAFE=sinθ;sinFEA=sinβ sinFEA=sin(135−θ)=sin135cosθ−sinθcos135 ((√(10))/(10))+((√2)/2)×((2(√5))/5)=((3(√(10)))/(10)) sin(DEC)=sin(90−β)=cosβ=(√(1−((90)/(100))))=((√(10))/(10)) ((sin(DEC))/(DC))=((sin90)/3)⇒DC=((3(√(10)))/(10)) EC=(√(ED^2 −CD^2 ))=(√(9−(9/(10))))=((√(810))/(10))=((9(√(10)))/(10)) ⇒AC=((13(√(10)))/(10))⇒area=(1/2)(((13(√(10)))/(10)))^2 =8.45

$${GF}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$$\frac{{sin}\mathrm{45}}{\mathrm{1}}=\frac{{sinAFE}}{{AE}}=\frac{\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\:\mathrm{5}}}{{AE}}\Rightarrow{AE}=\frac{\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{5}} \\ $$$${sinAFE}={sin}\theta;{sinFEA}={sin}\beta \\ $$$${sinFEA}={sin}\left(\mathrm{135}−\theta\right)={sin}\mathrm{135}{cos}\theta−{sin}\theta{cos}\mathrm{135} \\ $$$$\frac{\sqrt{\mathrm{10}}}{\mathrm{10}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$${sin}\left({DEC}\right)={sin}\left(\mathrm{90}−\beta\right)={cos}\beta=\sqrt{\mathrm{1}−\frac{\mathrm{90}}{\mathrm{100}}}=\frac{\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$$\frac{{sin}\left({DEC}\right)}{{DC}}=\frac{{sin}\mathrm{90}}{\mathrm{3}}\Rightarrow{DC}=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$${EC}=\sqrt{{ED}^{\mathrm{2}} −{CD}^{\mathrm{2}} }=\sqrt{\mathrm{9}−\frac{\mathrm{9}}{\mathrm{10}}}=\frac{\sqrt{\mathrm{810}}}{\mathrm{10}}=\frac{\mathrm{9}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$$\Rightarrow{AC}=\frac{\mathrm{13}\sqrt{\mathrm{10}}}{\mathrm{10}}\Rightarrow{area}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{13}\sqrt{\mathrm{10}}}{\mathrm{10}}\right)^{\mathrm{2}} =\mathrm{8}.\mathrm{45} \\ $$

Commented by mr W last updated on 30/Oct/23

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

Commented by AST last updated on 30/Oct/23

Answered by mr W last updated on 30/Oct/23

Commented by mr W last updated on 30/Oct/23

α+β=45° CB=(√2)×DB=(√2)×(√5)=(√(10)) BA=1×(3/( (√(10))))=((3(√(10)))/(10)) CA=(√(10))+((3(√(10)))/(10))=((13(√(10)))/(10)) area=(1/2)×(((13(√(10)))/(10)))^2 =((169)/(20))=8.45

$$\alpha+\beta=\mathrm{45}° \\ $$$${CB}=\sqrt{\mathrm{2}}×{DB}=\sqrt{\mathrm{2}}×\sqrt{\mathrm{5}}=\sqrt{\mathrm{10}} \\ $$$${BA}=\mathrm{1}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$${CA}=\sqrt{\mathrm{10}}+\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}}=\frac{\mathrm{13}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$${area}=\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{13}\sqrt{\mathrm{10}}}{\mathrm{10}}\right)^{\mathrm{2}} =\frac{\mathrm{169}}{\mathrm{20}}=\mathrm{8}.\mathrm{45} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply