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Question-187589




Question Number 187589 by Mingma last updated on 19/Feb/23
Commented by Mingma last updated on 19/Feb/23
Blue area=?
Answered by cortano12 last updated on 19/Feb/23
cos α=(((5(√2))^2 +(5(√5))^2 −5^2 )/(2.(5(√2))(5(√5))))   cos α=((50+125−25)/(50(√(10))))=(3/( (√(10))))   sin α=(1/( (√(10))))   ⇒cos β=cos β ; (x/(10))=((10)/(5(√5)))⇒x=4(√5)   blue area= (1/2).5(√2).4(√5) .(1/( (√(10))))=10
$$\mathrm{cos}\:\alpha=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{5}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{2}.\left(\mathrm{5}\sqrt{\mathrm{2}}\right)\left(\mathrm{5}\sqrt{\mathrm{5}}\right)} \\ $$$$\:\mathrm{cos}\:\alpha=\frac{\mathrm{50}+\mathrm{125}−\mathrm{25}}{\mathrm{50}\sqrt{\mathrm{10}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$$\:\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}} \\ $$$$\:\Rightarrow\mathrm{cos}\:\beta=\mathrm{cos}\:\beta\:;\:\frac{{x}}{\mathrm{10}}=\frac{\mathrm{10}}{\mathrm{5}\sqrt{\mathrm{5}}}\Rightarrow{x}=\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\:{blue}\:{area}=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{5}\sqrt{\mathrm{2}}.\mathrm{4}\sqrt{\mathrm{5}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}=\mathrm{10} \\ $$$$\: \\ $$
Commented by Mingma last updated on 19/Feb/23
Good!
Answered by mr W last updated on 19/Feb/23
blue =(5/2)×(5−((5/(5(√5))))^2 ×5)=10 ✓
$${blue}\:=\frac{\mathrm{5}}{\mathrm{2}}×\left(\mathrm{5}−\left(\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} ×\mathrm{5}\right)=\mathrm{10}\:\checkmark \\ $$
Commented by Mingma last updated on 19/Feb/23
Good!
Answered by HeferH last updated on 19/Feb/23
Commented by HeferH last updated on 19/Feb/23
(h/5) = (5/(5(√5)))   h = (5/( (√5))) = (√5);    10^2 = 5(√5)∙x   x = ((100)/(5(√5))) = ((20(√5))/5) = 4(√5)   A = ((h∙x)/2) = ((4(√5)∙(√5))/2) = 10u^2
$$\frac{{h}}{\mathrm{5}}\:=\:\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{5}}} \\ $$$$\:{h}\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{5}}}\:=\:\sqrt{\mathrm{5}};\: \\ $$$$\:\mathrm{10}^{\mathrm{2}} =\:\mathrm{5}\sqrt{\mathrm{5}}\centerdot{x} \\ $$$$\:{x}\:=\:\frac{\mathrm{100}}{\mathrm{5}\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{20}\sqrt{\mathrm{5}}}{\mathrm{5}}\:=\:\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\:{A}\:=\:\frac{{h}\centerdot{x}}{\mathrm{2}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{5}}\centerdot\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\mathrm{10}{u}^{\mathrm{2}} \\ $$

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