Menu Close

Question-100703




Question Number 100703 by bramlex last updated on 28/Jun/20
Commented by bramlex last updated on 28/Jun/20
(1) 2y^2  = 100−40a+4a^2   (2) 16 = (1/2)y^2  ⇒y^2 = 32  (1)∧(2) ⇒4a^2 −40a+100=64   4a^2 −40a+36=0 ; a^2 −10a+9=0  (a−1)(a−9)=0 → { ((a=1)),((a=9 (rejected))) :}  blue area = (a(√2))^2 = 2a^2  = 2  ratio blue area = 2%
$$\left(\mathrm{1}\right)\:\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{100}−\mathrm{40}{a}+\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{16}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} \:\Rightarrow{y}^{\mathrm{2}} =\:\mathrm{32} \\ $$$$\left(\mathrm{1}\right)\wedge\left(\mathrm{2}\right)\:\Rightarrow\mathrm{4}{a}^{\mathrm{2}} −\mathrm{40}{a}+\mathrm{100}=\mathrm{64}\: \\ $$$$\mathrm{4}{a}^{\mathrm{2}} −\mathrm{40}{a}+\mathrm{36}=\mathrm{0}\:;\:{a}^{\mathrm{2}} −\mathrm{10}{a}+\mathrm{9}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)\left({a}−\mathrm{9}\right)=\mathrm{0}\:\rightarrow\begin{cases}{{a}=\mathrm{1}}\\{{a}=\mathrm{9}\:\left({rejected}\right)}\end{cases} \\ $$$${blue}\:{area}\:=\:\left({a}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}{a}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$${ratio}\:{blue}\:{area}\:=\:\mathrm{2\%} \\ $$
Commented by bemath last updated on 28/Jun/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *