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Question Number 100703 by bramlex last updated on 28/Jun/20

Commented by bramlex last updated on 28/Jun/20
$(1) 2y^2 = 100−40a+4a^2 (2) 16 = (1/2)y^2 ⇒y^2 = 32 (1)∧(2) ⇒4a^2 −40a+100=64 4a^2 −40a+36=0 ; a^2 −10a+9=0 (a−1)(a−9)=0 → { ((a=1)),((a=9 (rejected))) :} blue area = (a(√2))^2 = 2a^2 = 2 ratio blue area = 2%$
$(1) 2 y^{2} = 100 - 40 a + 4 a^{2}$ $(2) 16 = \frac{1}{2} y^{2} \Rightarrow y^{2} = 32$ $(1) \land (2) \Rightarrow 4 a^{2} - 40 a + 100 = 64$ $4 a^{2} - 40 a + 36 = 0; a^{2} - 10 a + 9 = 0$ $(a - 1) (a - 9) = 0 \to {\begin{cases} a = 1 \\ a = 9 (r e j e c t e d) \end{cases}$ $b l u e a r e a = {(a \sqrt{2})}^{2} = 2 a^{2} = 2$ $r a t i o b l u e a r e a = 2 %$
Commented by bemath last updated on 28/Jun/20

$thank you sir$
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