Question-206275

Question Number 206275 by cortano21 last updated on 11/Apr/24

Answered by HeferH24 last updated on 11/Apr/24

CDEF = m ABFE = 3m ((4/6))^2 = (4/9)=((4k)/(9k)) 5k = 4m ((5k)/4)=m 4k+((5k)/4) = ((21k)/4) ((x/4))^2 = ((21k)/4) ∙ (1/(4k)) x = EF = (√(21))

$$\:{CDEF}\:=\:{m} \\ $$$$\:{ABFE}\:=\:\mathrm{3}{m} \\ $$$$\:\left(\frac{\mathrm{4}}{\mathrm{6}}\right)^{\mathrm{2}} =\:\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{4}{k}}{\mathrm{9}{k}} \\ $$$$\:\mathrm{5}{k}\:=\:\mathrm{4}{m} \\ $$$$\:\frac{\mathrm{5}{k}}{\mathrm{4}}={m} \\ $$$$\:\mathrm{4}{k}+\frac{\mathrm{5}{k}}{\mathrm{4}}\:=\:\frac{\mathrm{21}{k}}{\mathrm{4}} \\ $$$$\:\left(\frac{{x}}{\mathrm{4}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{21}{k}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{4}{k}} \\ $$$$\:{x}\:=\:{EF}\:=\:\sqrt{\mathrm{21}} \\ $$$$\: \\ $$

Commented by HeferH24 last updated on 11/Apr/24

Answered by mr W last updated on 11/Apr/24

EF=x [ABCD]=4[EFCD] ((4+6)/2)×h=4×((x+4)/2)×((x−4)/(6−4))×h x^2 =21 ⇒x=(√(21))

$${EF}={x} \\ $$$$\left[{ABCD}\right]=\mathrm{4}\left[{EFCD}\right] \\ $$$$\frac{\mathrm{4}+\mathrm{6}}{\mathrm{2}}×{h}=\mathrm{4}×\frac{{x}+\mathrm{4}}{\mathrm{2}}×\frac{{x}−\mathrm{4}}{\mathrm{6}−\mathrm{4}}×{h} \\ $$$${x}^{\mathrm{2}} =\mathrm{21} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{21}} \\ $$

Answered by cortano21 last updated on 11/Apr/24

$$\:\:\:\cancel{\underbrace{\:}} ^{} \\ $$

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Question-206275

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