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Question Number 209883 by Ismoiljon_008 last updated on 24/Jul/24

Commented by Ismoiljon_008 last updated on 24/Jul/24

   help , please

$$\:\:\:{help}\:,\:{please} \\ $$

Commented by Frix last updated on 25/Jul/24

I get x=18  AB=BD=CD=6(√5)  BE=12(√2)  CE=12

$$\mathrm{I}\:\mathrm{get}\:{x}=\mathrm{18} \\ $$$${AB}={BD}={CD}=\mathrm{6}\sqrt{\mathrm{5}} \\ $$$${BE}=\mathrm{12}\sqrt{\mathrm{2}} \\ $$$${CE}=\mathrm{12} \\ $$

Answered by mr W last updated on 25/Jul/24

Commented by mr W last updated on 25/Jul/24

a=6(√5)  x+12=(√5)a=30  ⇒x=18

$${a}=\mathrm{6}\sqrt{\mathrm{5}} \\ $$$${x}+\mathrm{12}=\sqrt{\mathrm{5}}{a}=\mathrm{30} \\ $$$$\Rightarrow{x}=\mathrm{18} \\ $$

Commented by Ismoiljon_008 last updated on 25/Jul/24

   Sorry, please interpret me that how     ∠BAD = 45^o

$$\:\:\:{Sorry},\:{please}\:{interpret}\:{me}\:{that}\:{how} \\ $$$$\:\:\:\angle{BAD}\:=\:\mathrm{45}^{{o}} \: \\ $$

Commented by mr W last updated on 25/Jul/24

ABDE are cyclic, see below.

$${ABDE}\:{are}\:{cyclic},\:{see}\:{below}. \\ $$

Commented by mr W last updated on 25/Jul/24

Commented by Ismoiljon_008 last updated on 25/Jul/24

  ok, I get it. Thank you very much

$$\:\:{ok},\:{I}\:{get}\:{it}.\:{Thank}\:{you}\:{very}\:{much} \\ $$

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