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Question Number 206554 by cortano21 last updated on 18/Apr/24

Answered by starsouf last updated on 18/Apr/24

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Answered by mr W last updated on 18/Apr/24

Commented by mr W last updated on 18/Apr/24

S=area of hexagon x+(4x−3)=(S/3)=(2x−1)+(3x−2) 2x−1+B=(S/3) ⇒B=3x−2 3x−2+A=(S/3) ⇒A=2x−1 (4x−3)−x=((ah)/2) ⇒6x−6=ah 2(3x−2)+(4x−3)−2(2x−1)−x=((2ah)/2) ⇒5x−5=ah 6x−6=5x−5 ⇒x=1 ✓

$${S}={area}\:{of}\:{hexagon} \\ $$$${x}+\left(\mathrm{4}{x}−\mathrm{3}\right)=\frac{{S}}{\mathrm{3}}=\left(\mathrm{2}{x}−\mathrm{1}\right)+\left(\mathrm{3}{x}−\mathrm{2}\right) \\ $$$$\mathrm{2}{x}−\mathrm{1}+{B}=\frac{{S}}{\mathrm{3}}\:\Rightarrow{B}=\mathrm{3}{x}−\mathrm{2} \\ $$$$\mathrm{3}{x}−\mathrm{2}+{A}=\frac{{S}}{\mathrm{3}}\:\Rightarrow{A}=\mathrm{2}{x}−\mathrm{1} \\ $$$$\left(\mathrm{4}{x}−\mathrm{3}\right)−{x}=\frac{{ah}}{\mathrm{2}}\:\Rightarrow\mathrm{6}{x}−\mathrm{6}={ah} \\ $$$$\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)+\left(\mathrm{4}{x}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)−{x}=\frac{\mathrm{2}{ah}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}{x}−\mathrm{5}={ah} \\ $$$$\mathrm{6}{x}−\mathrm{6}=\mathrm{5}{x}−\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{1}\:\checkmark \\ $$

Commented by A5T last updated on 16/May/24

2(3x−2)+(4x−3)−2(2x−1)−x=((2ah)/2) I guess this should be: 2(3x−2)+(4x−3)−2(2x−1)−x=2(((2ah)/2))=2ah ⇒5x−5=2ah⇒12x−12=5x−5 ⇒x=1

$$\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)+\left(\mathrm{4}{x}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)−{x}=\frac{\mathrm{2}{ah}}{\mathrm{2}} \\ $$$${I}\:{guess}\:{this}\:{should}\:{be}: \\ $$$$\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)+\left(\mathrm{4}{x}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)−{x}=\mathrm{2}\left(\frac{\mathrm{2}{ah}}{\mathrm{2}}\right)=\mathrm{2}{ah} \\ $$$$\Rightarrow\mathrm{5}{x}−\mathrm{5}=\mathrm{2}{ah}\Rightarrow\mathrm{12}{x}−\mathrm{12}=\mathrm{5}{x}−\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$

Commented by mr W last updated on 16/May/24

i think 2(3x−2)+(4x−3)−2(2x−1)−x=((2ah)/2) is right. base =2a hight =h area =((2a×h)/2)

$${i}\:{think} \\ $$$$\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)+\left(\mathrm{4}{x}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)−{x}=\frac{\mathrm{2}{ah}}{\mathrm{2}} \\ $$$${is}\:{right}. \\ $$$${base}\:=\mathrm{2}{a} \\ $$$${hight}\:={h} \\ $$$${area}\:=\frac{\mathrm{2}{a}×{h}}{\mathrm{2}} \\ $$

Commented by A5T last updated on 16/May/24

I understand that the area is ((2a×h)/2)=ah. But 2(3x−2)+4x−3−2(2x−1) is twice of that. If one applies this method to your latest question, one would get ?=3.

$${I}\:{understand}\:{that}\:{the}\:{area}\:{is}\:\frac{\mathrm{2}{a}×{h}}{\mathrm{2}}={ah}. \\ $$$${But}\:\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)+\mathrm{4}{x}−\mathrm{3}−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)\:{is}\:{twice}\:{of}\:{that}. \\ $$$${If}\:{one}\:{applies}\:{this}\:{method}\:{to}\:{your}\:{latest}\:{question}, \\ $$$${one}\:{would}\:{get}\:?=\mathrm{3}. \\ $$

Commented by mr W last updated on 16/May/24

you are right. it should be 2(3x−2)+(4x−3)−2(2x−1)−x=2×((2ah)/2)

$${you}\:{are}\:{right}.\:{it}\:{should}\:{be} \\ $$$$\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)+\left(\mathrm{4}{x}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)−{x}=\mathrm{2}×\frac{\mathrm{2}{ah}}{\mathrm{2}} \\ $$

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