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Question-210608

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Question Number 210608 by peter frank last updated on 13/Aug/24
Commented by peter frank last updated on 13/Aug/24
help (b)
$$\mathrm{help}\:\left(\mathrm{b}\right) \\ $$
Answered by mr W last updated on 13/Aug/24
Commented by mr W last updated on 13/Aug/24
the area of the parallelogram with diagonals a and b is A=((∣a∣∙∣b∣ sin θ)/2) ∣p∣=∣q∣=1 ∣a∣^2 =∣b∣^2 =1^2 +2^2 −2×1×2 cos 150°=5+2(√3) cos θ=((a∙b)/(∣a∣∣b∣)) =(((p+2q)∙(2p+q))/(∣a∣∣b∣)) =((2p∙p+2q∙q+5p∙q)/(∣a∣∣b∣)) =((2+2+5 cos 30°)/((5+2(√3)))) =((8+5(√3))/(2(5+2(√3))))=((9(√3)+10)/(26)) sin θ=((3(√(37−20(√3))))/(26)) ⇒A=(((5+2(√3))×3(√(37−20(√3))))/(52))=0.75 ✓
$${the}\:{area}\:{of}\:{the}\:{parallelogram}\:{with} \\ $$$${diagonals}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{is} \\ $$$${A}=\frac{\mid\boldsymbol{{a}}\mid\centerdot\mid\boldsymbol{{b}}\mid\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\mid\boldsymbol{{p}}\mid=\mid\boldsymbol{{q}}\mid=\mathrm{1} \\ $$$$\mid\boldsymbol{{a}}\mid^{\mathrm{2}} =\mid\boldsymbol{{b}}\mid^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{2}\:\mathrm{cos}\:\mathrm{150}°=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{cos}\:\theta=\frac{\boldsymbol{{a}}\centerdot\boldsymbol{{b}}}{\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\left(\boldsymbol{{p}}+\mathrm{2}\boldsymbol{{q}}\right)\centerdot\left(\mathrm{2}\boldsymbol{{p}}+\boldsymbol{{q}}\right)}{\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\boldsymbol{{p}}\centerdot\boldsymbol{{p}}+\mathrm{2}\boldsymbol{{q}}\centerdot\boldsymbol{{q}}+\mathrm{5}\boldsymbol{{p}}\centerdot\boldsymbol{{q}}}{\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}+\mathrm{2}+\mathrm{5}\:\mathrm{cos}\:\mathrm{30}°}{\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{8}+\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{9}\sqrt{\mathrm{3}}+\mathrm{10}}{\mathrm{26}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{3}\sqrt{\mathrm{37}−\mathrm{20}\sqrt{\mathrm{3}}}}{\mathrm{26}} \\ $$$$\Rightarrow{A}=\frac{\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)×\mathrm{3}\sqrt{\mathrm{37}−\mathrm{20}\sqrt{\mathrm{3}}}}{\mathrm{52}}=\mathrm{0}.\mathrm{75}\:\checkmark \\ $$
Commented by ajfour last updated on 14/Aug/24
(1/2)∣(p+2q)×(q+2p)∣ =(1/2){−(1/2)+4((1/2))}=(3/4)
$$\frac{\mathrm{1}}{\mathrm{2}}\mid\left({p}+\mathrm{2}{q}\right)×\left({q}+\mathrm{2}{p}\right)\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mr W last updated on 14/Aug/24
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Commented by peter frank last updated on 14/Aug/24
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 14/Aug/24
Commented by mr W last updated on 16/Aug/24
the area of the parallelogram with a and b as diagonals is half of the area of the parallelogram with a and b as sides. A=((∣a∣∣b∣ sin θ)/2)=((∣a×b∣)/2) =((∣(p+2q)×(2p+q)∣)/2) =((∣2p×p+4q×p+p×q+2q×q∣)/2) =((3∣q×p∣)/2)=((3×1×1× sin 30°)/2)=(3/4) ✓
$${the}\:{area}\:{of}\:{the}\:{parallelogram}\:{with}\:\boldsymbol{{a}} \\ $$$${and}\:\boldsymbol{{b}}\:{as}\:{diagonals}\:{is}\:{half}\:{of}\:{the}\:{area}\: \\ $$$${of}\:{the}\:{parallelogram}\:{with}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\: \\ $$$${as}\:{sides}. \\ $$$${A}=\frac{\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{b}}\mid\:\mathrm{sin}\:\theta}{\mathrm{2}}=\frac{\mid\boldsymbol{{a}}×\boldsymbol{{b}}\mid}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mid\left(\boldsymbol{{p}}+\mathrm{2}\boldsymbol{{q}}\right)×\left(\mathrm{2}\boldsymbol{{p}}+\boldsymbol{{q}}\right)\mid}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mid\mathrm{2}\boldsymbol{{p}}×\boldsymbol{{p}}+\mathrm{4}\boldsymbol{{q}}×\boldsymbol{{p}}+\boldsymbol{{p}}×\boldsymbol{{q}}+\mathrm{2}\boldsymbol{{q}}×\boldsymbol{{q}}\mid}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{3}\mid\boldsymbol{{q}}×\boldsymbol{{p}}\mid}{\mathrm{2}}=\frac{\mathrm{3}×\mathrm{1}×\mathrm{1}×\:\mathrm{sin}\:\mathrm{30}°}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}}\:\checkmark \\ $$
Commented by peter frank last updated on 14/Aug/24
thank you.
$$\mathrm{thank}\:\mathrm{you}. \\ $$

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