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Question Number 88886 by I want to learn more last updated on 13/Apr/20

Commented by I want to learn more last updated on 13/Apr/20

ABCD is a square, find AB. Please help

$$\mathrm{ABCD}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{square},\:\:\mathrm{find}\:\:\mathrm{AB}. \\ $$$$\mathrm{Please}\:\mathrm{help} \\ $$

Commented by I want to learn more last updated on 13/Apr/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by john santu last updated on 14/Apr/20

AF is the bisector of the angle DAE

$${AF}\:{is}\:{the}\:{bisector}\:{of}\:{the}\:{angle}\: \\ $$$${DAE} \\ $$

Commented by mr W last updated on 14/Apr/20

x=side length of square α+2β=(π/2) ⇒β=(π/4)−(α/2) tan α=((36)/x) tan β=((64)/x)=((16)/9) tan α ((1−tan (α/2))/(1+tan (α/2)))=((16)/9)×((2 tan (α/2))/(1−tan^2 (α/2))) 1−tan (α/2)=((16)/9)×((2 tan (α/2))/(1−tan (α/2))) 1−2 tan (α/2)+tan^2 (α/2)=((32)/9) tan (α/2) 9−50 tan (α/2)+9 tan^2 (α/2)=0 tan (α/2)=((25−4(√(34)))/9) ⇒tan α=((2×((25−4(√(34)))/9))/(1−(((25−4(√(34)))/9))^2 ))=((9(√(34)))/(136)) x=((36×136)/(9(√(34))))=16(√(34))≈93.295

$${x}={side}\:{length}\:{of}\:{square} \\ $$$$\alpha+\mathrm{2}\beta=\frac{\pi}{\mathrm{2}}\:\Rightarrow\beta=\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{36}}{{x}} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{64}}{{x}}=\frac{\mathrm{16}}{\mathrm{9}}\:\mathrm{tan}\:\alpha \\ $$$$\frac{\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{16}}{\mathrm{9}}×\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}} \\ $$$$\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{16}}{\mathrm{9}}×\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\mathrm{1}−\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}+\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{32}}{\mathrm{9}}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{9}−\mathrm{50}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}+\mathrm{9}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{25}−\mathrm{4}\sqrt{\mathrm{34}}}{\mathrm{9}}\:\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{2}×\frac{\mathrm{25}−\mathrm{4}\sqrt{\mathrm{34}}}{\mathrm{9}}}{\mathrm{1}−\left(\frac{\mathrm{25}−\mathrm{4}\sqrt{\mathrm{34}}}{\mathrm{9}}\right)^{\mathrm{2}} }=\frac{\mathrm{9}\sqrt{\mathrm{34}}}{\mathrm{136}} \\ $$$${x}=\frac{\mathrm{36}×\mathrm{136}}{\mathrm{9}\sqrt{\mathrm{34}}}=\mathrm{16}\sqrt{\mathrm{34}}\approx\mathrm{93}.\mathrm{295} \\ $$

Commented by I want to learn more last updated on 17/Apr/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by john santu last updated on 14/Apr/20

AE = 100 then AB = (√(100^2 −36^2 )) = (√(136×64)) = 8(√(136)) =16(√(34)) ← the answer

$${AE}\:=\:\mathrm{100} \\ $$$${then}\:{AB}\:=\:\sqrt{\mathrm{100}^{\mathrm{2}} −\mathrm{36}^{\mathrm{2}} } \\ $$$$=\:\sqrt{\mathrm{136}×\mathrm{64}}\:=\:\mathrm{8}\sqrt{\mathrm{136}} \\ $$$$=\mathrm{16}\sqrt{\mathrm{34}}\:\leftarrow\:{the}\:{answer}\: \\ $$

Commented by I want to learn more last updated on 17/Apr/20

How is AE 100 sir

$$\mathrm{How}\:\mathrm{is}\:\mathrm{AE}\:\:\mathrm{100}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 18/Apr/20

Commented by mr W last updated on 18/Apr/20

ΔEAF ′ is isosceles. AE=EF ′=36+64=100

$$\Delta{EAF}\:'\:{is}\:{isosceles}. \\ $$$${AE}={EF}\:'=\mathrm{36}+\mathrm{64}=\mathrm{100} \\ $$

Commented by I want to learn more last updated on 19/Apr/20

I appreciate sir.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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