determiner-h-CD-20-AB-30-h1-25-

Question Number 209980 by a.lgnaoui last updated on 27/Jul/24

$$\mathrm{determiner}\:\mathrm{h}\:? \\ $$$$\boldsymbol{\mathrm{CD}}=\mathrm{20}\:\:\:\:\boldsymbol{\mathrm{AB}}=\mathrm{30} \\ $$$$\boldsymbol{\mathrm{h}}\mathrm{1}=\mathrm{25} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 27/Jul/24

Commented by mr W last updated on 27/Jul/24

is CD^(⌢) semi−circle? is AB^(⌢) circular arc?

$${is}\:\overset{\frown} {{CD}}\:{semi}−{circle}? \\ $$$${is}\:\overset{\frown} {{AB}}\:{circular}\:{arc}? \\ $$

Commented by mr W last updated on 27/Jul/24

r=radius of CD^(⌢) R=radius of AB^(⌢) r=((CD)/2)=((20)/2)=10 2R sin^(−1) ((15)/R)=πr R sin^(−1) ((15)/R)=5π ⇒R=30 h=10−(30−(√(30^2 −15^2 )))=15(√3)−20≈5.98

$${r}={radius}\:{of}\:\overset{\frown} {{CD}} \\ $$$${R}={radius}\:{of}\:\overset{\frown} {{AB}} \\ $$$${r}=\frac{{CD}}{\mathrm{2}}=\frac{\mathrm{20}}{\mathrm{2}}=\mathrm{10} \\ $$$$\mathrm{2}{R}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{15}}{{R}}=\pi{r} \\ $$$${R}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{15}}{{R}}=\mathrm{5}\pi\:\Rightarrow{R}=\mathrm{30} \\ $$$${h}=\mathrm{10}−\left(\mathrm{30}−\sqrt{\mathrm{30}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }\right)=\mathrm{15}\sqrt{\mathrm{3}}−\mathrm{20}\approx\mathrm{5}.\mathrm{98} \\ $$

Commented by a.lgnaoui last updated on 28/Jul/24

Please ,What will be the value of h if lignes AB and CD are hyperboliques?

$$\boldsymbol{\mathrm{Please}}\:,\boldsymbol{\mathrm{What}}\:\boldsymbol{\mathrm{will}}\:\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{h}}\:\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{lignes}}\:\boldsymbol{\mathrm{AB}}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{CD}} \\ $$$$\boldsymbol{\mathrm{are}}\:\:\boldsymbol{\mathrm{hyperboliques}}? \\ $$

Commented by a.lgnaoui last updated on 28/Jul/24

$$\mathrm{thank}\:\mathrm{you}\: \\ $$

Commented by mr W last updated on 28/Jul/24

$${if}\:{the}\:{curves}\:{are}\:{parabolas}\:{or} \\ $$$${hyperbolas}\:{there}\:{is}\:{no}\:{solution} \\ $$$${possible}. \\ $$

Commented by Frix last updated on 28/Jul/24

2 similar ellipses are possible. ⇒ AB^(⌢) =CD^(⌢) and we have h_2 =h_1 −((∣AB∣)/2)=10 h=h_1 −((∣CD∣)/2)=15

$$\mathrm{2}\:\mathrm{similar}\:\mathrm{ellipses}\:\mathrm{are}\:\mathrm{possible}. \\ $$$$\Rightarrow\:\:\overset{\frown} {{AB}}=\overset{\frown} {{CD}}\:\:\mathrm{and}\:\mathrm{we}\:\mathrm{have} \\ $$$${h}_{\mathrm{2}} ={h}_{\mathrm{1}} −\frac{\mid{AB}\mid}{\mathrm{2}}=\mathrm{10} \\ $$$${h}={h}_{\mathrm{1}} −\frac{\mid{CD}\mid}{\mathrm{2}}=\mathrm{15} \\ $$

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