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Question Number 178884 by mr W last updated on 22/Oct/22

Commented by mr W last updated on 22/Oct/22

a goat is tied at a point on the foot of a hill which has the shape of a sphere with radius 60m and height 20m as shown. the length of the rope is 20m. find the total area where the goat can graze.

$${a}\:{goat}\:{is}\:{tied}\:{at}\:{a}\:{point}\:{on}\:{the}\:{foot}\:{of} \\ $$$${a}\:{hill}\:{which}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{sphere}\: \\ $$$${with}\:{radius}\:\mathrm{60}{m}\:{and}\:\:{height}\:\mathrm{20}{m}\:{as}\: \\ $$$${shown}.\:{the}\:{length}\:{of}\:{the}\:{rope}\:{is}\:\mathrm{20}{m}. \\ $$$${find}\:{the}\:{total}\:{area}\:{where}\:{the}\:{goat}\: \\ $$$${can}\:{graze}.\: \\ $$

Commented by mr W last updated on 23/Oct/22

give a try sir?

$${give}\:{a}\:{try}\:{sir}? \\ $$

Commented by Rasheed.Sindhi last updated on 23/Oct/22

Sorry sir!

$$\mathrm{Sorry}\:\mathrm{sir}! \\ $$

Answered by mr W last updated on 23/Oct/22

Commented by mr W last updated on 23/Oct/22

a part of the area where the goat can graze in on the ground (A_1 ). the other part is on the hill (sphere surface) (A_2 ).

$${a}\:{part}\:{of}\:{the}\:{area}\:{where}\:{the}\:{goat}\:{can} \\ $$$${graze}\:{in}\:{on}\:{the}\:{ground}\:\left({A}_{\mathrm{1}} \right). \\ $$$${the}\:{other}\:{part}\:{is}\:{on}\:{the}\:{hill}\:\left({sphere}\right. \\ $$$$\left.{surface}\right)\:\left({A}_{\mathrm{2}} \right). \\ $$

Commented by mr W last updated on 23/Oct/22

Commented by aleks041103 last updated on 23/Oct/22

A_1 is comprised of 2 shapes. 1 is half a circle and 2 is some strange shape the rope traces out when the touching point between the rope and the hill moves.

$${A}_{\mathrm{1}} \:{is}\:{comprised}\:{of}\:\mathrm{2}\:{shapes}. \\ $$$$\mathrm{1}\:{is}\:{half}\:{a}\:{circle} \\ $$$${and} \\ $$$$\mathrm{2}\:{is}\:{some}\:{strange}\:{shape}\:{the}\:{rope}\:{traces}\:{out} \\ $$$${when}\:{the}\:{touching}\:{point}\:{between}\:{the}\:{rope} \\ $$$${and}\:{the}\:{hill}\:{moves}. \\ $$

Commented by mr W last updated on 23/Oct/22

cos α=((20)/(2×60))=(1/6) β=π−2α sin β=sin 2α=2×(1/6)×((√(35))/6)=((√(35))/(18)) A_1 =(1/2)(2π−2α)×20^2 −2×((60^2 )/2)×(β−sin β) A_1 =(π−α)×20^2 −60^2 ×(π−2α−((√(35))/(18))) =(π−cos^(−1) (1/6))×20^2 −60^2 ×(π−2 cos^(−1) (1/6)−((√(35))/(18))) =(2×60^2 −20^2 ) cos^(−1) (1/6)−(60^2 −20^2 )π+60^2 ((√(35))/(18)) ≈672.887 m^2

$$\mathrm{cos}\:\alpha=\frac{\mathrm{20}}{\mathrm{2}×\mathrm{60}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\beta=\pi−\mathrm{2}\alpha \\ $$$$\mathrm{sin}\:\beta=\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{6}}×\frac{\sqrt{\mathrm{35}}}{\mathrm{6}}=\frac{\sqrt{\mathrm{35}}}{\mathrm{18}} \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi−\mathrm{2}\alpha\right)×\mathrm{20}^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{60}^{\mathrm{2}} }{\mathrm{2}}×\left(\beta−\mathrm{sin}\:\beta\right) \\ $$$${A}_{\mathrm{1}} =\left(\pi−\alpha\right)×\mathrm{20}^{\mathrm{2}} −\mathrm{60}^{\mathrm{2}} ×\left(\pi−\mathrm{2}\alpha−\frac{\sqrt{\mathrm{35}}}{\mathrm{18}}\right) \\ $$$$\:\:\:=\left(\pi−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}\right)×\mathrm{20}^{\mathrm{2}} −\mathrm{60}^{\mathrm{2}} ×\left(\pi−\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{35}}}{\mathrm{18}}\right) \\ $$$$\:\:\:=\left(\mathrm{2}×\mathrm{60}^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} \right)\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}−\left(\mathrm{60}^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} \right)\pi+\mathrm{60}^{\mathrm{2}} \:\frac{\sqrt{\mathrm{35}}}{\mathrm{18}} \\ $$$$\:\:\:\approx\mathrm{672}.\mathrm{887}\:{m}^{\mathrm{2}} \\ $$

Commented by mr W last updated on 23/Oct/22

Commented by mr W last updated on 23/Oct/22

(R−20)^2 +60^2 =R^2 ⇒R=((20^2 +60^2 )/(40))=100 m AC^(⌢) =AB^(⌢) =AD^(⌢) =20 m=rope length .......

$$\left({R}−\mathrm{20}\right)^{\mathrm{2}} +\mathrm{60}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{20}^{\mathrm{2}} +\mathrm{60}^{\mathrm{2}} }{\mathrm{40}}=\mathrm{100}\:{m} \\ $$$$\overset{\frown} {{AC}}=\overset{\frown} {{AB}}=\overset{\frown} {{AD}}=\mathrm{20}\:{m}={rope}\:{length} \\ $$$$....... \\ $$

Commented by Tawa11 last updated on 23/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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