4.3 Area of a Sector. Area of a Sector Section
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1 ea of a Secto Section ea of a Secto In geomety you leaned that the aea of a cicle of adius is π 2. We will now lean how to find the aea of a secto of a cicle. secto is the egion bounded by a cental angle and its intecepted ac, such as the shaded egion in Figue Let be a cental angle in a cicle of adius and let be the aea of its secto. Simila to ac length, the atio of to the aea of the entie cicle is the same as the atio of to one evolution. In othe wods, again using adian measue, aea of secto aea of entie cicle = secto angle one evolution Solving fo in the above equation, we get the following fomula: π 2 = 2π. Figue 4..1 In a cicle of adius, the aea of the secto inside a cental angle is whee is measued in adians. Example 4.8 = 1 2 2, (4.) Find the aea of a secto whose angle is π ad in a cicle of adius 4 cm. Solution: Using = π and =4 in fomula (4.), the aea of the secto is Example 4.9 = = 1 2 (4)2 π = 8π cm2. Find the aea of a secto whose angle is 11 in a cicle of adius. m. Solution: s with ac length, we have to make sue that the angle is measued in adians o else the answe will be way off. So conveting =11 to adians and using =. in fomula (4.) fo the aea of the secto, we get = 11 = π = ad = = 1 2 (.)2 (2.042) = 12.1 m 2. Fo a secto whose angle is in a cicle of adius, the length of the ac cut off by that angle is s=. Thus, by fomula (4.) the aea of the secto can be witten as: = 1 2 s (4.6) Note: The cental angle that intecepts an ac is sometimes called the angle subtended by the ac. In some texts this fomula is taken as a esult fom elementay geomety and then used to pove fomula (4.).
2 96 hapte 4 Radian Measue 4. Example 4.10 Find the aea of a secto whose ac is 6 cm in a cicle of adius 9 cm. Solution: Using s=6 and =9 in fomula (4.6) fo the aea, we get = 1 2 s = = 1 2 (9)(6) = 2 cm2. Note that the angle subtended by the ac is = s = 2 ad. Example 4.11 Find the aea K inside the belt pulley system fom Example 4. in Section 4.2. Solution: Recallthatthebeltpulleyshaveadiiofcmand8cm, andtheicentesae1cmapat. We showed in Example 4. that EF = =6 6, DE =1. ad, and GF =1. ad. We see fom Figue 4..2 that, by symmety, the total aea K enclosed by the belt is twice the aea above the line DG, that is, K = 2((ea of secto DE) + (ea of ectangle EF) + (ea of tiangle ) + (ea of secto GF)). F 6 6 E D 1 G Figue 4..2 elt pulleys with adii cm and 8 cm Since EF is a ectangle with sides and 6 6, its aea is 0 6. nd since is a ight tiangle whose legs have lengths and 6 6, its aea is 1 2 ()(6 6)=9 6. Thus, using fomula (4.) fo the aeas of sectos DE and GF, we have ( K = 2 (ea of secto DE) ) 6 + (ea of secto GF) ( = ()2 (1.) (1.)) (8)2 = 8.9 cm 2.
3 ea of a Secto Section 4. 9 So fa we have dealt with the aea cut off by a cental angle. How would you find the aea of a egion cut off by an inscibed angle, such a as the shaded egion in Figue 4..? In this pictue, the cente of the cicle is inside the inscibed angle, and the lengths a and b of the two chods ae given, as is the adius of the cicle. Dawing line segments b fom the cente of the cicle to the endpoints of the chods indicates how to solve this poblem: add up the aeas of the two tiangles and the secto fomed by the cental angle. The aeas and angles of the two Figue 4.. tiangles can be detemined (since all thee sides ae known) using methods fom hapte 2. lso, ecall (Theoem 2.4 in Section 2.) that a cental angle has twice the measueof any inscibed angle which intecepts the same ac. In the execises you will be asked to solve poblems like this (including the cases whee the cente of the cicle is outside o on the inscibed angle). nothe type of egion we can conside is a segment of a cicle, whichistheegionbetweenachodandtheacitcutsoff. InFigue 4..4 the segment fomed by the chod is the shaded egion between the ac and the tiangle O. y fomula (2.2) in Section 2.4 fo the aea of a tiangle given two sides and thei included angle, we know that O aea of O = 1 2 ()()sin = sin. Thus, since the aea K of the segment is the aea of the secto O minus the aea of the tiangle O, we have Figue 4..4 aea K of segment = sin = ( sin ). (4.) Note that as a consequence of fomula (4.) we must have >sin fo 0< π (measued in adians), since the aea of a segment is positive fo those angles. Example 4.12 Find the aea of the segment fomed by a chod of length in a cicle of adius 2. Solution: Figue 4.. shows the segment fomed by a chod of length in a cicle of adius =2. We can use the Law of osines to find the subtended cental angle : 2 2 cos = (2)(2) = 0.12 = ad Thus, by fomula (4.) the aea K of the segment is: Figue 4.. K = ( sin ) = 1 2 (2)2 (1.696 sin 1.696) = 1.408
4 98 hapte 4 Radian Measue 4. Example 4.1 The centes of two cicles ae cm apat, with one cicle having a adius of cm and the othe a adius of 4 cm. Find the aea K of thei intesection. Solution: In Figue 4..6(a), we see that the intesection of the two cicles is the union of the segmentsfomedbythechodd ineachcicle. Thus,oncewedeteminetheangles D and D we can calculate the aea of each segment and add those aeas togethe to get K. 4 4 D α β (a) = 1 2 D, = 1 2 D (b) Tiangle Figue 4..6 y symmety, we see that = 1 2 D and = 1 2 D. So let α= and β=, as in Figue 4..6(b). y the Law of osines, we have cos α = ()() cos β = ()(4) = α = 0.94 ad D = 2(0.94)=1.188 ad = 0.14 β = 0. ad D = 2(0.)=1.0 ad Thus, the aea K is K = (ea of segment D in cicle at ) + (ea of segment D in cicle at ) = 1 2 ()2 (1.188 sin 1.188) (4)2 (1.0 sin 1.0) =.66 cm 2. Execises Fo Execises 1-, find the aea of the secto fo the given angle and adius. 1. =2.1 ad, =1.2 cm 2. = π ad, =. ft. =8, =6 m 4. The centes of two belt pulleys, with adii of cm and 6 cm, espectively, ae 1 cm apat. Find the total aea K enclosed by the belt.. In Execise 4 suppose that both belt pulleys have the same adius of 6 cm. Find the total aea K enclosed by the belt.
5 ea of a Secto Section Find the aea enclosed by the figue eight in Execise 8 fom Section 4.2. Fo Execises -9, find the aea of the secto fo the given adius and ac length s.. = cm, s=2 cm 8. =a, s=a 9. =1 cm, s=π cm Fo Execises 10-12, find the aea of the segment fomed by a chod of length a in a cicle of adius. 10. a=4 cm, =4 cm 11. a=1 cm, = cm 12. a=2 cm, = cm 1. Find the aea of the shaded egion in Figue Figue 4.. Execise 1 Figue 4..8 Execise 14 Figue 4..9 Execise Find the aea of the shaded egion in Figue (Hint: Daw two cental angles.) 1. Find the aea of the shaded egion in Figue The centes of two cicles ae 4 cm apat, with one cicle having a adius of cm and the othe a adius of 2 cm. Find the aea of thei intesection. 1. Thee cicles with adii of 4 m, 2 m, and 1 m ae extenally tangent to each othe. Find the aea of the cuved egion between the cicles, as in Figue (Hint: onnect the centes of the cicles.) Figue Execise 1 Figue Execise Showthatthetotalaeaenclosedbytheloopaoundthetheeciclesofadius infigue4..11 is (π+6+ ) Fo a fixed cental angle, how much does the aea of its secto incease when the adius of the cicle is doubled? How much does the length of its intecepted ac incease?
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