LESSON 11: TRIANGLE FORMULAE
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1 . THE SEMIPERIMETER OF TRINGLE LESSON : TRINGLE FORMULE In wht follows, will hve sides, nd, nd these will e opposite ngles, nd respetively. y the tringle inequlity, nd..() So ll of, & re positive rel numers. The perimeter of is + + nd its semiperimeter is s. This quntity plys very importnt role in lultions, s we shll presently see. It is esily seen tht s s..() s nd y the tringle inequlities, ll these re positive. In this lesson, we shll lulte three importnt onstnts ssoited with tringle, nmely, its re, the rdius of its insried irle (lled the inirle) nd the rdius of the (unique) irle tht psses through the three verties, tht is, the irumirle of. ll these will e lulted in terms of, nd. The symol R is used to denote rdius of the irumsried irle of.. THE SIN RULE The sin rule for tringle sys tht the rtio of side of tringle to the sin of the ngle opposite it, is the sme, regrdless of the ngle seleted. More preisely, O sin sin sin tully is sys little more, nmely, tht this onstnt numer is tully the dimeter of the irumsried irle of. D
2 Refer to the figure longside. The irumentre is O, nd the onstrution is to drw the dimeter D through. So D = R. Now, dimeter of irle sutends right ngle t the irumferene, so ngle D = 90. So D is right ngled tringle nd sin D. So R. ut ngles nd D re R sin D oth sutended y the sme hord, nmely,. Therefore D nd R = dimeter of sin irumirle. In like mnner, the other two rtios n lso e shown to e equl to the dimeter. So.3 THE OS RULE THE SIN RULE: R.(3) sin sin sin There is lso osine rule. It is rule wherey the osine of ny ngle of tringle n e epressed in terms of the three sides. The onstrution here is to drw the ltitude D. Let = D. Then D =. Using the theorem of Pythgors, we hve the following: D ( ) - D Hene. ut os so os, from whih follows the osine rule: os For eh side of the tringle, there is osine rule: os os os () The sin nd the osine rules hold lso for otuse ngled tringles tringles, lthough our proofs were only for ute ngled tringles.. RE FORMUL (): We re fmilir with the formul tht gives the re of tringle s one hlf the produt of its se nd height. Tht is,
3 h h h, so sin h. Sustitute to get: sin. This gives the re of the tringle, given two sides nd Now sin n inluded ngle. There is nothing speil out ngle ; there re three suh formul; sin sin sin. D sin sin sin, (5).5 RE OF TRINGLE () : HERON S FORMUL Wht is the re of tringle whose three sides re given? Heron s formul nswers this question. HERON S FORMUL: The re of tringle whose sides re, nd is given y: s( s )( s )( s )...(6) where s is its semiperimeter. Proof: From eqution 5, so sin sin ( os ) ( os )( os )...(from eqution ) 3
4 ( )( ) 6 ( ) ( ) 6 ( ) ) ( ) 6 ( )( )( ) ( ) 6 s( s )( s )( s )...from eqution () The result now follows..6 RDII OF THE IRUMSRIED ND INSRIED IRLES.6. The rdius of the irumsried irle Rell tht R Hene sin...from eqution (3) R sin sin...from eqution (5) R=...(7) s( s - )( s - )( s - )
5 .6. The rdius of the insried irle The three isetors of the ngles of tringle meet t one point, whih is usully nmed I. If, from I, perpendiulrs re drwn to the three sides, they ll hve the sme length, sy, r. The irle entred t I nd hving rdius r, touhes ll three sides the sides re tngent to this irle. It is this r tht we shll now lulte. r k I r r If denotes the re of the tringle, we see tht is the sum of the res of three tringles, whih re swept out when I is joined to the three verties. If, nd re tken to e the ses of these tringles, then they ll hve the sme height, nmely, r. We n onlude: r r r r sr Hene sr r s s( s )( s )( s )...(8) s using Heron s formul..7 THE TRINGLE INEQULITY Finlly, the quntities s, s, nd s turn out to e preisely the distne etween the verties nd the points of ontt with the insried irle. More preisely: Let, y nd z e the lengths of the tngents from the verties of the tringle, to the points of ontt. (See digrm elow). The two tngents drwn from point outside the irle hve the sme lengths. From the digrm, the perimeter of the tringle is + y + z. ut the perimeter is lso twie the semiperimeter, tht is s. Hene s = + y + z, nd s = +y + z. From the digrm gin, it is ler tht y y z z y y z z 5
6 It therefore follows tht s - y - s - z s - ( y z) yz 0, y 0, z 0 (9) The lst set of inequlities is just nother wy of writing the tringle inequlities. So our digrm eomes; s- s- s- s- s- s- 6
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