The e of Tingle tkhlid June 1, 015 1 Intodution In this tile we will e disussing the vious methods used fo detemining the e of tingle. Let [X] denote the e of X. Using se nd Height To stt off, the simplest method fo finding the e of tingle is to multiply the se of the tingle y the height to tht se, nd then divide the esult y. Fo exmple, onside the following tingle: h The e of would then e = h. How do we pove this? Well the e of this tingle is simply hlf the e of etngle with the sme sidelength nd height. Let s tke look t the digm elow: 1
h s you n see, we hve two opies of eh of the ight tingles, nd togethe they mke up the etngle. So only one opy of eh would podue hlf the e of the etngle. We know the e of etngle is l w, so the e of the tingle would e l w. ut l is the se nd w is the height. Thus we hve the e is se height. Howeve, we e not lwys given the height of the tingle, nd tht n e polemti if we e still sked to find the e..1 etemining the Height Let s sy we wee only given the lengths of the sides of the tingle. s of this moment, we don t know how to find the e. ut we do know tht the only wy we n find the e is y fist finding the height. Fo this, we e going to hve to ll ou old fiend, the Pythgoen Theoem. h x y onside the following digm: We let the ltitude split into two segments of length x nd y s shown. Using the pythgoen theoem nd the simple segment ddition postulte, we hve (1) x +h = () y +h = (3) x+y =
Sutting (1) fom () nd using diffeene of two sques we get (y x)(y +x) = Plugging in (3) we hve dding this to (3) esults in Using () we get (y x) = y x = y = + y = + h = y h = ( y)(+y) ( h + )( + + ) = ( h ( ) )( (+) ) = h = ( +)(+ )(+ )(++) 4 Thus, tking the sque oot we hve h = ( +)(+ )(+ )(++) 3 Heon s Fomul We mnged to detemine the height of the tingle solely in tems of the side lengths, whih mens we n find the e solely in tems of the side lengths. ut insted of witing out tht nsty long expession ll the time, we n find wy to mke it look muh nie, whih will not only sve time, ut will lso look moe visully ppeling. We define the semipeimete of the tingle to e hlf of the peimete, o ++. Now notie tht s = ++ (s ) = + (s ) = + (s ) = + Pluggin these in to the expession fo the height we get 16s(s )(s )(s ) h = Multiplying y the se nd dividing y esults in = s(s )(s )(s ) [ ] = s(s )(s )(s ) 3
4 e of Tingle With Sine Just like Heon s, this method of omputing tingle e is lso n extemely useful tool. In ft, this omined with Lw of Sines ws essentilly ll tht ws equied fo n IMO question in 007! h α Let s efe to this tingle gin. Let = α. Now ell tht Howeve we hve tht Pluggin this in we get tht [ ] = h sinα = h sinα = h [ ] = sinα 5 e With Indius The inile of tingle is defined s the ile tht is tngent to eh of the sides of the tingle. euse it is tngent to the sides, if we onnet the ente of the ile to the tngeny points, we will otin ight ngles. Hee is digm: 4
I F Now to find the e of in tems of, we n simply dw lines fom I to points,, nd. We will then septely lulte the e of the thee tingles fomed nd dd them. I F Using the simple e of tingle (se nd height definition) we hve the following: nd dding ll of these togethe gives us [ I] = [ I] = [ I] = ( ) ++ [ ] = 5
ut ell tht ++ = s, whee s is the semipeimete. So we hve. [ ] = s 6 e With iumdius O To find the e with the iumdius we e going to hve to use the Lw of Sines. s shown ove, we stt y dwing line though one vetex nd the ente of the ile. We then tke the intesetion point of this line nd onnet it to nothe vetex. This foms 90 ngle, euse the fist line is the dimete of the ile. In ddition, the ngle fomed etween the dimete nd segment is equl to one of the vetex ngles. Looking t we n sy tht sin = R, o sin = R. Plugging in =, we get = R. omputing sin similly fo the othe sides esults in sin = sin = sin = R. Now ell tht the e of this tingle my e witten s sin. Using the Lw of Sines we n mnipulte to find sin. Moe speifilly we hve sin =. Plugging this in will esult in R. [] = 4R 7 Othe Useful Tehniques Sometimes we n find the e of tingle y detemining the tio etween tht e nd given e of nothe tingle. Fo exmple, let s tke look t this digm: 6
Let s sy we wee given tht [] = 40 nd tht = 5, nd we wee sked to find the e of. Rthe 3 thn to use one of ou elie lened methods to find the e, we n simply do the following. Fist note tht oth tingles hve the sme height, so the tio of thei es is in ft just the tio of thei ses! We n now set up popotion [] [] = Thus we hve. 40 [] = 5 3 [] = 4 Fom this ide of tios we n ome up with nifty yet esy fomul. onside the following digm: 7
We hve the following eqution: [] = [] Ty to pove it youself! 8 Polems 1. In = 49 nd the height fom to hs length 0. Find the e of.. Let e the point on side of suh tht. Let = 17, = 5, nd = 8. ompute the e of. 3. In we hve = 13, = 14, nd = 15. () Find the length of the height fom to. () ompute []. 4. In, = 4, = 015, nd = 30. ompute []. 5. In, = 13, = 14, nd = 15. () ompute the inddius of. () ompute the iumdius of the tingle. 6. Points nd e hosen on segments nd espetively of 6 units wy fom point. Given tht = 1 nd = 18, find the tio of the e of qudiltel to tht of. 7. quiltel hs side length 1, nd sques, HI, FG lie outside the tingle. Wht is the e of hexgon FGHI? F G I H (014 M 10 #13) 8. Tpezoid hs ses nd nd digonls inteseting t K. Suppose tht = 9, = 1, nd the e of K is 4. Wht is the e of tpezoid? (008 M 10 #0) 9. In, side nd the pependiul iseto of meet in point, nd isets. If = 9 nd = 7, wht is the e of? (00 M 1 #3) 8
10. onvex qudiltel hs = 9 nd = 1. igonls nd inteset t, = 14, nd nd hve equl es. Wht is? (009 M 1 #0) Thee s unh of e polems, poly t lest one on evey M o IM you n think of, so if this wsn t enough, feel fee to om ops (o nywhee on the intenet fo tht mtte) fo pst omepeitions in ode to quenh you thist! 9