MENSURATION. 



MENSURATION Of PLAXC 



Rectilineal plane figures mar he resolved into tri- 

 anglea ; therefore the mensuration of their side* and 



mar be rrfrrml to PUne 'I rignno 



The 



agon nrj. 



of their areas firm* the sabject of this 



PWOSLCM. 



To find the area of a parallelogram. 



> the base and perpendicular at* given. 

 thf AJJT by tkt perpendicular ftigkl, 

 Ike prttimet m It lie the Of fa. 



n.| All, fa rectangle ABCD, be 



each eqaai to the lineal unit by 



v are meswurrd. and let lines be drawn 



of division parallel to the sides : 



le the figure into equal square*, each 



etjual to the eoperfit And since there will be 



a row in the dim linn of either 



aide as there are uniu in that aide, and aa many rows 



as there are units ia the other aide, the whole number 



of aqwarea. or tfce area, wiO be Ike prodwct of the asflav 



ben which express the lineal aiaaiaiai of the sides. 



For example, if the lineal unit be contained four tiaaea 



in one side, and three time* in the other tide, the area 



of the rectangle will be 4 x 3 = U', that is. it will 



contain the niperftcial unit IS times. 



Binaa every paraililqgiaia ia equal to a rectangle 

 having the same base and altitude. (See GIOMI 

 Sect. TV. Prop. I ) iu area w 01 be the product of the 

 base by the perpendicular height. 



I >... A rectangular board U 5 feet 6 inches 

 long aad 9 inche* broad, what U iu area? 



e the base is5/i6i. = (i6 ia., and the perpendi- 

 cular height 9 <*: The area =fti X 9=594*? = 

 ; f = j .9 /. Or, by vulgar fraction*, 

 Sf.Gn if = 'Jf. and 9 



X i = = r 



e by decimal fractions, since 3 /. 6 = 



and 9m =.73/. the area = 5.5 x "' = 4 l*j / / 



red the area of a square ABCD, whose 

 Ida AB U |o^ inches. 



Here I0| x I0| = 10.3 x 105 = 110 ?5 square 

 inches ia the area. 



I < i. Find the area of a paraJleiagram ABCD, whose 

 length AB = 37 fi, and breadth DE - .'.[. or 

 8Mk Here 37 x .5 = lyi U square l..t = 21.50) 

 square yards is the area. 



C.t II. When the two adjacent aides and the an- 

 gle they contain are given. 



RVLR. Kadiut u lo Ike pntJutt of Ike ndet mt the 



the angU lk>j coolnim io ike area. 

 In the parallelogram ABCD draw the perpendicular 



Then rad :.in. A . \|) |. 

 MRTRT); but v AB x 



t TUT, 3d Prop, of 4th Sr vRulel! 



AB x DE U the area, therefore Rad. : Sin. A : AB 

 X AD. area. 



The aide* of a rhomboid are 12 feet 4 inches 

 and 15 feet, and the angle between then is 4* 

 what ia iu area ? 



Bad. 1000OOO 



la, Sin. *T 1 3' 

 * = I.' ,>l 10, 



13 1.17609 



PROBLEM II. 



To find the area of a triangle. 



( iti I. \' !.i-n the bwe and perpendicular are given. 



Kite. Multiply tkf bait by tke perpendicular, and 

 half tke product mill be Ike area. Cto. ' 



The base of a triangle is 250 yards, and the per. 

 pendicular 52 yards 2 feet : Find the area. 



Reducing the feet to the fraction of a yard, we hare 



, f l58 TV- *50 X158 



Jr- */ 



' _ 



aquare yards. 



>. II. \N hi-n the two aides and the included angle 

 of a triangle are given to find the are*. 



RVLK. lindtia u to Mr tint of tke included angle at 

 the product of tht tidet to tmnct tke area. 



This rule follows immediately from Rule 8. for find- 

 ing the area of a parallelocrmm. 



tx. Two sides of a triangle are 14.38 and 12.9 

 chaina, and the included angle i* 7* 20*, what U iu 



l:>> 



10.00000 



U.M 

 189 



1.15776 

 1.11059 



Twice the 

 The area, . 



! - (-.' 



A. R. 



= 3 



7- c*. 

 P. V. 

 81 12.1 



the three aide* of a Uianirle ara 



Cc III When 

 given to in.1 the ar 



Rttr. From kalf tit nm of the tJtrt 

 lie aW amraatjr. Jtf mlttplg tke tmlf m 



mm* U* 



aW ike three 

 rtoi mf 



Ike ImttpnAta mriU be Ime trea. 



Let ABC be a triangle, produce AB one of it* tide*, Pig. 4. 

 and take BD and B I each equal to 



to BC; j,,i., 



and C < and through A draw a line parallel to BC, 

 meeting CD and C J producrd in E and e. The angle 

 AM) will be equal to BCD. (Gao. 21. 1.) which i 

 equal to the angle BDC. (t*.l.) or ADE; therefore 

 AE=AI> ( 13.1.) In like manner, became the angle 

 A e d u equal to the angle BC it. (Sl.l) that u to 

 B rf C (l*.i; or A d t, (41.1). therefore A e=AD 

 (1S.1.) 



From A aa a cmtri>. with AD or AE ai radiua, 

 detcribc a circle meeting AC in F and (i , anil from 

 the ame centre, with A a* or A t a* a radius, dncriba 

 another circle meeting AC in / and *. and take 



From thai dtipotkion of the line* it it manifctt that 

 ( - \( =AB+BC + AC=pertinUr, 



I =l)rf=*BC, 



F^F^+^g^S Brf+2a- A=2 BA. 

 In order to abridge, let p represent half the perime- 

 ter. and consequently * p the whole perimeter ; also 

 let m, 6. c denote the ides opposite to the angles A, B, 

 C respectively ; then it follows that 



( |- p 



CG = Fo'=CF-CO=2/>_* b, 



'- 



The area . . 124.38 jo./rf. 



r>ra A (wrpnidicutar to CD and C d, 



nd I . B a* are equal, tin i. in 



a cirrle. of . -the dia- 



meter, therefore CD ml f* d are 1) II and A, 



(G2) and the angle DC d U a right angle, 



