Pythagoras Theorem and Its Applications

1 Chapter 1 Pythagoras Theoremand Its Pythagoras Theorem and its Pythagoras TheoremThe lengthsa b<cof the sides of a right triangle satisfy the relationa2+b2= Converse TheoremIf the lengths of the sides of a triangles satisfy the relationa2+b2=c2,thenthe triangle contains a right : Euclidean a triangle withBC=a,CA=b,andAB=csatisfy-inga2+b2=c2 . Consider another triangleXY ZwithYZ=a, XZ=b,6 XZY=90 .By the Pythagorean Theorem ,XY2=a2+b2=c2,sothatXY= the triangles4 ABC 4XY Zby the SSS test. This means that6 ACB=6 XZYis a right Dissect two given squares into triangles and quadrilaterals and re-arrange the pieces into a equilateral triangles inside a linesAXandAYare extended to intersectBCandCDrespec-tively atPandQ.

2 Show that(a)AP Qis an equilateral triangle;(b)4AP B+4 ADQ= : Euclidean a triangle with a right angle atC. If the median on the sideais the geometric mean of the sidesbandc, show thatc= (a) Supposec=a+kbfor a right triangle with legsa,b, and <k<1, anda:b:c=1 k2:2k:1+k2.(b) Find two right triangles which are not similar, each satisfyingc=34a+ a triangle with a right angle the geometric mean of the sidesaandb, show that one of the acuteangles is 15 .6. LetABCbe a right triangle with a right angle at a square withPon the hypotenuse, andX,Yon the that the lengthtofasideofthissquareisgivenby1t=1a + + 1/ b = 1/t.

3 1/a^2 + 1/b^2 = 1/ d^ :b:c=12:35:37or12:5:13. Moregenerally,forh k,thereis,uptosimilarity, a unique right triangle satisfyingc=ha+kbprovided(i)h<1 k,or(ii) 22 h=k<1, or(iii)h, k >0,h2+k2= are two such right triangles if0<h<k<1,h2+k2> : Euclidean Geometry47. LetABCbe a right triangle with sidesa,band height of on the hypotenuse, show that1a2+1b2= (Construction of integer right triangles) It is known that every righttriangle of integer sides (without common divisor) can be obtained bychoosing two relatively prime positive integersmandn, one odd, oneeven, and settinga=m2 n2,b=2mn,c=m2+n2.

4 (a) Verify thata2+b2=c2.(b) Complete the following table tofindallsuch right triangles withsides<100:mna=m2 n2b=2mnc=m2+n2(i)21345(ii)32(iii)41(iv)4 3(v)52(vi)54(vii)61(viii)65(ix)72(x)74(x i)76(xii)81(xiii)83(xiv)85(xv)92(xvi)946 57297 YIU: Euclidean Euclid s Proof of Pythagoras Euclid s Application: construction of geometric meanConstruction 1 Given two segments of lengtha<b,markthreepointsP,A,Bon a linesuch thatPA=a,PB=b,andA,Bare on thesameside ofP. Describea semicircle withPBas diameter, and let the perpendicular throughAintersect the semicircle PB, so that the length ofPQis the geometric mean = a, PB = b; PQ^2 = : Euclidean Geometry6 Construction 2 Given two segments of lengtha,b,markthreepointsA,P,Bon a line(PbetweenAandB)suchthatPA=a,PB=b.

5 Describe a semicirclewithABas diameter, and let the perpendicular throughPintersect thesemicircle PB, so that the length ofPQis thegeometric mean ^2 = a(a+b) = a^2 + ab,y ^2 = a^2 + x^ ore, ab = x^ cut a given rectangle of sidesa<binto three pieces that can be re-arranged into a ||==This construction is valid as long asa and Fisher, : Euclidean Geometry7 Exercise1. The midpoint of a chord of length 2ais at a distancedfrom themidpoint of the minor arc it cuts out from the circle. Show that thediameter of the circle isa2+ Two parallel chords of a circle has lengths 168 and 72, and are at adistance 64 apart.

6 Find the radius of the A crescent is formed by intersecting two circular arcs of qual The centralline intersects the arcs at two pointsPandQat a the radius of the Qis a rectangle constructed on the hypotenuse of a right the intersections : The distance from the center to the longer chord is 13. From this, the radiusof the circle is 85. More generally, if these chords has lengths 2aand 2b,andthedistancebetween them isd, the radiusrof the circle is given byr2=[d2+(a b)2][d2+(a+b)2] : Euclidean Geometry8 YXPQACBYXQPACB(a) IfABP Qis a square, show thatXY2=BX AY.

7 (b) IfAB= 2 AQ, show thatAX2+BY2= Construction of regular Equilateral triangle, regular hexagon, and square||==||====||==Given a circle of radiusa,wedenotebyznZnthe length of a side ofan inscribeda circumscribedregularn 3a, Z3=2 3a;z4= 2a, Z4=2a;z6=1,Z6=23 : Euclidean a chord of length 2 in a circleO(2).Cis the midpoint of theminor arcABandMthe midpoint of the that (i)CM=2 3; (ii)BC= 6 thattan 15 =2 3,sin 15 =14( 6 2),cos 15 =14( 6+ 2).YIU: Euclidean The regular pentagon and its The regular pentagonXQPBAQPZYXEDACBS inceXB=XCby symmetry, the isosceles trianglesCABandXCBaresimilar.

8 From this,ACAB=CXCB,andAC CB=AB Division of a segment into the golden ratioSuch a pointXis said to divide the segmentABin thegolden ratio,andcan be constructed as follows.(1) Draw a right triangleABPwithBPperpendicular toABand halfin length.(2) Mark a pointQon the hypotenuseAPsuch thatPQ=PB.(3) Mark a pointXon the segmentABsuch thatAX= the golden ratio, namely,AX:AB= : Euclidean Geometry11 Exercise1. IfXdividesABinto the golden ratio, thenAX:XB= :1,where =12( 5+1) .Show also thatAXAB=12( 5 1) = 1=1 .2. If the legs and the altitude of a right triangle form the sides of anotherright triangle, show that the altitude divides the hypotenuse into thegolden an isosceles triangle with a pointXonABsuch thatAX=CX=BC.

9 Show that(i)6 BAC=36 ;(ii)AX:XB= LetEbe the midpoint of the sideAC. Show thatXE=14q10 + 2 thatcos 36 = 5+14,sin 36 =12q10 2 5,tan 36 =q5 2 an isosceles triangle withAB=AC= a point onABsuch thatAX=CX= the midpoint length ofAD, and deduce thatsin 18 = 5 14,cos 18 =14q10 + 2 5,tan 18 =15q25 10 : Euclidean Construction of a regular pentagon1. Divide a segmentABinto the golden ratio Construct the circlesA(X)andX(B) to intersect Construct a circle centerC,radiusAB,tomeetthetwocirclesA(X) andB(AX) ,ACBEDis a regular Justify the following construction of an inscribed regular The cosine formula and its The cosine formulac2=a2+b2 2abcos.

10 YIU: Euclidean Geometry13 cbacaECABEACB'Exercise1. Show that the (4,5,6) triangle has one angle equal to twice of If =2 , show thatc2=(a+b) Find a simple relation between the sum of the areas of the three squaresS1,S2,S3, and that of the squaresT1,T2, a triangle witha=12,b+c= 18, and cos = + E688, Piz a. Here,b=9 5, andc=9+ : Euclidean Stewart s TheoremIfXis a point on the sideBC(or its extension) such thatBX:XC= : ,thenAX2= b2+ c2 + a2( + ) the cosine formula to compute the cosines of the anglesAXBandAXC,andnotethatcosABC= Apollonius TheoremThe lengthmaof themedianADis given bym2a=14(2b2+2c2 a2).