**GEOMETRY & MENSURATIONConcepts To be rememberedThe lines are parallel and a transversal cuts these lines.

If Angle 1 = , Angle2 = [180 ]

**GEOMETRY & MENSURATIONConcepts To be rememberedLines AB & CD are parallel. E is a point such that BAE = 450 and ECD = 300. Find AEC Draw a line GF passing through E parallel to AB

FG450300 AEG = 450

GED = 300 AEC = 750

**GEOMETRY & MENSURATIONConcepts To be rememberedLines AB & CD are parallel. AE and CE are internal angular bisectors . Find AECLet FAB = 2x0 ACD = 2x0. BAC = 2y02x+2y = 1800 CAE = y0 ACE = x0X + Y = 90From AEC, AEC = 900

2x

What is the sum of all internalangles in the figure?

There are 8 triangles and hence the sum of all internal angles is 1440.**GEOMETRY & MENSURATIONConcepts To be rememberedWhat is theSum Of all internalangles inFigureA,B,C & D?A = 720B = 540 C = 360 D = 900

**Sum of all internal angles = (n-2)

Sum of all External angles= 2

In a regular n sided figure, Each internal angle = (n-2) /nEach external angle = 2 /n

GEOMETRY & MENSURATION

NSum of all internal AnglesAverage 318060436090554010867201207900900/78108013591260140101440144

In a polygon the measure (in degrees) of each angle is a distinct integer. If the largest angle is 1450 and the polygon has maximum number of sides possible, then its largest exterior angle is (1)500 (2) 470 (3) 450 (4) 520 (5) 480In the above question what is the number of sides of Polygon?9 (2) 7 (3) 10 (4) 8 (5) 11*GEOMETRY & MENSURATIONIf largest angle is 1450, let us find the maximum number of sides of the polygon. The sum of all internal angles is (n-2) . Let us assume it is a decagon. Total sum is 1405.Hence it cannot be decagon.Let us assume it is a nonagon. The total sum of internal angles is 1260.The maximum number of sides is 9. Its largest exterior angle is 520

114514521442893143432414257451417256140865713910048138114291371269101361405

114514521442893143432414257451417256140865713910048138114291281260

The internal angles of a convex polygon are in arithmetic Progression. The smallest angle measures 500 and the common difference is 100 . The polygon has maximum number of sides n. n=1) 3 2) 24 3) 3 0r 24 4) none of these.

** Sn = n/2 {100+10 (n-1)} =(n-2)180 50n+5n(n-1) = 180(n-2) 10n+n(n-1) = 36(n-2) n2+9n = 36n -72 n2 - 27n+72 = 0 n = 3 Or 24Answer : 3 Why?GEOMETRY & MENSURATION

The internal angles of a polygon are in arithmetic Progression. The smallest angle measures 650 and the common difference is 180 . The number of sides of the polygon is 1) 8 2) 9 3) 10 4) 7 5) 11

** Sn =

=

As the value on the right hand side is an integer, n or (n-1) should be divisible by 8. Hence substituting n= 9 , we get 1260 as the sum of all internal angles of a 9 sided figure.

GEOMETRY & MENSURATION= (n-2)180= (n-2)180

For the side ED there can be only one triangle. For each side there will be only one triangle.Totally there are 5 such triangles**In a Pentagon how many triangles , using the vertices of Pentagon, can be formed such that only one side of the triangle is same as one side of the Pentagon. GEOMETRY & MENSURATION

In a Hexagon how many triangles , using the vertices of Hexagon, can be formed such that only one side of the triangle is same as one side of the hexagon. For the side ED there can be 2 triangles. For each side there will be 2 triangles.Totally there are 12 such triangles**GEOMETRY & MENSURATIONIn a Heptagon how many triangles , using the vertices of Heptagon, can be formed such that only one side of the triangle is same as one side of the Heptagon. For side ED there can be 3 triangles.For each side there will be 3 triangles.Totally there are 21 such triangles

In a Decagon how many triangles , using the vertices of a Decagon can be formed such that only one side of the triangle is same as one side of the Decagon. For the side AB there can be 6 triangles. For each side there will be 6 triangles.Totally there are 60 triangles.**In a n sided figure ,how many triangles, using the vertices of the n sided figure, can be formed such that only one side of the triangle is same as one side of the n sided figure. For each side there will be (n 4 ) triangles. Totally there will be n ( n 4) triangles.GEOMETRY & MENSURATION

In a Pentagon how many triangles , using the vertices of Pentagon, can be formed such that two sides of the triangle are same as two sides of the Pentagon. There are 5 such triangles.**GEOMETRY & MENSURATIONIn a hexagon how many triangles , using the vertices of hexagon, can be formed such that two sides of the triangle are same as two sides of the hexagon. There are 6 such triangles.*

In a heptagon how many triangles , using the vertices of heptagon, can be formed such that two sides of the triangle are same as two sides of the heptagon. There are 7 such triangles.**GEOMETRY & MENSURATIONIn a Decagon, how many triangles , using the vertices of Decagon, can be formed such that two sides of the triangle are same as two sides of the Decagon. There are 10 such triangles.In a n sided figure, how many triangles , using the vertices of n sided figure, can be formed such that two sides of the triangle are same as two sides of the n sided figure. There are n such triangles.

In a n sided figure, Let x represent the no. of triangles that can be formed, using the vertices of n sided figure, such that two sides of the triangle are same as two sides of the n sided figure. Let Y represent the no. of triangles that can be formed, using the vertices of n sided figure, such that only one side of the triangle is same as one side of the n sided figure. If x + y = 28. What is the value of n? X + Y = n + n(n-4) = 28 n2 3n 28 =0 (n 7) (n +4) = 0n = 7**GEOMETRY & MENSURATION

**GEOMETRY & MENSURATIONCertain Basic Notations & Facts.BC = a , CA = b & AB = ca + b > c b + c > a c + a > bABC is called a right angled , if either A or B or C equals 900.b2 = a2 + c2 ABC is called an acute angled , if A , B & C are less than 900.a2 < b2 + c2b2 < c2 + a2c2 < a2 + b2ABC is called an obtuseangled , if either A or B or C is more than 900. a2 > b2 + c2 or b2 > c2 + a2 or c2 > a2 + b2

**GEOMETRY & MENSURATIONHow many obtuse angled can be drawn if two of the sides are 8 & 15 respectively.Let the third side be x. 8 x 22.Case 1: 15 is not the longest side

x2 > 82 +152 > 289 :: x2 > 289 x > 17 :: x < 23X can take values18,19,20,21,22 5 values.Case 2 : 15 is the longest side

225 > 82 + x2 x2 < 161 x < 13 x > 7X can take values8,9,10,11& 12 5 valuesTotally 10 triangles can be drawn.

**GEOMETRY & MENSURATIONHow many obtuse angled les can be drawn if two of the sides are 6 & 10 respectively.Let X be the third side. 5 x 15.Case 1: 10 is the longest side 100 > 62 + x2 x2 < 64x < 8 x > 4X can take values 5,6 & 7 3 valuesCase 2: 10 is not the longest side x2 > 62 +102 > 136 x > 11 x < 16X can take values 12,13,14&15 4 values.Totally 7 triangles can be drawn.

**GEOMETRY & MENSURATION ABC & DEF are similaras A = D :: B = E :: C = FThenAB/DE = BC/EF = CA/FD ABC & DEF are similaras AB/DE = BC/EF = CA/FD = Then A = D :: B = E :: C = F

** ABC & DEF are right angled triangles and are similar.AB/DE = BC/EF = CA/FD = 2/1Perimeter of ABC/ Perimeter of DEF = 2/1In radius of ABC/ In radius of DEF = 2/1Circum radius of ABC/ Circum radius of DEF = 2/1Area of ABC/ Area of DEF = 4/1Area = 24 :: Perimeter =24In radius = 2 Circum radius = 5Area = 6 :: Perimeter =12In radius = 1 Circum radius =2. 5

**GEOMETRY & MENSURATION ABC & DEF are similaras A = D :: B = E :: C = FAB/DE = BC/EF = CA/FD = m/nPerimeter of ABC/ Perimeter of DEF = m/nIn radius of ABC/ In radius of DEF = m/nCircum radius of ABC/ Circum radius of DEF = m/nArea of ABC/ Area of DEF = m2/n2

**GEOMETRY & MENSURATION

ABC & EFD are similar. AB = 12 , EF = 8 . If BC = 18 , Find DF. ABC and EFD are similar.AB/EF = AC/DE = BC/DF 12/8 = 18/DF DF = 18*8/12 DF = 12ABC and ADE are similar and ADE = area of quadrilateral DECB implies ADE / ABC = AD/AB = 1/2. AD = 62DB = AB - AD = 12 - 62 = 6 (2 - 2).In ABC, let D & E be points on AB & AC such that ABC and ADE are similar and ADE = area of quadrilateral DECB. If AB = 12 , Find DB.

**

GEOMETRY & MENSURATIONABD =2(BD) (Alt. from A to BC)ADC =2(DC) (Alt. from A to BC) ABD/ADC = BD/DC = m/nIf the base is divided in the ratio m : n, area of the 2les so formed are in the ratio m : n.1600If area of triangle ABC is 100 sq units, what is the area of triangle ADE given that BD = 3 AB & CE =4 AC? Let us join DC. Consider ACD. AB:BD:: 1 : 3 BCD = 300 sq units ACD = 400 sq units Consider CDE. AC : CE :: 1 : 4 CDE= 1600 Sq.units ADE=2000 Sq.units

**

GEOMETRY & MENSURATIONIf area of triangle ABC is 20 sq units, what is the area of triangle ADE given that BD = 9 AB & CE =19AC? Let us join DC. Consider ACD.AB:BD::1:9 BCD= 180 sq units. ACD =200 sq units AC: CE:: 1:19 CDE= 3800 Sq.units ADE=4000 Sq.units If area of triangle ABC is x sq units, what is the area of triangle ADE given that BD = k AB & CE =m AC? Let us join DC. Consider ACD.AB:BD::1:k BCD = k x sq units ACD = x + kx sq units AC: CE:: 1:m CDE= m(x+kx )Sq.units ADE=(k+1)(m+1)x

In a ABC,AD is median from A to BC. ABD = ADC (Why?)BE is median from B to CA.CF is median from C to AB The point of concurrence is Centroid (G) Centroid divides the median in the ratio 2:1 AGF = BGF = BGD CGD = CGE = AGE = XAB2+AC2=2(AD2+BD2)BA2+BC2=2(BE2+CE2)CA2+CB2=2(CF2+AF2)

ABCEGFDGEOMETRY & MENSURATION**

GEOMETRY & MENSURATION**AB2 + AC2 = 2(AD2 +BD2)162 + 82 = 2(AD2 +62)AD2 = 124In ABC, AB = 16,BC= 12, AC = 8. Find AD, BE & CF (Medians)BC2 + BA2 = 2(BE2 +CE2)162 + 122 = 2(BE2 +42)BE2 = 184CB2 + CA2 = 2(CF2 +BF2)122 + 82 = 2(BE2 +82)CF2 = 40BE > AD > CFBE2 + AD2 + CF2 = 348 AB2 + BC2 + CA2 = 464[BE2 + AD2 + CF2] /[AB2 + BC2 + CA2] =

GEOMETRY & MENSURATION**AD2 = 62 + 42 = 52CF2 = 82 + 32 = 73BE2 = 25

In a right angled ABC, AB = 6 BC = 8 & AC = 10. Let AD , BE & CF be medians. CF > AD > BEBE2 + AD2 + CF2 = 150 AB2 + BC2 + CA2 =200[BE2 + AD2 + CF2] /[AB2 + BC2 + CA2] =

**BC = 10 ABC . CA = 26.The given triangle is divided into 2 triangles whose areas are equal. What is the maximum possible sum of perimeter of the 2 triangles so formed? The given triangle can be divided into 2 triangles whose areas are equal if we draw a median . As maximum perimeter of the 2 triangles so formed is sought, the longest median is to be drawn. Longest Median is to the shortest side. AF = 601. Maximum possible perimeter of the 2 triangles so formed is 60+2 601

ACBGEOMETRY & MENSURATION55F24601

In the triangle given alongside, AD, BE & CF are medians. GECD is a cyclic quadrilateral, find the ratio of AE:GD.GECD is a cyclic quadrilateral implies that the vertices G, E, C & D lie on a circle.Therefore AG * AD = AE * ACIf GD = Y AG = 2Y & AD = 3YIf AE = X ,EC = X , AC = 2XX * 2 X = 2Y *3 YX= 3 Y. The ratio is 3 : 1

**ABCEGFDGEOMETRY & MENSURATION

In the triangle given alongside, D,E & F are midpoints of sides BC, CA & AB. AK is r from A to BC. Find FKD + FED.**

ABKDFECXXXXGEOMETRY & MENSURATIONConsider AFE & ABCL AFE = L ABC = X Consider CED & ABCLFED = L AFE = XConsider ABK, FK is median to Hyp AB. Hence FK = FB.Hence LFKB = XL FKD = (180 -X)L FED + L FKD = 180

Medians Points to rememberA median divides a triangle into 2 triangles whose areas are equalThree medians divide the triangle into 6 triangles whose areas are equal. The point of concurrence is called centroid.Centroid divides the median in the ratio 2:1AB2+AC2=2(AD2+BD2) BA2+BC2=2(BE2+CE2)CA2+CB2=2(AF2+BF2)where AD,BE & CF are MediansMedian to hypotenuse in a right angled is half the hypotenuse.Shortest median is to the longest side.Longest Median is to the shortest side.Ratio of Sum of squares of medians to sum of squares of sides is 3:4

**

GEOMETRY & MENSURATION**ABCIDEFAI, BI & CI are internal angular bisectors of A B & C. The point of concurrence is called the in-centre. ID,IE & IF are in radii Therefore in-centre is equidistant from the sides. AID need not be a Straight line.BIE need not be a Straight line.CIF need not be a Straight line.AB, BC & CA are tangents to in-circle. Length of tangents from a point outside the circle are equal.BD = BF :: AF = AE :: CE = CD

BIC = 2 (a)(r) CIA = 2 (b)(r) AIB = 2 (c)(r) ABC = 2 (a + b + c)(r) ABC = (s) (r)

GEOMETRY & MENSURATION**ABCIDBI & CI are internal angular bisectors of B & C respectively. I is the in-centre & L BAC = 700 , find the angle BIC.Consider ABC, 2X+2Y = 110X+Y = 55Consider BIC, +X+Y = 180 = 1250 :: = 90+ A 70AI, BI & CI are internal angular bisectors of A B & C. BIC = 90+ 2 A CIA = 90+ 2 B AIB = 90+ 2 C

GEOMETRY & MENSURATION**ABCIDEF AI, BI & CI are internal angular bisectors of A B & C. ID,IE & IF are in radii. If AB=20, AC=22 &BC= 24. Find BD.AB=20 AC=22 BC= 24.Let BD = X :: BF = XAF =AB BF = 20 XAE = 20 X CE = AC CE = X + 2CD= X+2BD+CD = 2X+2 = 242X = 22X =11 BD = BF = [BC + BA AC] /2 CD = CE = [CB + CA AB] /2AF = AE = [AB + AC BC] /2

**GEOMETRY & MENSURATIONIn a ABC, AB = 17, BC = 25, CA = 28, find in-radius S = [17 + 25 + 28]/2 = 35 35*18*10*7 = 35*r35*18*10*7=352 * r2 r2 = 36. r = 6

oIn the given AB = 6, BC = 8 & AC = 10 and angle B is 900. Find radius of the circle inscribed in the Triangle.Area of Triangle ABC = 2 (6)(8) = 24 sq. unitsSemi perimeter = 12 unitsArea of Triangle ABC = (12)(r) = 24 sq. units :: r= 22r+4 = 8 implies r = 2 units

ABCD is a square. E is the midpoint of BC. Find the ratio of in-radius of the circle inscribed in the DCE, to the side of the square.AB = BC = CA=DA = 2a In radius of circle = r.Consider DCE.CE = a ::CD = 2a :: DE = 5a s = (3+ 5 ) a/2 DCE = (2a)(a) DCE = (3+5)a r r/2a = 1/(3+5)**2aaAB5 aEcD2arGEOMETRY & MENSURATION

ABC,AB=16,BC=18 &AC = 20. If AD is an internal angular bisector, Find BD ABC, AB=16, BC=18 AC=20.Let CE be parallel to AD ABD IIIr EBCBD/DC = AB/AC= 16/20BD = 4K CD = 5KBD = 4K = 8**ABCEDABC, internal angular bisector divides the BC in the ratio in the ratio of sides. BD/DC = AB/ACGEOMETRY & MENSURATION

**EBCDAGiven ABC is any triangle.AD is external angular bisector of LEAC. BD is internal angular bisector of LABC. What is the value of LADB - LACB?Let LABC = 2X LACB = 2Y LBAC = 2Z.2x+2y+2z = 180LEAC = 2X+2Y LDAC = X+Y.From ADB,+x+2z+x+y = 180 - y = 0LADB - LACB = 0xx2z2yx + yx + yGEOMETRY & MENSURATION

ABC is a right angled triangle LB = 900. AB = 24. BC = 7. AD is external angular bisector, D is a point on BC Extended. Find CD.Let AD be the external angular bisector. Draw BE l to ADLAEB = LABE = XAE = AB = 24 EC = 1 CEB |||r CAD CE /EA = CB/BD BD = 168 CD = 175

*716812424*ABcDEGEOMETRY & MENSURATION

ABC is a right angled triangle LB = 900. AB = 15. BC = 8. AD is external angular bisector, D is a point on BC Extended. Find CD.Let AD be the external angular bisector. Draw BE parallel to ADLAEB = LABE = XAE = AB = 15 EC = 2 CEB |||r CAD CE /EA = CB/BD BD = 60 CD = 68

*86021515*ABcDEGEOMETRY & MENSURATION

GEOMETRY & MENSURATION

*yxy/(z-x)z -xxx*ABcDEABC is a right angled triangle LB = 900. AB = x. BC = y. AD is external angular bisector, D is a point on BC Extended. Find CD.Let AD be the external angular bisector. Draw BE parallel to ADLAEB = LABE = xAE = AB = x EC = z -x CEB |||r CAD CE /EA = CB/BD BD = x y/(z-x)

ABC is a right angled triangle LB = 900. AB = 8. BC = 6. AD is external angular bisector, D is a point on BC Extended. Find AD.Let AD be the external angular bisector. Draw BE parallel to ADLAEB = LABE = XAE = AB = 8 EC = 2 CEB |||r CAD CE /EA = CB/BD BD = 24 AD = 640

*624288*ABcDXEGEOMETRY & MENSURATION640

Area of ABCLet AD be the altitude from A to BC. Sin C= AD/AC = AD/bAD = b Sin CArea of ABC = (BC) (AD) = (a) (AD) = a b Sin C

**ABCDGEOMETRY & MENSURATIONABC is a triangle with AB = 7 & AC = 5. Given LBAD= LCAD = 600. Find AD. ABD = (7) (AD) Sin 60 ADC = (5) (AD) Sin 60 ABC = (7) (5) Sin 120 ABC = ABD+ ADC = {7(AD)Sin60+5(AD)Sin60} = (7) (5) Sin 120 AD = 35/12

ABC is a triangle with AB = & AC =.Given LBAD= LCAD = X0. Find AD. ABD = () (AD) Sin 60 ADC = () (AD) Sin 60 ABC = () () Sin 120 ABC = ABD+ ADC = {(AD)Sin x+ (AD)Sin x} = () () Sin 2x AD = Sin 2x /[ + ] Sin x**ABCDxxGEOMETRY & MENSURATION

GEOMETRY & MENSURATION**ABCIDEFAI, BI & CI are internal angular bisectors of A B & C. ID,IE & IF are in radii. If BF=6, CE=8 and ID = 4. Find AB+AC. ABC = s(s-a)(s-b)(s-c) ABC = (14+x)(x)(8)(6) ABC = s r s = 14+x ABC = (14+x) (4) (14+x)(x)(8)(6) = (14+x)2 (16) 48x = (14+x) (16) 3x = (14+x) 2x = 14 :: x = 7 AB+AC = 286868xx

In a ABC,AD is altitude from A to BC.BE is altitude from B to CA.CF is altitude from C to AB The point of concurrence is orthocentre (H) AFHE is a cyclic quadrilateralDHEC is a cyclic quadrilateralDHFB is a cyclic quadrilateralGeometry & MensurationABCEHFD If FAE = 70o, what is BHC?FHEA is a Cyclic Quadrilateral.FAE = 70oFHE = 110oBHC = 110o**

In a triangle ABC, one of the side is 10 units .The longest side is 20 units. Area is 80 sq units.Find the third side.Geometry & MensurationABCDLet AD be the altitudeABC = (20) (AD) = 80 AD = 8 unitsFrom ABD, BD2 = AB2 AD2 = 102 82 = 36 = 6 DC = 14From ADC , AC2 = AD2 + DC2 = 82 + 142 = 260 AC = 260148106260**

In a triangle ABC, one of the sides is 10 units & another side is 20 units. Area is 80 sq units. Find the longest side.Geometry & MensurationABCDLet AD be the altitudeABC = (20) (AD) = 80 AD = 8 unitsADB, DB2 = AB2 AD2 = 102 82 = 36 DB = 6 :: DC = 26ADC , AC2 = AD2 + DC2 = 82 + 262 = 740 AC = 740208106740**

In a ABC, AB = AC = 100 units, the area of the triangle is not less than 4800 sq units. What is the difference between maximum and minimum perimeter of such a triangle?Geometry & MensurationABCDyx100y100Let AD be the altitude & AD = X. The altitude bisects the base in an isosceles triangle. Let BC = 2Y x = 1002 y2 { (2y) (x) } 4800 y {1002 y2} (4800) y2 ( 1002 y2 )2 (4800)2 y4 10000y2 + (4800)2 0 (y2 3600) (y2 6400) 03600 y2 6400 :: 60 y 8060 y 80Min perimeter 320. Max perimeter 360 .Diff 40.**

ABC is a right angled triangle. AB = 6 BC = 8 CA = 10 . BE is a median & BD is r from B to AC. Find DE.Let BD & BE be altitude & Median respectively. BE =5.0(6)(8) = (10)(BD) BD = 4.8 unitsDE={(5.0)2- (4.8)2} = 1.96 = 1.4

**ABC108 6 DE5 4.8 Geometry & MensurationABC is a right angled triangle. B = 90. BD is r from B to AC. AE = 4 EC = 9 . Find BE.Consider AEB & BEC.AEB = 90 =BEC ABE = =ECB.The other angle must be equal AEB |||r BEC.BE/AE = EC/BE.BE 2 = (AE) (EC)=4*9 :: BE = 6

**ABCEGeometry & MensurationABC is a right angled triangle.B = 90. BE is r from B to AC. Consider AEB & BEC.AEB = 90 =BEC. ABE = =ECB.The other angle must be equal AEB |||r BEC.BE/AE = EC/BE :: BE 2 = (AE) (EC)ABC is a right angled triangle.B = 90. BE is r from B to AC. Consider AEB & ABC.AEB = 90 =BEC. ABE = =ACB.The other angle must be equal AEB |||r ABC.AE/AB = AB/AC = BE/BCABC is a right angled triangle.B = 90. BE is r from B to AC. Consider CEB & ABC.CEB = 90 =ABC. CBE = =CAB.The other angle must be equal AEB |||r ABC.BE/AB = BC/AC = CE/BC

**ABCEGeometry & MensurationABC is a right angled triangle.B = 90. BE is r from B to AC.We get 3 sets of similartriangles. AEB |||r BEC AEB |||r ABC BEC |||r ABCIn ABC, D, E & F are midpoints of BC, CA & AB. Let L1, L2, L3 represent r passing through D, E, F.L1, L2, L3 are r bisectors of BC, CA & AB.The point of concurrence is s, circumcentreSA SB = SCCircumcentre is equidistant from vertices

Area of ABC = (b) (c) Sin A. Let BD be the diameter of the circum circle Then LA = LD and BD = 2R,where R is the circum radius.From BDC, Sin D = a/2R = Sin AArea of ABC = abc/4R

Geometry & MensurationacbArea of ABC = (b) (c) Sin A. Let BD be the diameter of the circum circle Then LA = LD and BD = 2R,where R is the circum radius.From BDC, Sin D = a/2R = Sin ASimilarly a/SinA= b/SinB = c/SinC = 2R **

Area of ABC = (base) (height)Area of ABC =s(s - a) (s - b) (s - c)Area of ABC = r sArea of ABC = b c Sin AArea of ABC = c a Sin B Area of ABC = a b Sin CArea of ABC = abc/4Ra/SinA= b/SinB = c/SinC = 2R

COS A = [b2 + c2 - a2]/2bcCOS B = [c2 + a2 - b2]/2caCOS C = [a2 + b2 - c2]/2abArea of ABC = (3/4) (a2) if ABC is equilateral triangle of side a.

Geometry & Mensuration**

Area of ABC = (base) (height)Area of ABC =s(s - a) (s - b) (s - c)Area of ABC = r sArea of ABC = b c Sin AArea of ABC = c a Sin B Area of ABC = a b Sin CArea of ABC = abc/4RGeometry & MensurationLet AB = ACAD is Median AD is Internal Angular BisectorAD is AltitudeAD is r bisector. ABD = ADCTherefore, in centre, circumcentre, orthocentre , & centroid lie on AD.ABC**

Geometry & MensurationLet ABC be a right angled triangleABC = 90 . AD is Median.BE is Median. AE = EC.

B is orthocentre.E is circumcentreCentroid lies on BE.G is Centroid. BG/GE = 2/1. In any triangle,Centroid divides the line joining Orthocentre and circumcentre in the ratio 2:1 . The proof is indicative .

Let ABC be an Equilateral triangle. AD, BE & CF are Medians, Altitudes ,r bisectors, & internal angular bisectors .

G is Centroid ,incentre , Circumcentre and Orthocentre

Let AB = AC . AD is Median . AD is Internal Angular Bisector . AD is Altitude AD is r bisector. ABD = ADC Therefore, in centre, circumcentre, orthocentre , & centroid lie on AD.**

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