geometry & mensuration 1

04/13/23 1

GEOMETRY & MENSURATION

Concepts To be remembered

The lines are parallel and a transversal cuts these lines.

2

2

2

2

2

2

2

2

If

Angle 1 = Ѳ,

Angle2 = [180 – Ѳ]

1

1

1

1

1

1

1

1

04/13/23 2

GEOMETRY & MENSURATION

Concepts To be remembered

Lines AB & CD are parallel. E is a point such that Ľ BAE = 450 and Ľ ECD = 300. Find Ľ AEC

A

D

B

C

E

450

300

Draw a line GF passing through E parallel to AB

FG

450

300

Ľ AEG = 450 Ľ GED = 300 Ľ AEC = 750

04/13/23 3

GEOMETRY & MENSURATION

Concepts To be remembered

Lines AB & CD are parallel. AE and CE are internal angular bisectors . Find Ľ AEC

Let ĽFAB = 2x0

Ľ ACD = 2x0. Ľ BAC = 2y0

2x+2y = 1800

Ľ CAE = y0

Ľ ACE = x0

X + Y = 90From ∆ AEC,Ľ AEC = 900

A

D

B

C

E

F

2xy

y

xx

What is the sum of all internal

angles in the figure?

There are 8 triangles and hence the sum of all internal angles is 1440.

04/13/23 4

GEOMETRY & MENSURATION

Concepts To be remembered What is the

Sum Of all

internalangles inFigureA,B,C &

D?A = 720B = 540 C = 360 D = 900

A

B

C

D

A

J

I

HG F

E

D

C

B

1

67

85

43

2

1

43

2

1

32

1

2

1

54

3

2

04/13/23 5

Sum of all internal angles = (n-2)π

Sum of all External angles= 2 πIn a regular n sided figure, Each internal angle = (n-2) π /nEach external angle = 2 π /n

N Sum of all internal Angles

Average

3 180 60

4 360 90

5 540 108

6 720 120

7 900 900/7

8 1080 135

9 1260 140

10 1440 144

GEOMETRY & MENSURATION

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) 480

In the above question what is the number of sides of Polygon?(1) 9 (2) 7 (3) 10 (4) 8 (5) 11

6

GEOMETRY & MENSURATION

1 145

145

2 144

289

3 143

432

4 142

574

5 141

725

6 140

865

7 139

1004

8 138

1142

9 137

1269

10

136

1405

If largest angle is

1450, let us find the

maximum number of

sides of the polygon.

The sum of all

internal angles is

(n-2) π. 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

1 145

145

2 144

289

3 143

432

4 142

574

5 141

725

6 140

865

7 139

1004

8 138

1142

9 128

1260

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.

04/13/23 7

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 18¾0 . The number of sides of the polygon is 1) 8 2) 9 3) 10 4) 7 5) 11

04/13/23 8

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[130 +

75

4(n -1)]

[65 n +758

n(n -1) ]

= (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

04/13/23 9

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

04/13/2310

GEOMETRY & MENSURATION

A B

F C

E D

A

B

F

C

E

DG

In 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.

04/13/23 11

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

A B

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.

04/13/23 12

GEOMETRY & MENSURATION

In 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.

A

B

C

DE

04/13/23

A

B C

D

EF

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.

04/13/23 13

GEOMETRY & MENSURATION

In 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) = 0

n = 7

04/13/23 14

GEOMETRY & MENSURATION

04/13/23 15

GEOMETRY & MENSURATION

Certain Basic Notations & Facts.BC = a , CA = b & AB = ca + b > c b + c > a c + a > b

ABC is called a right angled ∆ , if either Ĺ A or Ĺ B or Ĺ C equals 900.b2 = a2 + c2

C

A

B a

bc

a

bc

C

A

B

ABC is called an acute angled ∆ , if Ĺ A , Ĺ B & Ĺ C are less than 900.a2 < b2 + c2

b2 < c2 + a2

c2 < a2 + b2a

bc

C

A

B

ABC is called an obtuseangled ▲ , if either Ĺ A or Ĺ B or Ĺ C is more than 900. a2 > b2 + c2 or b2 > c2 + a2

or c2 > a2 + b2

C

A

Bc

b

a

04/13/23 16

GEOMETRY & MENSURATION

C

A

B8

15

How many obtuse angled ∆ can be drawn if two of the sides are 8 & 15 respectively.

C

A

B8

15

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 values

Totally 10 triangles can be drawn.

04/13/23 17

GEOMETRY & MENSURATION

C

A

B6

10

How 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 values

C

A

B6

10

Case 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.

04/13/23 18

GEOMETRY & MENSURATION

C

A

Bǿ

δ

θF

D

Eǿ

δ

θ

∆ ABC & ∆ DEF are similaras Ĺ A = Ĺ D :: Ĺ B = Ĺ E :: ĹC = Ĺ FThenAB/DE = BC/EF = CA/FD

∆ ABC & ∆ DEF are similaras AB/DE = BC/EF = CA/FD = ¾

Then Ĺ A = Ĺ D :: Ĺ B = Ĺ E :: ĹC = Ĺ F

C

A

B

6

8

12

F

D

E

3

4

6

04/13/23 19

F

D

E

C

A

B

3

4

5

6

8

10

∆ 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 = 2/1

Area of ∆ ABC/ Area of ∆ DEF = 4/1

Area = 24 :: Perimeter =24In radius = 2 Circum radius = 5

Area = 6 :: Perimeter =12In radius = 1 Circum radius =2. 5

04/13/23 20

GEOMETRY & MENSURATION

C

A

Bǿ

δ

θF

D

Eǿ

δ

θ

∆ ABC & ∆ DEF are similaras Ĺ A = Ĺ D :: Ĺ B = Ĺ E :: ĹC = Ĺ 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/n

Area of ∆ ABC/ Area of ∆ DEF = m2/n2

04/13/23 21

GEOMETRY & MENSURATION

C

A

B

ǿ

δθ

F

D

Eǿ

δ

θ

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 = 12

E

C

A

B

D

ǿ

δθ

δθ

ABC and ADE are similar and ADE = area of quadrilateral DECB implies ADE / ABC = ½AD/AB = √1/√2. AD = 6√2DB = AB - AD = 12 - 6√2 = 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.

04/13/23 22

GEOMETRY & MENSURATION

▲ABD =⅟2(BD) (Alt. from A to BC)

▲ADC =⅟2(DC) (Alt. from A to BC)

▲ABD/▲ADC = BD/DC = m/n

If the base is divided in the ratio m : n, area of the 2▲les so formed are in the ratio m : n.

A

B CDm n

A

B C

D E

3

100

1

300

1600

4

1

If 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

04/13/23 23

GEOMETRY & MENSURATION

180

9

201

If 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

A

B C

D E

19

1

3800

A

B C

DE

1 1

K

m

x

kx

m(x+kx)

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)

A

B C

EG

F

D

GEOMETRY & MENSURATION

04/13/23 24

GEOMETRY & MENSURATION

04/13/23 25

AB2 + AC2 = 2(AD2 +BD2)

162 + 82 = 2(AD2 +62)

AD2 = 124

D

G

A

B

EF

C

In ∆ ABC, AB = 16,BC=

12, AC = 8. Find AD, BE &

CF (Medians)

BC2 + BA2 = 2(BE2 +CE2)

162 + 122 = 2(BE2 +42)

BE2 = 184

CB2 + CA2 = 2(CF2 +BF2)

122 + 82 = 2(BE2 +82)

CF2 = 40

BE > AD > CF

BE2 + AD2 + CF2 = 348

AB2 + BC2 + CA2 = 464

[BE2 + AD2 + CF2] /[AB2

+ BC2

+ CA2] = ¾

GEOMETRY & MENSURATION

04/13/23 26

AD2 = 62 + 42 = 52

CF2 = 82 + 32 = 73

BE2 = 25

D

G

A

B

EF

C

In a right angled ABC, AB = 6 BC = 8 & AC = 10. Let AD , BE & CF be medians.

CF > AD > BE

BE2 + AD2 + CF2

= 150

AB2 + BC2 + CA2

=200[BE2 + AD2 + CF2] /[AB

2 + BC

2 + CA

2] = ¾

04/13/23 27

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

A

CB

GEO

METR

Y &

MEN

SU

RA

TIO

N

55F

24√601

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 : 104/13/23 28

A

B C

E

G

FD

GEOMETRY & MENSURATION

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

04/13/23 29

A

B K D

F E

C

XX

X

X X

GEOMETRY & MENSURATION

Consider ∆AFE & ∆ ABC

L AFE = L ABC = X

Consider ∆CED & ∆ ABC

LFED = L AFE = X

Consider ∆ABK, FK is

median to Hyp AB. Hence

FK = FB.

Hence LFKB = X

L 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

04/13/23 30

GEOMETRY & MENSURATION

04/13/23 31

A

B C

I

D

E

F

AI, BI & CI are internal angular bisectors of ĹA ĹB & ĹC. 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 = ⅟2 (a)(r)▲ CIA = ⅟2 (b)(r)▲ AIB = ⅟2 (c)(r)▲ ABC = ⅟2 (a + b + c)(r)▲ ABC = (s) (r)

GEOMETRY & MENSURATION

04/13/23 32

A

B C

I

D

BI & CI are internal angular bisectors of ĹB & ĹC respectively. I is the in-centre & L BAC = 700 , find the angle BIC.Consider ∆ ABC, 2X+2Y = 110X+Y = 55Consider ∆ BIC, Ø+X+Y = 180Ø = 1250 :: Ø = 90+ ½ Ĺ A

70

X Y

X Y

ø

AI, BI & CI are internal angular bisectors of ĹA ĹB & ĹC. ĹBIC = 90+ ⅟2 A

ĹCIA = 90+ ⅟2 B

ĹAIB = 90+ ⅟2 C

GEOMETRY & MENSURATION

04/13/23 33

A

B C

I

D

EF AI, BI & CI are internal angular bisectors of ĹA ĹB & ĹC. 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

X

X

20 - x 2

0 - x

2+

x

2+ x

BD = BF = [BC + BA – AC] /2 CD = CE = [CB + CA – AB] /2

AF = AE = [AB + AC – BC] /2

04/13/23 34

GEOMETRY & MENSURATION

In 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

B

A

C

r

r r

r o

6-r

6-r

r+4

r+4

In 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 = ⅟2 (6)(8) = 24 sq. unitsSemi perimeter = 12 unitsArea of Triangle ABC = (12)(r) = 24 sq. units :: r= 22r+4 = 8 implies r = 2 units

B

A

C

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 = ½ (2a)(a)∆ DCE = ½ (3+√5)a r r/2a = 1/(3+√5)

04/13/23 35

2a

a

A B

√5 a

E

cD2a

r

GEOMETRY & 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= 16/20

BD = 4K CD = 5KBD = 4K = 8

04/13/23 36

A

B C

ø

øø

ø

E

D∆ABC, internal angular bisector divides the BC in the ratio in the ratio of sides. BD/DC = AB/AC

GEOMETRY & MENSURATION

04/13/23 37

E

B C

D

A

Given ABC is any triangle.AD is external angular bisector of LEAC. BD is internal angular bisector of LABC. What is the value of LADB - ½LACB?Let LABC = 2X LACB = 2Y LBAC = 2Z.2x+2y+2z = 180LEAC = 2X+2Y LDAC = X+Y.From ∆ ADB,ǿ+x+2z+x+y = 180ǿ - y = 0LADB - ½LACB = 0

xx

2z

2y

x + yx + y ǿ

GEOMETRY & 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

04/13/23

7168124

24

38

A

B cD

X

X

X

X E

GEOMETRY & 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

04/13/23

860215

15

39

A

B cD

X

X

X

X E

GEOMETRY & MENSURATION

GEOMETRY & MENSURATION

04/13/23

yxy/(z-x)z -x

x

x

40

A

B cD

øø

ø

øE

ABC 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

04/13/23

6242

8

8

41

A

B cD

X

X

X

X E

GEOMETRY & MENSURATION

√640

Area of ∆ ABCLet AD be the altitude from A to BC. Sin C= AD/AC = AD/bAD = b Sin CArea of ∆ ABC = ½ (BC) (AD) = ½ (a) (AD) = ½ a b Sin C

04/13/23 42

A

B CD┐

GEOMETRY & MENSURATION

A

B D C 5

7

60 60

ABC is a triangle with AB = 7 & AC = 5. Given LBAD= LCAD = 600. Find AD.∆ ABD = ½ (7) (AD) Sin 60∆ ADC = ½ (5) (AD) Sin 60∆ ABC = ½ (7) (5) Sin 120 ∆ ABC = ∆ ABD+ ∆ ADC = ½{7(AD)Sin60+5(AD)Sin60} = ½ (7) (5) Sin 120 AD = 35/12

ABC is a triangle with AB = α & AC =β.

Given LBAD= LCAD = X0. Find AD.

∆ ABD = ½ (α) (AD) Sin 60

∆ ADC = ½ (β) (AD) Sin 60

∆ ABC = ½ (α) (β) Sin 120

∆ ABC = ∆ ABD+ ∆ ADC

= ½{α(AD)Sin x+ β(AD)Sin x}

= ½ (α) (β) Sin 2x

AD = α β Sin 2x /[α + β] Sin x

04/13/23 43

A

B CDβ

α

xx

GEOMETRY & MENSURATION

GEOMETRY & MENSURATION

04/13/23 44

A

B C

I

D

EFAI, BI & CI are internal angular bisectors of ĹA ĹB & ĹC. 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 = 28

6 8

6 8

x x

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 quadrilateral

Geometry & Mensuration A

B C

EH

F

D

If ĹFAE = 70o, what is Ĺ BHC?FHEA is a Cyclic Quadrilateral.ĹFAE = 70o

ĹFHE = 110o

ĹBHC = 110o

04/13/23 45

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 & Mensuration

A

B CD

Let AD be the altitudeABC = ½ (20) (AD) = 80 AD = 8 unitsFrom ABD, BD2 = AB2 – AD2

= 102 – 82

= 36 = 6 DC = 14From ADC , AC2 = AD2 + DC2

= 82 + 142 = 260 AC = √260

14

8

10

6

√260

04/13/23 46

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 & MensurationA

B CD

Let AD be the altitudeABC = ½ (20) (AD) = 80 AD = 8 unitsADB, DB2 = AB2 – AD2

= 102 – 82 = 36 DB = 6 :: DC = 26ADC , AC2 = AD2 + DC2

= 82 + 262 = 740 AC = √ 740

20

8 10

6

√740

04/13/23 47

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 & Mensuration

A

B CD y

x

100

y

100

Let AD be the altitude & AD = X. The altitude bisects the base in an isosceles triangle. Let BC = 2Y x = √ 1002 – y2

½ { (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 ≤ 80

60 ≤ y ≤ 80Min perimeter 320. Max perimeter 360 .Diff 40.

04/13/23 48

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 =5.0½(6)(8) = ½(10)(BD) BD = 4.8 unitsDE=√{(5.0)2- (4.8)2} = √1.96 = 1.4

04/13/23 49

A

B C

10

8

6 D

E

5

4.8

Geometry & Mensuration

ABC is a right angled triangle. ĹB = 90. BD is ┴r from B to AC. AE = 4 EC = 9 . Find BE.Consider ∆ AEB & ∆ BEC.ĹAEB = 90 =ĹBEC ĹABE = ø =ĹECB.The other angle must be equal ∆ AEB |||r ∆ BEC.BE/AE = EC/BE.

BE 2 = (AE) (EC)=4*9 :: BE = 6ø

ø

A

B C

9

4

E

04/13/23 50

A

B C

E

┐┘

Geometry & MensurationABC is a right angled triangle.ĹB = 90. BE is ┴r from B to AC. Consider ∆ AEB & ∆ BEC.ĹAEB = 90 =ĹBEC. ĹABE = ø =ĹECB.The other angle must be equal ∆ AEB |||r ∆ BEC.BE/AE = EC/BE :: BE 2 = (AE) (EC)

ø

ø

ABC is a right angled triangle.ĹB = 90. BE is ┴r from B to AC. Consider ∆ AEB & ∆ ABC.ĹAEB = 90 =ĹBEC. ĹABE = ø =ĹACB.The other angle must be equal ∆ AEB |||r ∆ ABC.

AE/AB = AB/AC = BE/BC

ABC is a right angled triangle.ĹB = 90. BE is ┴r from B to AC. Consider ∆ CEB & ∆ ABC.ĹCEB = 90 =ĹABC. ĹCBE = ø =ĹCAB.The other angle must be equal ∆ AEB |||r ∆ ABC.

BE/AB = BC/AC = CE/BC

04/13/23 51

A

B C

E

┐┘

Geometry & Mensuration

ø

ø

ABC is a right angled triangle.ĹB = 90. BE is ┴r from B to AC.We get 3 sets of similartriangles. ∆ ∆ AEB |||r ∆ BEC ∆ AEB |||r ∆ ABC ∆ BEC |||r ∆ ABC

In 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

A

B

F

D C

E

S

Area of ABC = ½ (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 A

Area of ABC = abc/4R

Geometry & MensurationA

B

D

C

S

a

cb

Area of ABC = ½ (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 A

Similarly a/SinA= b/SinB = c/SinC = 2R

A

B

D

C

S

04/13/23 52

Area of ABC = ½ (base) (height)Area of ABC =√s(s - a) (s - b) (s - c)Area of ABC = “r” “s”Area of ABC = ½ b c Sin AArea of ABC = ½ c a Sin B Area of ABC = ½ a b Sin C

Area of ABC = abc/4Ra/SinA= b/SinB = c/SinC = 2R

COS A = [b2 + c2 - a2]/2bc

COS B = [c2 + a2 - b2]/2ca

COS C = [a2 + b2 - c2]/2ab

Area of ABC = (√3/4) (a2) if ABC is equilateral triangle of side “a”.

Geometry & Mensuration

A

B C

bc

a

04/13/23 53

Area of ABC = ½ (base) (height)Area of ABC =√s(s - a) (s - b) (s - c)Area of ABC = “r” “s”Area of ABC = ½ b c Sin AArea of ABC = ½ c a Sin B Area of ABC = ½ a b Sin C

Area of ABC = abc/4R

Geometry & Mensuration

Let AB = ACAD is Median AD is Internal Angular BisectorAD is AltitudeAD is r bisector. ┴ ABD = ADCTherefore, in centre, circumcentre, orthocentre , & centroid lie on AD.

A

B C

A

BD C

04/13/23 54

Geometry & MensurationLet ABC be a right angled triangleĹABC = 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 .

A

B

G

CD

E

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

G is Centroid ,incentre , Circumcentre and Orthocentre

A

B

FE

D C

G

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.

A

B D C04/13/23 55