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Day 2: Session 2

Module : Quadrilaterals, Circles and other Polygons, mensuration for 2D & 3D: Area, Surface

and Volume

Topics Covered : Concept Fundamentals, Formulae & their application for Quadrilaterals, Circles

and other Polygons, mensuration for 2D & 3D: Area, Surface and Volume.

Framing of Session : The trainer should frame the chapter along the following lines :

1. Explain the concept of Quadrilaterals, Circles and other Polygons, mensuration for 2D & 3D:Area, Surface and Volume.

In The Following Figure : ABCD is a quadrilateral. Sum of 4 angles of a quadrilateral = 3600. For a cyclic Quadrilateral :

(AB X CD) + (BC X DA) = AC X BD.

A B

D C

Area = X DB X (AE + FC). For a cyclic Quadrilateral :

Sum of opposite angles = 1800. Exterior angle = Interior opposite angle.

F

E

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Problem:

The angles of a Quadrilateral ABCD are in the ratio 2:3:6:7 . What are the actual measurement o

these angles.

1400

1200

400

600

Solution:

We know that : Sum of 4 angles of a quadrilateral = 3600.

Therefore: 2x + 3x + 6x + 7x = 3600

18x = 3600

x = 200.

Thus , we have angles as : 400, 60

0, 120

0& 140

0.

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In The Following Figure :ABCD is a parallelogram. Its Characteristics are:

Opposite sides are equal and parallel. Diagonals bisect each other. Opposite angles are equal. Sum of any 2 adjacent angles = 1800. Area = X b X h. OR Area = BC X DC X Sin .

A B

D C

Problem: If ABCD is a square whose area is 100, CDEF is a parallelogram, and the perimeter of ABCFED is 80, wha

s the length of CF? E

A D

F

B C

Solution: Since ABCD is a square, AB = CD = BC = DA.

Since CDEF is a parallelogram, CD = EF and CF = DE. Thus, AB = EF.

For the perimeter of ABCDEF, there are only two different lengths:

(1) AB = AD = BC = CD = EF(2) CF = DE

In calculating the perimeter of ABCFED, AB is used four times while CF is used twice.

P = 80 = 4*AB + 2*CF.

The area of square ABCD is (AB)2 where AB is the length of the side AB.

Since the area of square ABCD is 100, the length of side AB is the square root of 100, which is 10.

Substitute 10 in for AB in the equation P = 4*AB + 2*CF = 4*10 + 2*CF = 80.

Solve for CF: P = 40 + 2CF = 80

2CF = 80 - 40 = 40

CF = 40/2 = 20

h

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In The Following Figure :ABCD is a Trapezium. Its Characteristics are:

One pair of Opposite sides are parallel but other is not. PQ = (AD + BC). Area = X (AD + BC) X AE. Say RS is parallel to AD & BC And divides the distance between them in ratio m:n as shown

Then:

RS = { [n / (m + n)] X Longer side (BC) + [m / (m + n)] X Shorter Side (AD)}

D A

P Q h

R S

C E B

Problem : In trapezium ABCD,DC=2AB,EF parallel to AB, cuts AD at F and BC at E such that BE/EC=3/4, Prove that

7EF=10AB.

Solution: extend sides DA and CB to intersect at a common point V (above AB)* This will create 3 similar 's: VAB, VFE, and VDC

All corresponding sides will be in the same ratio.

Let x represent the length of segment AB, and DC = 2AB & AB : DC = 1 : 2,

then length of DC = 2x

BE : EC = 3 : 4, let 3y represent the length of segment BE, and 4y the length of segment EC

If CE = 4y, and EB = 3y, then CB = 7y

AB : DC = 1 : 2, corresponding sides of VAB and VDC will also be in ratio 1 : 2

VB : VC = 1 : 2

VC = 2VB = 2BC = 2(7y) = 14y

VE = VB + BE

VE = 7y + 3y = 10y

VB : VE = 7y : 10y = 7 : 10 V

7 : 10 = AB : EF7EF = 10AB as required

A B

F E 3:4

D C

m

n

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In reference of The Following Figure :We Say :

PA . PB = PC . PD = PT2

SECANT

B

A

D

C

P T

TANGENT

Problem: In the figure given below, find the length of PO, given PT = 14cm & OT = 10.5cm

T

P

S

Solution:

Since Tri. PTO is a right angle triangle with Angle T = 900

;

PO2

= PT2

+ OT2

= (14)2

+ (10.5)2

PO = 17.5cm

O

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In The Following Figure : *

B

A

We Say :

A perpendicular drawn from the centre of the circle to the chord, bisects the chord & CONVERSELY

.

Two Chords in a circle if equal in length are equidistant from the centre & CONVERSELY. Two tangents can be drawn to a circle from a point outside the circle and they are also

equal.

Angles subtended at the centre by equal chords are equal.

O

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In The Following Figure : *

ACB is Minor Arc & ADB is Major ARC.

We Say: The angle subtended by an Arc / Chord at the centre is twice the angle subtended by the

Arc / Chord on the remaining part of the Circle.

Angle (AOB) = 2 X Angle (AXB) = 2 X Angle (AYB).

D

X Y

A B

C

O

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In reference of The Following Figure : *

We can see that:

If the diameter subtends an angle on the circumference of the circle , That Angle = 900. If the diameter subtends an angle at a point in the interior of the circle , That Angle > 900. If the diameter subtends an angle at a point in the exterior of the circle , That Angle < 900.

C

900

D

= 90

0

900

A B

E

O

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In reference of The Following Figure :We Say:

Angle subtended by a chord in major segment = Angle b/w tangent & chord in minor segment.

THIS IS ALTERNATE SEGMENT THEOREM

S

R

Angle Subtended by

Chord in Major Segment

CHORD

TANGENT

T

Angle between Tangent & Chord in Minor Segment

B

Problem: Prove the alternate segment theorem

C

x

A

Solution: A tangent makes an angle of 90 degrees with the radius of a circle,

so we know that OAC + x = 90.The angle in a semi-circle is 90, so BCA = 90.The angles in a triangle add up to 180, so BCA + OAC + y = 180Therefore 90 + OAC + y = 180 and so OAC + y = 90But OAC + x = 90, so OAC + x = OAC + yHence x = y

y

O

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In reference of The Following Figure : *We can see different tangents,

Length of :

Direct Common Tangent =[ (Distance b/w the centres)2 (R1 R2)2]

Transverse Common Tangent =[ (Distance b/w the centres)2 (R1 + R2)2]

P Q

Direct Common Tangent

R S

C B

A D

Transverse common tangent

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In reference of The Following Figures : *

CYLINDER :

H

Volume = . R2 . H . Curved Surface Area = 2 . . R . H . Total surface Area = 2 . . R (H + R)

= 2 . . R . H + 2 . . R2

.

CONE

H L

L2 = R2 + H2 . Volume = 1/3 ( R

2

H) .

Curved Surface Area = R L . Total Surface area = R L + R2 .

R

R

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SPHERE

Volume = 4/3 ( R3) . Surface Area = 4 R2 .

R

R

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PolygonsIn a Regular Polygon of n sides :

Total no. of diagonals = [n(n 3)] / 2. No. of diagonals from a single vertex = n 3. No. of triangles created by above said (n3) diagonals from a particular vertex = n2. Sum of interior angles = (n 2) X 1800 OR (2n 4) X 900. Sum of exterior angles = 3600. Each exterior angle = 3600 / n. If each interior angle is D0, then D = [(2n 4)/4] X 900. Area of any regular polygon with each side = a & n sides

A = [(n.a2)/4].cot(180

0/n)

REGULAR POLYGON EACH ANGLE

Pentagon 1080

Hexagon 1200

Octagon 1350

Problem: The sum of the interior angles of a polygon is 16200. How many sides does polygon have

Solution: We have:

Sum of the interior angles of polygon = (2n 4) X 900 = 16200

2n 4 = 1620/90 = 18

Therefore, 2n = 22

n = 11.

Thus the polygon has 11 sides

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Learning Outcome : At the end of the session students should know :

The concept of Quadrilaterals, Circles and other Polygons, mensuration for 2D & 3D: Area,Surface and Volume.

The Fundamental of Area, Surface and Volume. Methods, rules and shortcuts for solving the problems.

Assignments : Handouts and practice tests will be given to students.