Pedagogy of Mathematics (Part II) - Geometry, Geometry, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Quadrilaterals, polygons, concave polygon, convex polygon, special names for some quadrilaterals, Types of quadrilaterals, properties of quadrilaterals,

1. PEDAGOGY OF
MATHEMATICS –
PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

2. Std IX - Ex 4.2

33. Solution :
The four angles are 2x, 4x, 5x and 7x.
Sum of interior angles of quadrilaterals = 360
2x + 4x + 5x + 7x = 360
18x = 360
x = 360/18 = 20
x = 20
2x = 2(20) = 40
4x = 4(20) = 80
5x = 5(20) = 100
7x = 7(20) = 140
Hence the required angles are 40, 80, 100 and 140.

34. Solution:
<A + <C = 180
72 + <C = 180
<C = 180 - 72
<C = 108
<B + <D = 180
2x - 10 + x + 4 = 180
3x - 6 = 180
3x = 186
x = 186/3
x = 62
2x - 10 = 2(62) - 10 = 124 - 10 = 114
x + 4 = 62 + 4 = 66
Hence the required angles are 72, 108, 114 and 66.

35. Every angle in a rectangle is 90 degree.
Diagonals of rectangle cut each other into their half.
So, AO = BO
<OBA = <OAB
<OBA = 46° ------- (1)
<ABC = <OBA + <OBC
90° = <46° + <OBC
<OBC = 44°
Solution :

36. Solution :
In rhombus diagonals cut each other at right angles.
In triangle OAB,
AB2 = OA2 + OB2
AB2 = 62 + 82
AB2 = 36 + 64
AB2 = 100
AB = 10
Hence the side length of rhombus is 10 cm.

37. In triangle ADS,
<SAD = (1/2)<A ----(1)
<ADS = (1/2) <D ----(2)
(1) + (2)
<SAD + <ADS = (1/2) [<A + <D]
<SAD + <ADS = (1/2) (180)
<SAD + <ADS = 90
Solution :

38. In triangle ADS,
<SAD + <ASD + <SDA = 180
<SAD + <SDA + <ASD = 180
90 + <ASD = 180
<ASD = 180 - 90
<ASD = 90, <PSR = 90 (Vertically opposite angle)
Hence PQRS is a rectangle.

39. Solution :
Let ΔABP and a parallelogram ABCD be on the same
base AB and between the same parallels AB and PC.
To Prove : ar( ΔPAB ) = (1/2)ar( ABCD)
Draw BQ ||AP to obtain another parallelogram.ABQP
and ABCD are on the same base AB and between the
same parallels AB and PC.
Therefore, ar(ABQP) = ar(ABCD)
But ΔPAB ≅ ΔBQP( Diagonals PB divides parallelogram
ABQP into two congruent triangles.
So ar (PAB) = ar(BQP) -----------(2)
ar (PAB) = (1/2)ar(ABQP) -----------------(3)
This gives ar (PAB) = (1/2)ar(ABCD)

41. Solution :
(i) Angle between b and c is 30.
(ii) Angle between d and e :
75 + angle between d and e = 180
Angle between d and e = 180 - 75
= 105
(iii) Angle between d and f :
In a parallelogram opposite angles will be equal.
Angle between d and is also 75.
(iv) Angle between c and f :
In a parallelogram opposite angles will be equal.
Angle between c and f is 105.

42. Solution :
Since BO and CO are bisectors of ABC and ACB.
<OBC = 58/2 = 29
In triangle ABC,
<ABC + <BAC + <BCA = 180
58 + 64 + <BCA = 180
<BCA = 180 - 122

43. <BCA = 58
<y = 58
In triangle OBC,
<OBC + <BOC + <BCO = 180
29 + <BOC + 29 = 180
<BOC = 180 - 58
x = <BOC = 122

44. Solution :
Here we have three triangles, Δ AEC, Δ BCF and Δ BDC.
Triangle BCD which is similar to ACE.
In triangle AEC,
AC = AB + BC = 2 + 6 = 8 cm
AE = 6 cm and EC = 7 cm
a = 8, b = 6 and c = 7

45. Using a formula for area of scalene triangle
= √s(s - a) (s - b) (s - c)
= √10.5(2.5) (4.5) (3.5)
= √413.43
= 20.33 ----(1)
In triangle BFC
BF = 8 cm , BC = 6 cm, and CF = 7 cm
s = (8 + 6 + 7)/2 = 10.5
= √s(s - a) (s - b) (s - c)

46. = √10.5(2.5) (4.5) (3.5)
= √413.43
= 20.33 ----(1)
Triangle BCD which is similar to ACE.
And BCF
To gets lengths BD and CD for triangle BCD we will use linear scale factor.
We get lsf from lengths AC and BC as follows :
8/6 = 4/3
8(BF) /BD = 4/3

47. 24 = 4BD
BD = 24/4 = 6
CE/CD = 8/6
7/(CD) = 8/6
42 = 8CD
CD = 42/8
= 5.25
Area of BCD :
S = (6 + 6+, 5.25) /2 = 8.625
A = √8.625(8.625 - 6)(8.625 - 6)(8.625 - 5.25)
= √200.58
= 14.16
Area of CDF :
S = (5.25 + 7 + 2) / 2 = 7.125
A = √7.125(7.125 - 5.25) (7.125 - 7)(7.125 - 2)

48. = √8.558 = 2.925
= 2.925
Area of ABDE.
= 20.33 - 14.16 = 6.17
The ratio of ABDE : CDF
= 6.17 : 2.925
Approximately : 6 : 3
= 2 : 1

53. Solution :
[QPO] = [QAM] + [PBM] + [AMONB] -----(1)
[QPO] = [AMONB] + [MDC] + [NCD] -----(2)
From (2),
= [AMONB] + [MDC] + [NOC] + [DOC]
= [ABCD] + [DOC]
Let [ABCD] = K
= K + [DOC]

54. [DCNM ] is also a parallelogram.
So,
[DOC] = 1/4 [DCNM] , 2[ABCD] = [DCNM]
= 1/8 [ABCD]
= k/8
[QPO] = K + K/8
[ [QPO] = 9k/8 ]
Hence proved.