Pedagogy of Mathematics (Part II) - Geometry, Geometry, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, shapes, types of angles, complementary angles, supplementary angles, adjacent angles, Linear pair of angles, vertically opposite angles, Transversal, triangles, congruent triangles,

1. BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

2. Std IX - Ex 4.1

20. (i) Solution :
Let us draw a line passing through T and parallel to AB and CD.
The lines AB and TS are parallel and TA is a
transversal.
∠ ATS + ∠ TAB = 180
∠ ATS + 140 = 180
∠ ATS = 180 - 140 = 40 --(1)
In a same way
∠ TCD + ∠ CTS = 180
150 + ∠ CTS = 180
∠CTS = 180 - 150 = 30 ---(2)
(1) + (2) ==> 40 + 30 = 70
Hence the value of x is 70.

21. (ii) Solution : ∠ PRB = 180 - 48
∠ PRB = 132 ----(2)
In the same way,
∠ PRD + ∠ CDR = 180
∠ PRD + 24 = 180
∠ PRD = 180 - 24 = 156 ---(1)
(1) + (2) ==> 132 + 156 = 288
Hence the required angle is 288.
Now we have drawn a line passing
through x and it is parallel to AB and CD.
∠ PRB + ∠ ABR = 180
∠ PRB + 48 = 180

22. Solution :
∠ BAD = ∠ ADC (Alternative angles)
∠ ADC = 53
In triangle ECD,
∠ ECD + ∠ CDE + ∠ DEC = 180
38 + 53 + ∠ DEC = 180
∠ DEC = 180 - 91
∠ DEC = 89
Hence the value of x is 89.

23. Solution :
The angles of triangle are x, 2x and 3x.
Sum of the interior angles of triangle = 180
x + 2x + 3x = 180
6x = 180
x = 180/6
x = 30
Hence the required angles are 30, 60 and 90.

25. (i) Solution :
In triangle ABC, and triangle PQR
AB = QP
BC = RQ
From the given information, the triangles are not congruent. If AC and RP were equal, then both the
triangles will be congruent by using SSS criterion.

26. Solution :
In triangle ABD, in triangle BDC
AB = DC (S)
AD = BC (S)
DB = DB (S)
Hence the triangle ABD and BDC are congruent using the criterion SSS.

27. Solution :
In triangle TXP and PXZ
YX = ZX (S)
YP = PZ (S)
PX = PX (S)
Hence the triangles TXP and PXZ are congruent using the
criterion SSS.

28. Solution :
In triangle ABO and ODC
AO = OC (S)
<ABO = <ODC (A)
<AOB = <DOC (A)
By using the criterion ASA triangle ABO and ODC are
congruent.

29. Solution :
In triangle AOB, triangle ODC
BO = DO (S)
AO = OC (S)
<AOB = <ODC (A)
By using the criterion SAS the triangles are AOB and ODC.

30. Solution :
In triangle ABM and AMC
AB = AC (S)
<AMB = <AMC (A)
AM = AM (S)
Hence the triangles ABM and AMC are congruent.

31. Solution :
AB = EF (S)
∠ ABC = ∠ EDF (A)
∠ BCA = ∠ DEF (A)
Hence the triangles are congruent.

32. ∠ ACB = ∠ ABC + ∠ BAC
4x - 15 = 2x - 5 + x + 35
4x - 2x - x = 35 - 5 + 15
x = 45
4x - 15 = 4(45) - 15 = 165
2x - 5 = 2(45) - 5 = 85
x + 35 = 45 + 35 = 80
Hence the required angles are 165, 85 and 80.