Pedagogy of Mathematics (Part II) - Geometry, Geometry, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, circumradius, acute triangle, obtuse triangle, right triangle, incircle of triangle, incentre,

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 4.6

15. Solution:
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of (AB
and BC) any two sides and let them, meet at S
which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as
radius draw the circumcircle to passes through A, B
and C.
Circum radius = 4.3 cm.

16. Solution:
Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct the perpendicular bisector of (PQ and
PR) any two sides and let them meet at S which is the
circumcenter.
Step 3:With S as centre and SP = SQ = SR as radius draw
the circumcircle to passes through P, Q and R.
Circum radius = 3.8 cm.

17. Solution:
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any
two sides (AB and BC) and let them meet at S which
is the circumcenter.
Step 3: With S as centre and SA = SB = SC as
radius draw the circumcircle to passes through A, B
and C.
Circum radius = 4.3 cm.

18. Solution:
Given PQ = PR
∴ ∠R = 50° (opposite angles are equal)
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any
two sides (QR and PR) and let them meet at S. S is
the circumcenter of ΔPQR.
Step 3: With S as centre SP = SQ = SR as radius.
Draw the circumcircle.
Circum radius = 3.5 cm.

19. Solution:
Steps for construction:
Step 1: Draw the ΔABC with the each side measure
6.5 cm.
Step 2: Construct the angles bisectors of any two
angles (A and B) and let them meet at I. Then I is
the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the
side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the
circle. This circle touches all the sides of the triangle
internally.

20. Solution:
Steps for construction:
Step 1: Draw the ΔABC with AB = 8cm, BC = 10 cm
and ∠A = 90°.
Step 2: Construct the angle bisectors of any two
angles (∠B and ∠C) and let them meet at I. Then I is
the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the
side to meet AB at D.
Step 4: With I as centre and ID as radius draw the
circle. This circle touches all the sides of the triangle
internally.

21. Solution:
Step 1: Draw the ΔABC with AB = 9cm, ∠B = 40°
and ∠A = 115°.
Step 2: Construct the angle bisectors of any two
angles (A and B) and let them meet at I. Then I is
the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the
side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the
circle. This circle touches all the sides of the triangle
internally.

22. Solution:
Steps for construction:
Step 1: Draw the ΔABC with AB = 6 cm, ∠B = 80° and BC
= 6 cm.
Step 2: Construct the angle bisectors of any two angles
(A and B) and let them meet at I.Then I is the incentre of
ΔABC.
Step 3: Draw perpendicular from I to any one of the side
(AB) to meet AB at D.
Step 4:With I as centre and ID as radius draw the circle.
This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.9 cm.