(i)\(\angle\)CAB = 25⁰
\(\therefore\) \(\angle\)BDC = 25⁰
(ii) \(\angle\)DBA = \(\angle\)DCA =180-(90+25⁰)
= 180⁰ - 115⁰
= 65
(iii)\(\angle\)COB = 2\(\angle\)CAB = 2\(\times\)25⁰ = 50⁰

Given ㄥABC = 30⁰
ㄥAOC = 2ㄥABC
(The angle subtended by an arc at the centre is double the angle at any point on the circle)
= 2 x 30⁰
= 60⁰

If a side is produced in a triangle, the angle formed is called an exterior angle.
Here BC, a side of ΔABC is produced, which lead to the formation of exterior angle ㄥACD.
Also BCD is a straight line ⇒ ㄥBCD is a straight angle.
ㄥBCD = 180⁰
ㄥc + ㄥd = 180⁰  ......(1)
Also in ΔABC ㄥa + ㄥb + ㄥc
From (1) and (2) we get,
ㄥa + ㄥb + ㄥc = ㄥc + ㄥd
∴ ㄥa + ㄥb =ㄥd
Hence, we conclude that the measure of exterior angle formed is equal to the sum of the measure of two opposite interior angles.

To draw the different types of quadrilateral by giving the instructions using their properties.
Example:
(i) It is four sides closed diagram.
(ii) Opposite sides are parallel.
(iii) Four sides are equal in size.
(iv) Diagonals bisect each other at 90°.
(v) Diagonals are not equal.
The diagram represents as a rhombus
Like this for hexagon, parallelogram, Trapezium etc.
Any one must give the clues to draw the diagram.
At last they have to find what it represent.

ABCD is a parallelogram.
OA = OC and
OB = OD (∴ Diagonals bisect each other)
now OB + BQ = OD + DP
OQ = OP and OA = OC
APCQ is a parallelogram.

In an equilateral \(\triangle \) on side = 6 cm,

Construction :
Step 1: Draw ΔABC with equal sides of 6 cm length.
Step 2: Draw perpendicular bisectors of any two sides (BC and AC) to find the mid points of BC and AC.
Step 3: Draw medians BD and CE. Let them meet at G.
Step 4: Draw angle bisectors of any two sides (ㄥB and ㄥC): Let them meet at I.
Step 5: G and I are concurrent point and G is the centroid, I is the incentre.
Step 6: In an equilateral triangle centroid and incentre lie on the same point.

In PQR,
PQ = 5 cm,
PR = 6 cm
\(\angle \)QPR = 60°

Construction :
Step 1: Draw △ PQR with the given measurement
Step 2: Draw perpendicular bisectors of any two sides (PQ and QR) to find the mid points of PQ and QR.
Step 3: Draw medians PD and RE. Let them meet at G.
Step 4: G is the centroid of the given △PQR.

In \(\triangle \)ABC, AB = 6 cm, LB= 110°, AC = 9 cm

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 6 cm, ∠B =110°, AC = 9 cm
Step 2: Draw perpendicular bisectors of any two sides (BC and AB) to find the mid points of BC and AB.
Step 3: Construct medians AD and CE. Let them meet at G.
Step 4: G is the centroid of the given \(\triangle \)ABC.

In \(\triangle \)ABC,
AB = 4 cm, AC = 3 cm, \(\angle\)A =  90°

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 4 cm, AC = 3 cm, \(\angle\)A = 90°
Step 2: Draw perpendicular bisectors of any two sides (AB and AC) to find the mid points of AB and AC.
Step 3: Draw the medians CD and BE. Let them meet at G.
Step 4: G is the centroid of the given triangle.

In \(\triangle \)LMN
LM = 7.5 cm,
MN = 5 cm,
LN = 8 cm

Construction :
Step 1: Draw \(\triangle \)LMN with LNM = 8 cm, MN = 5 cm, LM = 7.5 cm
Step 2: Construct perpendicular bisectors for any two sides (LN and MN) to find the mid points of LM and MN.
Step 3: Draw the medians LD, ME. Let them meet at G.
Step 4: G is the centroid of the triangle LMN.