Mathematics [5 marks sum]

i. Steps of construction:
1. Draw BC = 6.5 cm using a ruler.
2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
3. With Q as center and same radius, cut the previous arc at P.
4. Join BP and extend it.
5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
6. Join AC to obtain ΔABC.

ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.

iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.

iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)

Steps of Construction:
i) Draw a line segment BC = 6.4 cm
ii) At B, draw a ray BX making an angle of 75° with BC and cut off BA = 5 cm.
iii) Join AC.
ΔABC is the required triangle.
iv) Draw the perpendicular bisector of BC.
v) Draw the angle bisector of angle ACB which intersects the perpendicular bisector of BC at P.
vi) Join PB and draw PL ⊥ AC.
Proof: In and ΔPBQ and ΔPCQ
PQ = PQ (Common)
∠AQB = ∠PQC (Each = 90°)
BQ = QC (PQ is the perpendicular bisector of BC)
∴ By side Angle side criterion of congruence
ΔPBQ ≅ ΔPCQ (SAS Postulate)
The Corresponding parts of the congruent triangle are congruent
∴ PB = PC (CPCT)
Hence, P is equidistant from B and C.
∠PQC = ∠PLC ( Each = 90°)
∠PCQ = ∠PCL (Given)
PC = PC (Common)
Again in ΔPQC and ΔPLC ∴ By Angle – Angle side criterion of congruence,
ΔPQC ≅ ΔPLC (AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴PQ = PL (CPCT)
Hence, P is equidistant from AC and BC.

Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
ΔPQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Proof: In ΔLTQ and ΔMTQ
∠TLQ = ∠TMQ (Each = 90°)
∠LQT = ∠TQM (QT is angle bisector)
QT = QT (Common)
∴ By Angle – Angle – side criterion of congruence,
∴ΔLTQ ≅ ΔMTQ (AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴ TL = TM (CPCT)
Hence, T is equidistant from PQ and QR.

Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC . Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.