Mathematics [4 marks sum]

1. i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
5. ii) On measuring AB = 5.7 cm

Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.

Steps of Construction:
i) Plot the points A(1,1), B(5,3) and C(2,7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm.

Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.

Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm

Steps of Construction: i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC. ΔABC is the required triangle.
v) Again with centre B and C and radius greater than 12 BC, draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB. Proof: In ΔDBE and ΔDCE BE = EC (LM is bisector of BC) ∠DEB = ∠DEC (Each = 90°) DE = DE (Common)
∴ By side angle side criterion of congruence, we have ΔDBE ≅ ΔDCE (SAS postulate)
The corresponding parts of the congruent triangle are congruent
∴ DB = DC (CPCT) Hence, D is equidistant from B and C.

Construction: From X, draw XL ⊥ AC and XM ⊥ AB. Also join YC.
Proof:
(i) In Δ AXL and ΔAXM,
∠XAL = ∠XAM (Given)
AX = AX (Common)
∠XLA = ∠XMA (Each = 90°)
∴ By Angle side angle criterion of congruence,
ΔAXL ≅ ΔAXM (ASA Postulate)
The corresponding parts of the congruent triangles are congruent
∴ XL = XM (CPCT)
Hence, X is equidistant from AB and AC
(ii) In ΔYTA and ΔYTC,
AT = CT (∵ PQ is a perpendicular bisector of AC)
∠YTA =∠YTC (Each = 90°)
YT = YT (common)
∴ By side – Angle – side criterion of congruence,
∴ ΔYTA ≅ ΔYTC (SAS postulate)
The corresponding parts of the congruent triangle are congruent.
∴ YA = YC (CPCT)
Hence, Y is equidistant from A and C.

Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC = ABC = 75°
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC.
Since D lies on the angle bisector BP of ∠ABC.
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.

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.