Mathematics [3 marks sum]

The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.

The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.

Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A

Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.

Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.

Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.

Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.

(i)Draw a line segment AB = 8 cm.
(ii)Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii)Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv)Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.

Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴ A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M

Construction: From P, draw PL ⊥ AB and PM ⊥ BC
Proof: In ΔPLB and ΔPMB
∠PLB = ∠PMB (each = 90°)
∠PBL = ∠PBM (Given)
PB = PB (Common)
∴ By Angle – angle side criterion of congruence,
ΔPLB ≅ ΔPMB (AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴ PL = PM (CPCT)
Hence, P is equidistant from AB and BC

Construction: Join PQ which meets AB in D.

Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.

Construction: From P, draw PL ⊥ AB and PM ⊥ CB
Proof: In ΔLPC and ΔMPC,
∠PLC = ∠PMC (Each = 90°)
∠PCL = ∠MCP (Given)
PC = PC (Common)
∴ By angle- side angle criterion of congruence,
ΔLPC ≅ ΔMPC (AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴ PL = PM (CPCT)
Hence, P is equidistant from AC and AB

On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.

On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.

Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.

Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.

Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.

Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB. Since, P lies on the right bisector of line AB. Therefore, P is equidistant from A and B.
i.e. PA = PB Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.

Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC .... (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD .... (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.

Construction: Draw LF ⊥ AC
Proof: In Δ AFL and ΔAFE,
∠FEA = ∠FLA (Each = 90°)
∠LAF = FAE (AD bi sects ∠BAC)
AF = AF (common)
∴ By angle – Angle side criterion of congruence,
ΔAFL ≅ AFE (AAS postulate)
The corresponding parts of the congruent triangles are congruent.
∴ FE = FL (CPCT)
Hence, F is equidistant from AB and AC.

Construction: Join FB and FC
Proof: In ΔAFE and ΔFBE,
AE = EB (E is the mid-point of AB)
∠FEA = ∠FEB (Each = 90°)
FE = FE (Common)
∴ By side Angle side criterion of congruence,
ΔAFE ≅ ΔFBE (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ AF = FB (CPCT)
Hence, F is equidistant from A and B.

Steps of construction:
i) Draw line BC = 6 cm and an angle CBX = 60°. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
ii) Draw perpendicular bisector of BC and bisector of angle B
iii) Bisector of angle B meets bisector of BC at P.
⟹ BP is the required length, where, PB = 3.5 cm
iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.

PB = 3.6 cm