Mathematics [4 marks sum]

Draw angle bisector of AB and CD . Draw perpendio..ilars from AB and CD on angle bisector, say P. Pis the required point which is equidistant from AB and CD and at a distAnce of 1.8 cm from FF.

Steps of construction:
(i) Draw a line segment BC= 7.3 cm.
(ii) With Bas centre and radius 6 cm draw an arc.
(iii) With C as centre and radius 5.2 cm draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
(v) Draw perpendicuIar bisector of BC , AB and AC.
In triangIe ABC, P is the point of intersection of AB , AC and BC.
Therefore, PA = PB, PB = PC, PC = PA.
Thus, circum-centre of a triangle is the point which is equidistant from all its vertices.

Steps of construction:
(i) Draw a line segment BC = 3. 5 cm.
(ii) With Bas centre and radius 5 cm draw an arc.
(iii) With C as centre and radius 4 cm draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
(v) Dr aw perpendi cu I ar bi sector of BC.
(vi) Dr aw the angle bi sector of angle ABC which intersects the perpendicular bisector of BC at P.
Pis the required point which is equidistant from AB, BC, Band C.
The length of PB = 2.5 cm

OP bisects ∠ SPQ and OQ bisects ∠ PQR.
Draw OM perpendirular to RQ and OL perpendirular to SP
Now in Δ OQM and Δ OLP
∠ OLP = ∠ OMQ
∠ OPL = ∠OQM
OP= OQ
Therefore, Δ OQM and Δ OLP are oongruent.
Hence, OL = OM
O is equidistant from PS and QR. Proved.

Since I lies on bisector of ∠ R, I is equidistant from PR and QR.
Again I lies on the bisector of ∠ Q , I is equidistant from PQ and QR.
Hence , I is equi distant from alI sides of the triangle.
Therefore, I lies on the bisector of ∠ P i.e ∠ QPR .

Steps of Construction:
(i) Draw YZ = 5 cm
(ii) Draw an arc with angle Y= 120 ° and radius 4 cm.
(iii) Join XZ.
(iv) Draw perpendicular bisector of YZ.
(v) With X as centre and angle X as 90° , join X to the perpendicular bisectcr at T. T is the required point.
(vi) Measure TY. TY = 6.8 cm = TZ as T lies on perpendicular bisector of YZ.

Steps of oonstruction:
(i) Draw PQ = 5.5 cm
(ii) With P as centre and radius 4.8 cm draw an arc.
(iii) With Q as centre and radius 3.2 cm cut another arc which meets the first arc at R. Join PR and QR. PQR is the required triangle.
(iv) Draw perpendicular bisector of PR.
(v) Q as centre and radius as 2.5 cm, draw an arc which intersects the perpendicular bisector of PR at O.
O is the required point which is at a distance of 2.5 cm from Q.

Steps of Construction:
(i) Draw BC = 6 cm.
(ii) Draw AD = 8 on perpendicular to BC.
(iii) With B as centre draw arcs on AD.
(iv) With C as centre draw arcs on AD. ABCD is the required rhombus.
(v) Draw perpendicular bisectors of AB, and CD, which meet at 0.
(vi) Since AD and BC are diagonals of rhombus and meet at 0.
AO = OD
O is the point equidistant from AB, AD and C, D.

Steps of Construction:
(i) Draw AC= 6 cm.
(ii) With A as centre, draw two arcs of 5 cm on both sides of line AC.
(iii) With C as centre, draw two arcs of 5 cm on both sides of line AC.
(iv) All the arcs meet at Band D. Join AB, AD, BC and BD. ABCD is the required rhombus.
(v) On measuring, ∠ ABC = 78>.
(vi) Draw perpendicular bisector of BC meeting AD at R. R is the pdnt equidistant from Band C, hence RB = RC.
(vii) On measuring, R = 1.2 cm

Steps of construction:
(i) Draw a line AB = 6 cm.
(ii) Draw a ray making an angle of 45° with AB.
(iii) With a as centre, draw AD = 6 cm on the ray.
(iv) Draw an angle bisector of angle BAD.
(v) With Bas centre cut an arc BC = 3.6 cm on the angle bisector.
(vi) With Das centre cut an arc CD = 5 cm on the angle bisector. ABCD is the required quadrilateral.
(vii) Join BD.
(viii) Draw perpendicular bisectors of CD and BC which meet BD on P. Pis the required point.

Steps of Construction:
(i) ABC is the required triangle.
(ii) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from Band C.
(iii) Through A, draw a line m 11 BC.
(iv) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(v) Then triangle QBC is equal in area to triangle ABC. mis the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.

In Δ AMPand Δ DMP
MP = MP
AM = MD
∠ AMP = ∠ DMP = 90°
Therefore, Δ AMPand Δ DMP are congruent.
DP= AP ....... (i)
In Δ ANP and Δ BNP
NP= NP
AN= NB
∠ANP = ∠BNP = 90°
Therefore, Δ ANP and Δ BNP are congruent.
BP= AP ....... (ii)
From (i) and (ii)
BP= DP
Hence, proved.

Since O lies on the perpendirular bisector of AB, O is equidistant from A and B.
OA = OB ........ (i)
Again, O lies on the perpendirular bisector of AC, O is equidistant from A and C.
OA = OC ......... (ii)
From (i) and (ii)
OB= OC
Now in Δ OBM and Δ OCM,
OB = OC (proved)
OM=OM
BM =CM ( M is mid-point of BC)
Therefore, Δ OBM and Δ OCM are congruent.
∠ OMB= ∠ OMC
But BMC is a straight line, so
∠ OMB =∠ OMC = 90°
Thus, OM meets BC at right angles.

Join BQ and draw perpendicular bisector of AC cutting AC at L.
In Δ QBN and ΔQCN
QN = QN
BN =NC
∠ QNB = ∠ QNC = 90 degree.
Therefore, ∠ QBN and ∠.QCN are congruent .
Hence Q is equidistant from B and C.
In Δ QNC and Δ QLC
QC= QC
∠ QLC = ∠ QNC = 90 degree.
∠ QCL =∠ QCN (PC being angle bisector)
Therefore, .Δ QNC and Δ QLC are congruent.
Therefore, QL = QN.
Hence Q is equidistant from BC and AC.

Steps of Construction:
(i) Draw line segment PQ.
(ii) With P and Q as centre draw intersecting arcs at R.
(iii) Join PR and RQ.
(iv) Draw angle bisector of angle Q.
(v) Draw perpendicular bisectors of PQ and RQ which meet the angle bisector at S. S is the required point.
(vi) In Δ QSY and Δ QSX
SQ= SQ
∠ SQY = ∠ SQX
∠ SYQ = ∠ SXQ = 90 degrees.
Therefore, Δ QSY and Δ QSX are congruent.
Hence, SY = SX and therefore S is equidistant from PQ and RQ.

Steps of oonstruction:
(i) Draw a line segment AB of 9 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
Proof:
(i) Take any point on LM say P.
(ii) Join PA and PB.
Since, Plies on the right bisector of line AB.
Therefore, Pis 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.