Mathematics [4 marks sum]

Steps of Construction :
(i) Draw a line segment BC = 5.8 cm.
(ii) With Band C as centre and radius = 5.8 cm, draw two arcs which intersect each other at A.
(iii) Join AB and AC. ABC is the required triangle.
(iv) Dr aw the bi sectors AD, BE and CF of L.A, L. Band L. C respectively.
Hence, AD, BE and CF are the lines of symmetry of trianQle ABC.

Steps of construction:
(i) Plot the point A, B and Con the graph.
(ii) Point symmetric to A( 4, 1) about x = 7 is D( 10, 1)
(iii) Point symmetric to B(2,3) about x = 7 is E( 12, 3)
(iv) Point symmetric to C(S,6) about x = 7 is F(9,6)
(v) Join AB, AC, BC, AD, DE, DF, EF and CF.
The figure formed is a trapezium ADCF with two equal scalene triangles (ABC and DEF) attached to it.

Steps of construction:
(i) Plot the point A and B on the graph.
(ii) Point symmetric to A(2,2) about x-axis is C(2,-2)
(iii) Point symmetric to B(5,5) about x-axis is 0(5,-5)
(iv) Join AB, AC, CD, BD.
The figure formed is a trapezium.

Steps of construction :
(i) Plot the points P, Q and Ras per given data.
(ii) Point S symmetrical about QR is a point with vertices x= 12 and y = 1 i.e. 5 unit right of line RQ.
(iii) Plot S( 12, 1)
(iv) Join PR, PS and RS

Steps of oonstruction:
(i) Plot the points A, Band Casper given data.
(ii) Point D symmetrical about AB is a point with vertices x= -1 and y = 4 1. e. 3 units left of line AB.
(iii) Plot D( -1,4)
(iv) Join BC, AB, BD
The figure is an arrow.

Steps of oonstruction:
(i) Draw a line segment QR = 2.6 cm
(ii) At R draw a perpendicular to QR.
(iii) With Q as centre and radius 5.2 cm cut an arc on perpendicular to R at P.
(iv) Join PQ . Δ PQR is the required triangle.
(v) Produce QR to S such that RS = 2.6 cm
(vi) With S as centre and radius 5.2 cm cut an arc on perpendicular to Rat P.
(vii) Join PS. Δ PSR is the triangle which is the reflection of &PQR.
Δ PQS is the required triangle and is an equilateral triangle.

Steps of construction:
(i) Draw a line segment AC= 4.5 cm
(ii) With Band C as centres and 6 cm as radius 1 draw arcs which intersect each other at B.
(iii) Join AB and BC. Δ ABC is the required triangle.
(iv) Again with Band C as centres and 6 cm as radius 1 draw arcs which intersect each other at D.
(v) Join AD and DC. Δ ADC is the triangle which is the reflection of Δ ABC.
ABCD is the required quadrilateral and it is a rhombus.

Steps of construction:
(i) Draw a line segment BC= 8 cm
(ii) Draw its perpendicular bisector whidi intersects BC at D. With Das centre and BD or CD as radius, draw a semi-circle.
(iii) Produce the perpendicular bi sector of BC which intersects the ci rel e at A.
(iv) Join AB and AC. Triangle ABC is the required isosceles right-angled triangle.
The perpendicular bisector of hypotenuse BC is the line of symmetry.

Steps of construction :
(i) Draw a line segment AB = 5 cm
(ii) With A as centre and radius 8 cm, and 8 as centre and radius 5 cm, draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and C and radius 5 cm, draw arcs which intersect each other at D
(v) Join AD and CD. ABCD is the required rhombus.
(vi) Join BO.
Two diagonals AC and BO are the lines of symmetry .

Steps of construction :
(i) Draw a line segment AB = 3.8 cm
(ii) At A and B, draw rays making an angle of 120° eadi and cut off AF = BC = 3.8 cm
(iii) Again at F and C, draw rays making an angle of 120• each and cut off CD = FE= 3.8 cm
(iv) Join DE. ABCDEF is the required hexagon.
(v) Draw perpendicular bisectors of each of the opposite sides and al so join AD 1 BE and CF. These six lines are the lines of symmetry .

Steps of Construction :
(i) Draw a line segment AB = 4. 8 cm.
(ii) At A and B, draw perpendiculars AX and BY
(iii) From AX and BY, cut off AD = BC= 4. 8 cm
(iv) Join DC. ABCD is the required square.
(v) Now cir aw perpendi cu I ar bi sectors of AB and AD.
(vi) Also join the diagonals AC and BD.
The perpendicular bisectors and the diagonals are the lines of symmetry

Steps construction :
(i) Draw a line segment QR = 5.4 cm
(ii) At Q , draw a ray making an angle of 60 degrees with QR and OJt QP = SR = 6cm
(iii) Pas centre draw an arc equal to 5. 4 cm
(iv) Ras centre draw an arc equal to 6 cm which intersects the first arc at S.
(v) Join RS and PS. PQRS is the required parallelogram.
(vi) Join QS and PR which intersect each other at 0.
There is no line of symmetry of parallelogram PQRS but it has one point symmetry which is 0, the point of intersection of its diagonals.

Steps of construction:
>(i) Draw a line segment AB = 5 cm > >(ii) With A as centre, cut an arc of 7 cm on one side of line segment AB. > >(iii) With B as centre, cut an arc of 7 cm on same side of line segment AB. Let the point be C. > >(iv) Join AC and BC. ABC is the required triangle. > >(v) Draw angle bisector of angle C meeting AB at D. > >(vi) CD is perpendicular bisector of AB and AC=BC. Hence CD is the line of symmetry. > >Isosceles triangle has only one line of symmetry.