Construction Of Quadrilaterals

GIVEN : In quadrilateral ABCD ,

AB = 5.5 cm

BC = 3.5 cm

CD = 4 cm

AD = 5 cm

∠A = 45^o

PROCEDURE :

Step 1 : Draw a rough sketch of the required quadrilateral and mark the given measurements.

Step 2 : Draw ΔDAB using S.A.S property of construction , by taking AD = 5 cm , ∠ DAB = 45° and AB = 5.5 cm.

Step 3 : To locate the fourth vertex ‘C’ , draw an arc , with center D and radius 4cm (CD=4cm).

Draw another arc with center B and radius 3.5 cm (BC=3.5cm) which cuts the previous arc at C.

Step 4: Join DC and BC to complete the required quadrilateral ABCD.

GIVEN : In quadrilateral ABCD ,

BE = 2.9 cm

ES = 3.2 cm

ST = 2.7 cm

BT = 3.4 cm

∠B = 45^o

PROCEDURE :

Step 1 : Draw a rough sketch of the required quadrilateral and mark the given measurements.

Step 2 : Draw ΔTBE using S.A.S property of construction , by taking BT = 3.4 cm , ∠ TBE = 75° and BE = 2.9 cm.

Step 3 : To locate the fourth vertex ‘S’ , draw an arc , with center T and radius 2.7cm (TS=2.7cm).

Draw another arc with center E and radius 3.2 cm (ES=3.2cm) which cuts the previous arc at S.

Step 4: Join TS and ES to complete the required quadrilateral BEST.

GIVEN : In Parallelogram PQRS,

PQ = 4.5 cm

QR = 3 cm

∠PQR = 60^o

PROCEDURE :

Step 1: Draw a rough sketch of the parallelogram and mark the given measurements.

Here , we are given only 3 measurements. But as PQRS is a parallelogram , we can also write that RS = PQ = 4.5 cm and SP = QR = 3 cm.

(now we got 5 measurements in total)

Step 2 : Draw ΔPQR using the measures PQ = 4.5 cm , ∠PQR = 60°

and QR = 3 cm.

Step 3 : Locate the 4^th vertex ‘S’ using the other 2 measurements PS = 3 cm and RS = 4.5 cm. To locate the fourth vertex ‘S’ , draw an arc , with center P and radius 3cm (PS=3cm).

Draw another arc with center R and radius 4.5cm (RS=4.5cm) which cuts the previous arc at S.

Step 4 : Join RS and PS to complete the required parallelogram.

GIVEN : In Rhombus MATH,

AT = 4 cm

∠MAT = 120^o

PROCEDURE :

Step 1 : Draw a rough sketch of the required quadrilateral and mark the given measurements.

Here , we are given only 2 measurements. But as MATH is a rhombus , we can also write that MA = AT = TH = HM = 4cm.

(now we got 5 measurements in total)

Step 2 : Draw ΔMAT using S.A.S property of construction , by taking MA = 4 cm , ∠ MAT = 120° and AT = 4 cm.

Step 3 : To locate the fourth vertex ‘H’ , draw an arc , with center T and radius 4cm (TH=4cm).

Draw another arc with center M and radius 4 cm (MH=4cm) which cuts the previous arc at H.

Step 4: Join TH and MH to complete the required rhombus MATH.

GIVEN : In Rectangle FLAT,

FL = 5 cm

LA = 3 cm

∠FLA = 90^o

PROCEDURE :

Step 1: Draw a rough sketch of the parallelogram and mark the given measurements.

Here , we are given only 3 measurements. But as FLAT is a rectangle , we can also write that FL = AT = 5 cm and LA = TF = 3 cm.

(now we got 5 measurements in total)

Step 2 : Draw ΔFLA using the measures FL = 5 cm , ∠FLA = 90°

and LA = 3 cm.

Step 3 : Locate the 4^th vertex ‘T’ using the other 2 measurements TF = 3 cm and AT = 5 cm. To locate the fourth vertex ‘T’ , draw an arc , with center F and radius 3cm (FT=3cm).

Draw another arc with center A and radius 5cm (AT=5cm) which cuts the previous arc at T.

Step 4 : Join AT and FT to complete the required rectangle FLAT.

GIVEN : In Square LUDO,

LU = 4.5 cm

∠LUD = 90^o

PROCEDURE :

Step 1 : Draw a rough sketch of the required quadrilateral and mark the given measurements.

Here , we are given only 2 measurements. But as LUDO is a square , we can also write that LU = UD = DO = OL = 4.5 cm.

(now we got 5 measurements in total)

Step 2 : Draw ΔLUD using S.A.S property of construction , by taking LU = 4.5 cm , ∠ LUD = 90° and UD = 4.5 cm.

Step 3 : To locate the fourth vertex ‘O’ , draw an arc , with center L and radius 4.5cm (LO=4.5cm).

Draw another arc with center D and radius 4.5 cm (DO=4.5cm) which cuts the previous arc at O.

Step 4: Join LO and DO to complete the required square LUDO.

GIVEN : In quadrilateral ABCD ,

AB = 4.5 cm

BC = 5.5 cm

CD = 4 cm

AD = 6 cm

AC = 7 cm

PROCEDURE :

Step 1 : draw a rough sketch of the quadrilateral ABCD with the given measurements.

Step 2 : Construct ΔABC using SSS construction property with AB = 4.5 cm , BC = 5.5 cm and AC = 7 cm.

Step 3 : we have to locate the 4^th vertex ‘D’ . it would be on the other side of AC. So , with center A and radius 6 cm(AD = 6 cm) draw an arc and with center C and radius 4 cm (CD = 4 cm) draw another arc to cut the previous arc at D.

Step 4 : join A,D and C,D to complete the quadrilateral ABCD.

GIVEN : In quadrilateral PQRS ,

PQ = 3.5 cm

QR = 4 cm

RS = 5 cm

PS = 4.5 cm

QS = 6.5 cm

PROCEDURE :

Step 1 : draw a rough sketch of the quadrilateral PQRS with the given measurements.

Step 2 : Construct ΔPQS using SSS construction property with PQ = 4.5 cm , PS = 4.5 cm and QS = 6.5 cm.

Step 3 : we have to locate the 4^th vertex ‘R’ . it would be on the other side of QS. So , with center Q and radius 4 cm(QR = 4 cm) draw an arc and with center S and radius 5 cm (SR = 5 cm) draw another arc to cut the previous arc at R.

Step 4 : join S,R and Q,R to complete the quadrilateral PQRS.

GIVEN : In Parallelogram ABCD ,

AB = 6 cm

BD = 7.5 cm

BC = 4.5 cm

PROCEDURE :

Step 1 : draw a rough sketch of the Parallelogram ABCD with the given measurements.

Here , we are given only 3 measurements. But as ABCD is a parallelogram , we can also write that AB = CD = 6 cm and BC = AD = 4.5 cm.

Step 2 : Construct ΔABD with AB = 4.5 cm , AD = 4.5 cm and BD = 7.5 cm.

Step 3 : we have to locate the 4^th vertex ‘C’ . it would be on the other side of BD. So , with center B and radius 4.5 cm(BC = 4.5 cm) draw an arc and with center D and radius 6 cm (CD = 6 cm) draw another arc to cut the previous arc at C.

Step 4 : join C,B and C,D to complete the quadrilateral ABCD.

GIVEN : In Rhombus NICE,

NI = 4 cm

IE = 5.6 cm

PROCEDURE :

Step 1 : draw a rough sketch of the rhombus. Hence all the sides are equal , so , NI = IC = CE = NE = 4 cm and mark the given measurements.

Step 2 : draw ΔNIE using SSS construction with measures NI = 4cm , IE= 5.6 cm and EN = 4cm.

Step 3 : we have to locate the 4^th vertex ‘C’ . it would be on the other side of IE. So , with center I and radius 4 cm(IC = 4 cm) draw an arc and with center E and radius 4 cm (EC = 4 cm) draw another arc to cut the previous arc at C.

Step 4 : Join I,C and C,E to complete the required Rhombus NICE.

GIVEN : in Quadrilateral GOLD,

OL = 7.5 cm

GL = 6 cm

LD = 5 cm

DG = 5.5 cm

OD = 10 cm

PROCEDURE :

Step 1 : we first draw the rough sketch of the Quadrilateral GOLD.

Step 2 : draw ΔOLD using SSS construction property with measures OL = 7.5 cm , LD = 5 cm and OD = 10 cm.

Step 3 : with center L and radius 6 cm (LG = 6cm) and with center D and radius 5.5 cm (DG = 5.5 cm) , draw 2 arcs opposite to vertex L to locate G.

Step 4 : Join G,D , L,G and G,O to complete the Quadrilateral GOLD.

GIVEN : in Quadrilateral PQRS,

PQ = 4.2 cm

QR = 3 cm

PS = 2.8 cm

PR = 4.5 cm

QS = 5 cm

PROCEDURE :

Step 1 : we first draw the rough sketch of the Quadrilateral PQRS

Step 2 : draw ΔPQR using SSS construction property with measures PQ = 4.2 cm, QR = 3 cm and PR = 4.5 cm.

Step 3 : with center Q and radius 5 cm (QS = 5 cm) and with center P and radius 2.8 cm (PS = 2.8 cm) , draw 2 arcs opposite to vertex Q to locate S.

Step 4 : Join S,P , Q,S and S,R to complete the Quadrilateral PQRS.

GIVEN : In Quadrilateral HELP,

HE = 6cm

EL = 4.5 cm

∠H=60^o

∠E =105^o

∠P= 120^o

PROCEDURE :

Step 1: draw a rough sketch of the Quadrilateral HELP and mark the given measurements.

As we can see that ∠P is not between the given 2 sides , so we now find the ∠L that is between HE and EL using the property of sum of all angles of a quadrilateral ie. ∠L = 360° - (∠H + ∠E + ∠P) = 360° - (60° + 105° + 120°) = 75°.

∴ ∠L = 75°.

Step 2: construct ΔHEL using SAS property of construction model with HE = 6cm , ∠E = 105^o and EL = 4.5 cm.

Step 3 : Construct ∠H = 60° and draw HY .

( how to draw 60° angle ?

⇒an arc is drawn from H . let it intersect HE at H’ . with center H’ and with same radius draw 2 arcs to cut at 2 points A,B which gives 60° and 120° respectively.

So , draw a line from H which passes through A to get the required angle. )

Step 4 : construct ∠L = 75^o and draw LZ to meet HY at P.

HELP is the required quadrilateral.

GIVEN : In Parallelogram GRAM,

GR = AM = 5 cm

RA = MG = 6.2 cm

∠R = 85^o

PROCEDURE :

Step 1 : draw a rough sketch of the parallelogram GRAM and mark the given measurements.

Since the given measurements are not sufficient for construction , we shall find the required measurements using the properties of the parallelogram.

As opposite angles of parallelogram are equal so , ∠R = ∠M = 85^o and as the consecutive angles are supplementary so , ∠G = 180° - 85° = 95°.

Thus ∠G = ∠A = 95°.

Step 2 : construct ΔGRA using SAS property of construction model with GR = 5cm , ∠R = 85^o and RA = 6.2 cm.

Step 3 : construct ∠G = 95° and draw GY || RA.

Step 4 : construct ∠A = 95° and draw AN to meet GY at M.

GRAM is the required quadrilateral (ie. Parallelogram).

GIVEN : In Rectangle FLAG,

FL = 6 cm

LA = 4.2 cm

PROCEDURE :

Step 1 : draw a rough sketch of the Rectangle FLAG and mark the given measurements.

Since the given measurements are not sufficient for construction , we shall find the required measurements using the properties of the RECTANGLE.

As opposite sides of rectangle are equal so, FL = AG = 6cm and LA = GF = 4.2 cm and ∠F = ∠L = ∠A = ∠G = 90°.

Step 2 : construct ΔFLA using SAS property of construction model with FL = 6cm , ∠L = 90^o and LA = 4.2 cm.

Step 3 : construct ∠F = 90° and draw FY || LA.

Step 4 : construct ∠A = 90° and draw AN to meet FY at G.

FLAG is the required quadrilateral (ie. rectangle).

GIVEN : In Quadrilateral PQRS,

PQ = 3.6cm

QR = 4.5 cm

RS = 5.6 cm

∠PQR = 135^o and ∠QRS = 60^o

PROCEDURE :

Step1 : draw a rough sketch and mark the measurements given.

Step 2 : draw ΔPQR using SAS construction rule with measures PQ = 3.6cm, ∠PQR = 135^o and QR = 4.5 cm.

Step 3 : construct ∠R = 60° and draw RY.

Step 4 : with center ‘R’ and radius 5.6cm (RS = 5.6 cm) draw an arc to intersect RY at S. Join P,S . PQRS is the required quadrilateral.

GIVEN : In Quadrilateral LAMP,

AM = MP = PL = 5cm

∠M = 90^o and ∠P = 60^o.

PROCEDURE :

Step1 : draw a rough sketch and mark the measurements given.

Step 2 : draw ΔAMP using SAS construction rule with measures AM = 5cm, ∠M = 90^o and MP = 5 cm.

Step 3 : construct ∠P = 60° and draw PY.

Step 4 : with center ‘P’ and radius 5cm (PL = 5 cm) draw an arc to intersect PY at L. Join L,P . LAMP is the required quadrilateral.

GIVEN : In Trapezium ABCD,

AB || CD

AB = 8 cm

BC = 6cm

CD = 4cm

∠B = 60^o.

PROCEDURE :

Step1 : draw a rough sketch and mark the measurements given.

As it is given that AB || CD , so ∠B + ∠C = 180° (linear pair). So ∠C = 180° – 60° = 120°

Step 2 : draw ΔABC using SAS construction rule with measures AB = 8 cm, ∠B = 60^o and BC = 6 cm.

Step 3 : construct ∠C = 120° and draw CY.

Step 4 : with center ‘C’ and radius 4cm (CD = 4 cm) draw an arc to intersect CY at D. Join A,D . ABCD is the required quadrilateral (trapezium).

GIVEN : In rhombus CART,

CR = 6 cm

AT = 4.8 cm (diagonals)

PROCEDURE :

Step 1 : draw a rough sketch of rhombus CART and mark th given measurements.

The diagonals of a rhombus bisect each other perpendicularly.

CR and AT are diagonals of the rhombus CART which bisect each other at ‘O’ ie. ∠ COA = 90° and AO = OT = = = 2.4cm.

Step 2 : draw CR = 6cm (one diagonal of the rhombus CART) and draw a perpendicular bisector XY of it and mark the point of intersection as ‘O’.

Step 3 : as the other diagonal AT is perpendicular to CR , AT is a part of XY. So, with center ‘O’ and radius 2.4 cm (AO = OT = 2.4cm) draw 2 arcs on either sides of CR to cut XY at A and T.

Step 4 : join C,A ;A,R ; R,T ; C,T to complete the required rhombus CART.

GIVEN : In rhombus SOAP,

SA = 4.3 cm

OP = 5 cm (diagonals)

PROCEDURE :

Step 1 : Draw a rough sketch of rhombus SOAP and mark th given measurements.

The diagonals of a rhombus bisect each other perpendicularly.

SA and OP are diagonals of the rhombus SOAP which bisect each other at ‘B ie. ∠ SBO = 90° and OB = BP = = = 2.5cm.

Step 2 : Draw SA = 4.3cm (one diagonal of the rhombus SOAP) and draw a perpendicular bisector XY of it and mark the point of intersection as ‘B’.

Step 3 : As the other diagonal OP is perpendicular to SA , OP is a part of XY. So, with center ‘B’ and radius 2.5 cm (OB = BP = 2.5cm) draw 2 arcs on either sides of SA to cut XY at O and P.

Step 4 : Join S,O ;O,A ; A,P ; S,P to complete the required rhombus SOAP.

GIVEN : In square JUMP,

diagonal is 4.2 cm ie. JM = UP = 4.2cm

PROCEDURE :

Step 1 : draw a rough sketch of square JUMP and mark th given measurements.

The diagonals of a rhombus bisect each other perpendicularly.

JM and UP are diagonals of the square JUMP which bisect each other at ‘B ie. ∠ JBU = 90° and UB = BP = = = 2.1cm.

Step 2 : draw JM = 4.2cm (one diagonal of the square JUMP) and draw a perpendicular bisector XY of it and mark the point of intersection as ‘B’.

Step 3 : as the other diagonal UP is perpendicular to JM , UP is a part of XY. So, with center ‘B’ and radius 2.1 cm (OU = BP = 2.1cm) draw 2 arcs on either sides of JM to cut XY at U and P.

Step 4 : join J,U ;U,M ; M,P ; J,P to complete the required square JUMP.