The following steps will be helpful before drawing the actual figure.
Construction of a parallelogram whose area is equal to the area of given triangle when
(a) One angle (b) one side of the parallelogram are given:
(a) Construct a triangle ABC in which AB = 5cm, BC = 6cm and AC = 7cm and construct a parallelogram whose area is equal to the area of given triangle having one angle 60°.
Steps :
∴ PQRC is the required parallelogram.
(b) Construct a triangle ABC in which AB = 4cm, BC = 5cm and ∠B = 60° and then construct a parallelogram having a side 5.2 cm and equal area to the triangle.
Steps :
∴ PQRC is the required parallelogram.
Construct a triangle ABC in which AB = 6.3 cm, BC = 4.5cm and AC = 3.2ccm then construct a rectangle equal area to the triangle.
Steps:
∴ BPQR is the required rectangle.
Construct a triangle ABC in which AB = 6.3 cm, BC = 7.8 cm and AC = 7.2 cm and construct another triangle PBC equal area toΔABC.
Steps:
∴ PBC is the required triangle.
Construct a parallelogram ABCD in which AB = 5.5 cm, BC = 4.8 cm and∠ABC = 75° and construct another parallelogram equal area to the parallelogram ABCD.
Steps:
∴ BCQR is the required parallelogram.
Construct a quadrilateral ABCDin which AB = 2.8 cm BC = 3.6 cm, AC = 3 cm, CD = 1.7 cm and AD = 2.3 cm and construct a triangle equal area to the quadrilateral ABCD.
Steps:
∴ ABE is a required triangle.
Construct a triangle ABC in which a = 7.8cm b =7.2 cm and c = 6.3 cm and construct a quadrilateral having an equal area to the triangle ABC.
Steps:
∴ ABDE is the required quadrilateral.
Helpful steps to before drawing actual figure:
Construct:
Quadrilateral PQRS and \(\triangle\)PSZ
Results:
Area of quadrilteral PQRS = Areaof \(\triangle\)PSZ where;
PQ = 4.2 cm, QR = 6.5 cm, RS = 8 cm, SP = 5.3 cm, SQ = 7.4 cm.
To construct:
Quadrilateral PQRS and \(\\triangle\)PST
Result:
Area of quadrilateral PQRS = Area of \(\triangle\)PST
where, PQ = QR = 5.1 cm, PS = RS = 6.2 cm and QS= 5.6 cm.
To construct:
parallelogram ABCD and \(\triangle\)AEF
Result:
Area of pallelogram ABCD = Area of \(\triangle\)AEF
where, AB = 6 cm, BC = 4 cm, \(\angle\)BAD = 60° and one side of \(\triangle\)AEF = 7.5 cm.
To construct:
quadrilateral ABCD and \(\triangle\)ABE
Result:
Area of quadrilateral ABCD = Area of \(\triangle\)ABE
where,AB = BC = 5.6 cm, CD = AD = 4.9 cm and \(\angle\)BAD = 60°.
Construct:
quadrilateral ABCD and \(\triangle\)ADE
Results:
Area of quadrilateral ABCD = Area of \(\triangle\)ADE
where, Ab = 4.2 cm, BC = 4.8 cm, CD = 5.4 cm, DA = 5.8 cm.
To construct:
quadrilateral ABCD and rectangle ASTU
Result:
area of quadrilateral ABCD = Area of rectangle ASTU
where, AC = 6.6 cm, BD = 8 cm, AB = 5 cm.
To construct:
quadrilateral PQRS = \(\triangle\)PST
Results:
Area of quadrilateral PQRS = Area of (\triangle\)PST
where,PQ = QR = 5.9 cm, RS = Ps = 6.1 cm and \(\angle\)QPS = 75°.
To construct:
quadrilateral ABCD = \(\triangle\)ADE
Results:
Area of quadrilateral ABCD = Area of \(\triangle\)ADE
where,AB = 8 cm, BC = 3.5 cm, CD = 7 cm, DA = 6 cm and \(\angle\)BAD = 60°.
To construct:
\(\angle\)ABC and parallelogram BEFD
Results:
Area of \(\triangle\)ABC = Area of parallelogram BEFD
where a = 7.8 cm, b = 7.2 cm, c = 6.3 cm and an angle of parallelogram \(\angle\)DBC = 75°.
To construct:
\(\triangle\)PQR and rectangle ARCB
Results:
Area of \(\triangle\)PQR = Area of rectangle ARCB
in whichPQ = 6 cm, QR = 7 cm and RP = 4 cm.
To construct:
\(\triangle\)ABC and \(\triangle\)DBC
Results:
Area of \(\triangle\)ABC = Area of \(\triangle\)DBC
where, AB = 4.5 cm, BC = 6.5 cm, \(\angle\)C = 60°, DB = 8 cm.
Construct:
\(\triangle\)ABC and parallelogram PCQR
Resuts:
Area of \(\triangle\)ABC = Area of parllelogram PCQR
where, a = 7.8 cm, b = 7.2 cm, c = 6.3 cm and \(\angle\)RPC = 75°.
To construct:
\(\triangle\)PQR and parallelogram TQSU
Results:
Area of \(\triangle\)PQR = Area pf parallelogram TSQU
in whichPQ = 7.5 cm, QR = 6.8 cm and RP = 6 cm and TQ = 6.4 cm.
Construct:
\(\triangle\)ABC and parallelogram PBQR
Results:
Area of \(\triangle\)ABC = Area of parallelogram PBQR
where,a = 5 cm, b = 4.8 cm and c = 6.8 cm and \(\angle\) RPB = 45°.
Result:
Area of parallelogram PQRS = Area of \(\triangle\)PSA
where, QS = 8 cm, PR = 6 cm and PQ = 5 cm.
Results:
Area of \(\triangle\)ABC = Area of rectngle BDEF
where AB = 4 cm, BC = 6.8 cm and CA = 6.5 cm.
Results:
Area of quadrilateral ABCD = Area of \(\triangle\) ADE
where, AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm, AD = 4 cm and \(\angle\)ABC = 60°.
Results:
Area of quadrilateral PQRS = Area of \(\triangle\)QRT
Results:
Area of rectangle ABCD = Area of \(\triangle\)ABE
ASK ANY QUESTION ON Construction
You must login to reply
Ishwor
Construct a quadrilateral PQRS in which QR= 5 cm, RS = 5cm, PS=5.7, PQ = 6.2 cm and PR = 5.6, then constuct a triangle RQT equal in area to the quadrilateral PQRS.
Mar 18, 2017
0 Replies
Successfully Posted ...
Please Wait...
how to draw parallel line in quadrilateral
frgt
Mar 10, 2017
0 Replies
Successfully Posted ...
Please Wait...
sanchita
constuction of rectangle frm triangle
Mar 06, 2017
1 Replies
Successfully Posted ...
Please Wait...
construct a parallelogram ABCD such that AB=4.8cm ,BC=4cm and diagonal BD=5.4
Solve this problem
Feb 10, 2017
1 Replies
Successfully Posted ...
Please Wait...