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The following steps will be helpful before drawing the actual figure.

  1. Draw a rough sketch of a figure.
  2. Mark the given measurement in it.
  3. Analyze the figure and plan the steps.

1. Construction of a parallelogram and a triangle having equal area.

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:

SDF

(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 :

  1. DrawΔABC with AB = 5cm, BC = 6cm and AC = 7cm.
  2. Draw XY parallel to BC through the point A.
  3. Take P as mid-point of BC.
  4. Draw an angle of60° at P.
  5. Cut PC = QR and join the point R and C.
  6. Parallelogram PQRC andΔABC are equal in area.

∴ 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.

DSF

Steps :

  1. DrawΔABC with AB = 4cm, BC = 5cm and∠B = 60°.
  2. Draw XY parallel to BC through the point A.
  3. Take P as mid-point of BC.
  4. From P, cut PQ = 5.2cm on XY.
  5. CutPC = PQ and join the point R and C.
  6. Parallelogram PQRC andΔABC have equal area.

∴ PQRC is the required parallelogram.

2. Construction of rectangle equals in the area to given triangle.

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:

  1. DrawΔABC with AB = 6.3 cm BC = 4.5 cm and AC = 3.2 cm.
  2. Through A, draw XY//BC.
  3. Draw the perpendicular bisector PQ of BC.
  4. Draw BP = RQ and join the points R and B.
  5. Rectangle BPQR is the required rectangle equal toΔABC.

∴ BPQR is the required rectangle.

3. Construction of two triangles of equal area on the same base and between the same parallels.

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.

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Steps:

  1. Draw ΔABC withAB = 6.3 cm, BC = 7.8 cm and AC = 7.2 cm .
  2. Through A, draw XY//BC.
  3. Take any point P in XY and join P to B and C.
  4. ABC and PBC are the triangles of equal area.

∴ PBC is the required triangle.

4. Construction of two parallelograms of equal area on the same base and between the same parallels.

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.

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Steps:

  1. Draw a parallelogram ABCD having AB = 5.5 cm, BC = 4.8 cm and∠ABC =75°.
  2. Take two points R and Q in XY such that BC = RQ.
  3. Join R to B and Q to C.
  4. BCQR is a parallelogram equal in area to parallelogram ABCD.

∴ BCQR is the required parallelogram.

5. Construction of a triangle equal in area to the given quadrilateral.

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.

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Steps:

  1. Draw aquadrilateral ABCDin which BC = 3.6 cm, AB = 2.8 cm, AC = 3 cm,AD = 2.3 cm and CD = 1.7 cm.
  2. From D, draw DE parallel to AC.
  3. Produce BC to E.
  4. Join A to E.
  5. ABE is a triangle equal area to the quadrilateral ABCD.

∴ ABE is a required triangle.

6. Construction of a quadrilateral equal in area to the given 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.

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Steps:

  1. DrawΔABC in which BC = a = 7.8 cm, BA = c = 6.3 cm and AC = b = 7.2 cm.
  2. Take any point D on BC.
  3. Draw DA//CP.
  4. Take any point E on CP.
  5. ABDE is a quadrilateral equal area toΔABC.

∴ ABDE is the required quadrilateral.

Helpful steps to before drawing actual figure:

  1. Draw a rough sketch of the figure.
  2. Mark the given measurement in it.
  3. Analyze the figure and plan the step.
.

Very Short Questions

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.

figure

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.

figure

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.

figure

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°.

figure

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.

figure

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.

figure

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°.

figure

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°.

figure

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°.

figure

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.

figure

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.

figure

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°.

figure

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.

figure

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°.

figure
figure

Results:

Area of \(\triangle\)ABC = Area of rectngle BDEF

where AB = 4 cm, BC = 6.8 cm and CA = 6.5 cm.

figure
figure

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°.

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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.


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how to draw parallel line in quadrilateral

frgt


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sanchita

constuction of rectangle frm triangle


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construct a parallelogram ABCD such that AB=4.8cm ,BC=4cm and diagonal BD=5.4

Solve this problem


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