 ## Construction

Subject: Compulsory Mathematics

#### Overview

The theorems related to the area of a triangle is half of parallelogram standing on the same base and between same parallels, the area of parallelograms on the same base and between the same parallels are equal and so on.

##### Construction

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: (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. 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. 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. 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. 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. 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.

##### Things to remember

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.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Construction 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:

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.

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°.