Notes on Construction | Grade 10 > Compulsory Mathematics > Geometry | KULLABS.COM

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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: (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.

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 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.

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

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## ASK ANY QUESTION ON Construction

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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.

frgt

##### sanchita

constuction of rectangle frm triangle