Topics to be learn : Construction of quadrilaterals Parallelogram Rectangle Square Rhombus Trapezium Kite

Recall :

Construct the triangles with given measures.

(1) Δ ABC : l (AB) = 5 cm, l (BC) = 5.5 cm, l (AC) = 6 cm

Steps of Construction :

1. Draw base BC: Using a ruler, draw a straight line segment BC of length 5.5 cm.

2. Construct side AB: Place the compass pointer at point B. Open the compass to 5 cm. Draw an arc above the line segment BC.

3. Construct side AC: Place the compass pointer at point C. Open the compass to 6 cm. Draw another arc that intersects the previous arc.

4. Mark intersection point A: The intersection of the two arcs is point A. Join AB and AC with a ruler.

5. Triangle completed: The triangle ABC is now constructed with the required sides.

(2) Δ DEF : m ∠ D = 35°, m ∠ F = 100°, l(DF) = 4.8 cm

(3) Δ MNP : l(MP) = 6.2 cm, l(NP) = 4.5 cm, m ∠ P = 75°

(4) Δ XYZ : m ∠ Y = 90°, l(XY) = 4.2 cm, l(XZ) = 7 cm

Construction of a quadrilateral :

• Every quadrilateral has 4 angles, 4 sides and 2 diagonals, so there are 10 elements of each quadrilateral.
• A quadrilateral can be constructed, if the measures of some specific 5 elements out of 10 are known.

Specific Construction Scenarios :

The following table outlines the four primary methods for constructing a quadrilateral :

Scenario	Required Information
I	Lengths of four sides and one diagonal.
II	Lengths of three sides and two diagonals.
III	Two adjacent sides and three angles.
IV	Three sides and two angles included by those sides.

 

(I) To construct a quadrilateral if the lengths of four sides and a diagonal is given :

Example : Construct □ PQRS such that , l(PQ) = 5.6 cm , l(QR) = 5 cm, l(PS) = 4.3 cm, l(RS) = 7 cm, l(QS) = 6.2 cm

Solution :

Let us draw a rough figure and show the given information in it. From the figure we see that the sides of Δ SPQ and Δ SRQ are known.

So if we construct Δ SPQ and Δ SRQ of given measures, we get □ PQRS.

 

(II) To construct a quadrilateral if three sides and two diagonals are given.

Example : Construct □ WXYZ such that, l(YZ) = 4 cm, l(ZX) = 6 cm, l(WX) = 4.5 cm, l(ZW) = 5 cm, l(YW) = 6.5 cm

Solution:

Rough figure:

Let us draw a rough figure and show the given measures in it.

From the figure we see that all sides of Δ WXZ and Δ WZY are known.

So let us draw Δ WXZ and Δ WZY using given measures.

We will get □ WXYZ after drawing segment XY.

(III) To construct a quadrilateral if two adjacent sides and any three angles are given.

Example : Construct □ LEFT such that, l(EL) = 4.5 cm, l(EF) = 5.5 cm, m ∠ L = 60°, m ∠ E = 100°, m ∠ F = 120°

Solution:

We draw the rough figure and show all the given information in it.

• Seg LE of given length can be drawn.
• Angles of measure 60° and 100° can be drawn with the help of the protractor at vertex L and E respectively.
• Point F can be located at the distance of 5.5 cm at the angle drawn at vertex E thus ∠ LEF = 100° and l(EF) = 5.5 cm.
• Angle of measure 120° can be drawn with the help of protractor at vertex F.
• Thus, the point of intersection of angle drawn at vertex L and vertex F is T. □ ABCD is thus constructed.

(IV) To construct a quadrilateral, if three sides and two angles included by them are given :

Example : Construct □ PQRS such that l(QR)= 4.9 cm, l(RS) = 6.2 cm, l(SP) = 3.9 cm, m∠ R = 60° and m∠ S = 75°.

Solution :

We draw the rough figure and show all the given information in it.

• Seg QR of length 4.9 cm can be drawn.
• At point R, an angle of measure 60° can be drawn with the help of a protractor.
• We can locate point S on the angle such that l(SR) = 6.2 cm.
• A point S, an angle of measure 75° can be drawn and point P can be located such
• that l(PS) = 3.9 cm.
• On joining P and Q, □ PQRS is thus obtained.

Rectangle :

If all angles of a quadrilateral are right angles, it is called a rectangle.

In the figure, m∠ P = m∠ Q = m∠ R = m∠ S = 90°

∴ □ PQRS is a rectangle.

Properties of rectangle :

• Opposite sides of rectangle are congruent.

[In the figure, l(PQ) = l(SR) and l(PS) = l(QR)]

• Diagonals of rectangle are congruent and they bisect each other.

[In the figure l(PR) = l(QS) and l(PT) = l(TR) = \(\frac{1}{2}\)l(PR) and l(TQ) = l(TS) = \(\frac{1}{2}\)l(QS).

[Note : Since l(PS) = l(QS) therefore, l(TP) = l(TR) = l(TS) = l(TQ)]

Square :

If all sides and all angles of a quadrilateral are congruent, it is called a square.

In the figure, l(AB) = l(BC) = l(CD) = l(AD) and m∠ A = m∠ B = m∠ C = m∠ D = 90° ∴ □ ABCD is a square.

Properties of square :

• Diagonals of square are congruent.

In the figure, l(AC) = l(BD)

• Diagonals of square bisect each other.

In the figure, l(AE) = l(EC) = \(\frac{1}{2}\)l(AC) and l(BE) = l(ED) = \(\frac{1}{2}\)l(BD)

[Note : Since l(AC) = l(BD) therefore l(AE) = l(EC) = l(BE) = l(ED)]

• Diagonals of square bisect opposite angles.

In the figure, diagonal AC bisects ∠ BAD and ∠ BCD and diagonal BD bisects ∠ ABC and ∠ ADC.

Rhombus :

If all sides of a quadrilateral are of equal length (congruent), it is called a rhombus.

In the figure, l(AB) = l(BC) = l(CD) = l(AD) ∴ □ ABCD is a rhombus.

Properties of rhombus :

• Opposite angles of rhombus are congruent.

In the figure, ∠ ABC ≅ ∠ ADC and ∠ BAD ≅ ∠ BCD

Diagonals of rhombus are perpendicular to each other. In the figure, seg AC ⊥ seg BD at point M i.e. m∠ AMB = 90°.

Diagonals of rhombus bisect each other. In the figure, l(AM) = l(MC) = \(\frac{1}{2}\)l(AC) and l(BM) = l(MD) = \(\frac{1}{2}\)l(BD)

• Diagonals of rhombus bisect opposite angles.

In the figure, Diagonal BD bisects ABC i.e. it divides it into two angles of equal measures i.e. m∠ ABD = m∠ DBC and it also bisects ∠ ADC i.e. m∠ ADB = m∠ BDC.

Similarly, diagonal AC, bisects ∠ BAD and ∠ BCD i.e. m∠ BAC = m∠ DAC and m∠ BCA = m∠ DCA.

Parallelogram :

A quadrilateral is a parallelogram, if its opposite sides are parallel.

In the figure, side AB || side CD and side AD || side BC then □ ABCD is a parallelogram

Properties of parallelogram :

• Opposite sides of parallelogram are congruent.

In the figure, l(AB) = l(CD) and l(AD) = l(BC).

• Opposite angles of parallelogram are congruent.

In the figure, m∠ ABC = m∠ ADC and m∠ BAD = m∠ BCD.

• Diagonals of parallelogram bisect each other.

In figure l(AO) = l(OC) = \(\frac{1}{2}\)l(AC) and l(BO) = l(OD) = \(\frac{1}{2}\)l(BD).

Trapezium :

If only one pair of opposite sides of a quadrilateral is parallel then it is called a trapezium.

In □ WXYZ, only one pair of opposite sides, seg WZ and seg XY is parallel. So by definition □ WXYZ is a trapezium.

With the property of interior angles formed by two parallel lines and their transversal we get m∠ W + m∠ X = 180° and m∠ Y + m∠ Z = 180°.

• In a trapezium, out of four pairs of adjacent angles, two are supplementary.

Kite :

If one diagonal is the perpendicular bisector of the other diagonal then the quadrilateral is called a kite.

In the figure, diagonal BD is the perpendicular bisector of diagonal AC. ∴ □ ABCD is a kite.

Properties of Kite :

• Two pairs of adjacent sides are congruent.

In the figure, l(AB) = l(BC) and l(AD) = l(DC).

• One pair of opposite angles is congruent.

In the figure, m∠ BAD = m∠ BCD.