In the previous section we completed the discussion on construction of Triangles. In this section, we will discuss about the construction of Quadrilaterals. Quadrilaterals are closed plane figures having four sides. We will learn the construction of four quadrilaterals: Rhombus, Parallelogram, Trapezium, and Isosceles trapezium.
Following are the properties of a Rhombus:
• All sides are equal (as in a square). In the fig.12.1(a), AB = BC = CD = AD
• Opposite sides are parallel. In the fig (a), AB ∥ CD and BC ∥ AD
• Opposite angles are equal. In the fig.(b), ∠A = ∠C and ∠B = ∠D
• Sum of angles on the same side = 180o. In fig.(a), ∠A + ∠B = 180o, ∠B + ∠C = 180o, ∠C + ∠D = 180o, and ∠D + ∠A = 180o
• Diagonals are perpendicular bisectors of each other. In the fig.(c), ‘O’ is the midpoint of both AC and BD. And also AC and BD are perpendicular to each other.
• Consider any one diagonal. Let us take AC. This diagonal splits the rhombus into two isosceles triangles: ACD and ACB. Similarly, consider the other diagonal BD. It splits the rhombus into two isosceles triangles: BDA and BDCFor splitting a rhombus in this way, we must consider only one diagonal at a time.
• If we consider both the diagonals together, the rhombus is split into four right angled triangles.
Solved example 12.1: Construct a Rhombus ABCD with side 4 cm and ∠DAB = 40o
Solution: We use the properties of a rhombus to do the construction.
• We know that, for a rhombus, all the four sides will be equal. So, for our rhombus, each of the side will be equal to 4.0 cm. The rough sketch is shown in fig.12.2(a)
• Based on the rough sketch, we can proceed to do the construction
• First draw a horizontal line AB’ of any convenient length
• Draw AD’ of any convenient length at an angle 40o with AB’
• With A as center, draw two arcs (shown in green colour), each with 4 cm radius, cutting AB’ at B, and AD’ at D
• So we obtained three corners A, B and D. Now we want the remaining corner C.
• To draw CD: We know that CD is parallel to AB. So, through D, draw a line DC’, parallel to AB
• With D as center, and radius 4.0 cm, draw an arc (shown in yellow colour), cutting DC’ at C
• Join B and C. This gives the required rhombus ABCD
■ Fig.(c) shows the rhombus with sides equal to the same 4 cm, But ∠DAB changed to 30o
Solved example 12.2: Draw a rhombus PQRS whose diagonals are 6 cm and 4 cm
Solution: Here also we use the properties of a rhombus to do the construction
• We know that, the diagonals of a rhombus are perpendicular bisectors of each other
• The rough sketch is shown in the fig.12.3(a)
• First we draw a horizontal line PR of 6 cm length
• Next we draw the perpendicular bisector of PR. It will meet PR at O
• Thus PR is split into PO and OR, each with 3 cm length
• Mark S and Q on this perpendicular bisector in such a way that OS = OQ = 2 cm
• Now the first diagonal PR has become a perpendicular bisector of diagonal SQ
• Thus, PR and SQ have become perpendicular bisectors of each other. This is shown in fig.(b)
• Join P, Q, R and S. This is the required rhombus with diagonal PR = 6 cm, and QS = 4 cm
• After the construction, we can measure and check that all the four sides PQ, QR, RS and SP are equal
• Even without measuring, it can be proved theoretically by several methods. We will discuss two of them:
• Method 1: The two diagonals PR and SQ split the rhombus into four right angled triangles.
• For each of those four triangles, we have a combination: [Two sides and their included angle]
• This combination is same for all the four triangles: 2 cm, 3 cm, and included angle 90o
• So by SAS criterion all the four triangles are congruent to each other. Thus their fourth sides are all same
• Method 2: Each of the four sides is the hypotenuse of a right angled triangle
• Each of the four triangles have two legs the same: 2 cm and 3 cm. So the hypotenuse will also be the same
We have seen two methods for the construction of a Rhombus:
■ When the length of side and an angle is given
■ When the lengths of both the diagonals are given
In the next section, we will learn the construction of Parallelograms.
Construction of a Rhombus
A Rhombus is shown in fig.12. In the fig., (a), (b) and (c) shows the same Rhombus. They are shown separately only to avoid congestion of details. |
| Fig.12.1 |
• All sides are equal (as in a square). In the fig.12.1(a), AB = BC = CD = AD
• Opposite sides are parallel. In the fig (a), AB ∥ CD and BC ∥ AD
• Opposite angles are equal. In the fig.(b), ∠A = ∠C and ∠B = ∠D
• Sum of angles on the same side = 180o. In fig.(a), ∠A + ∠B = 180o, ∠B + ∠C = 180o, ∠C + ∠D = 180o, and ∠D + ∠A = 180o
• Diagonals are perpendicular bisectors of each other. In the fig.(c), ‘O’ is the midpoint of both AC and BD. And also AC and BD are perpendicular to each other.
• Consider any one diagonal. Let us take AC. This diagonal splits the rhombus into two isosceles triangles: ACD and ACB. Similarly, consider the other diagonal BD. It splits the rhombus into two isosceles triangles: BDA and BDCFor splitting a rhombus in this way, we must consider only one diagonal at a time.
• If we consider both the diagonals together, the rhombus is split into four right angled triangles.
Solved example 12.1: Construct a Rhombus ABCD with side 4 cm and ∠DAB = 40o
Solution: We use the properties of a rhombus to do the construction.
• We know that, for a rhombus, all the four sides will be equal. So, for our rhombus, each of the side will be equal to 4.0 cm. The rough sketch is shown in fig.12.2(a)
 |
| Fig.12.2 |
• First draw a horizontal line AB’ of any convenient length
• Draw AD’ of any convenient length at an angle 40o with AB’
• With A as center, draw two arcs (shown in green colour), each with 4 cm radius, cutting AB’ at B, and AD’ at D
• So we obtained three corners A, B and D. Now we want the remaining corner C.
• To draw CD: We know that CD is parallel to AB. So, through D, draw a line DC’, parallel to AB
• With D as center, and radius 4.0 cm, draw an arc (shown in yellow colour), cutting DC’ at C
• Join B and C. This gives the required rhombus ABCD
■ Fig.(c) shows the rhombus with sides equal to the same 4 cm, But ∠DAB changed to 30o
Solved example 12.2: Draw a rhombus PQRS whose diagonals are 6 cm and 4 cm
Solution: Here also we use the properties of a rhombus to do the construction
• We know that, the diagonals of a rhombus are perpendicular bisectors of each other
• The rough sketch is shown in the fig.12.3(a)
 |
| Fig.12.3 |
• Next we draw the perpendicular bisector of PR. It will meet PR at O
• Thus PR is split into PO and OR, each with 3 cm length
• Mark S and Q on this perpendicular bisector in such a way that OS = OQ = 2 cm
• Now the first diagonal PR has become a perpendicular bisector of diagonal SQ
• Thus, PR and SQ have become perpendicular bisectors of each other. This is shown in fig.(b)
• Join P, Q, R and S. This is the required rhombus with diagonal PR = 6 cm, and QS = 4 cm
• After the construction, we can measure and check that all the four sides PQ, QR, RS and SP are equal
• Even without measuring, it can be proved theoretically by several methods. We will discuss two of them:
• Method 1: The two diagonals PR and SQ split the rhombus into four right angled triangles.
• For each of those four triangles, we have a combination: [Two sides and their included angle]
• This combination is same for all the four triangles: 2 cm, 3 cm, and included angle 90o
• So by SAS criterion all the four triangles are congruent to each other. Thus their fourth sides are all same
• Method 2: Each of the four sides is the hypotenuse of a right angled triangle
• Each of the four triangles have two legs the same: 2 cm and 3 cm. So the hypotenuse will also be the same
We have seen two methods for the construction of a Rhombus:
■ When the length of side and an angle is given
■ When the lengths of both the diagonals are given
In the next section, we will learn the construction of Parallelograms.
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