In the previous section we learned the construction of Rhombus. In this section, we will learn the construction of Parallelograms.
Following are the properties of a Parallelogram:
• Opposite sides are equal. In the fig.12.4(a), AB = CD and BC = 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.(b), ∠A + ∠B = 180o, ∠B + ∠C = 180o, ∠C + ∠D = 180o, and ∠D + ∠A = 180o
• Diagonals bisect each other. In the fig.(c), ‘O’ is the midpoint of both AC and BD. Note that, though the diagonals bisect each other, unlike the rhombus, they are not perpendicular to each other.
Solved example 12.3
Construct a parallelogram ABCD, with AB = 6 cm, BC = 3 cm, and ∠DAB = 50o
Solution: First we draw a rough sketch as shown in the fig.12.5(a)
• Draw a horizontal line AB of length 6 cm
• Draw a line AD' at an angle 50o with AB
• With A as center, draw an arc of radius 3 cm, cutting AD' at D
• So we have fixed three corners A, B and D. We have to fix the remaining corner C
• Through D, draw a line DC' parallel to AB
• With D as center, draw an arc of radius 6 cm, cutting DC' at C
• Join B and C. Thus we get the required parallelogram ABCD
• The above steps and the completed parallelogram is shown in fig.(b)
• Fig.(c) shows a parallelogram with the same sides 6 cm and 3 cm. But the angle is changed to 60o. The method of construction is the same
Solved example 12.4
Construct a parallelogram PQRS with diagonals PR = 6 cm, SQ = 4 cm, and the angle between them 50o
Solution: First we draw a rough sketch as shown in the fig.12.6(a)
• In a parallelogram, the diagonals bisect each other. So PR will be split into PO and RO, each of 3 cm. Similarly, SQ will be split into SO and QO, each of 2 cm. This is shown in the rough sketch. We can now proceed to do the construction.
• Draw a horizontal line PR of 6 cm length, and draw it's perpendicular bisector P'R'.
• P'R' (shown in blue colour) will intersect PR at O. Note that P'R' is drawn only to obtain the midpoint of PR. It does not have any other purpose.
• Through O, draw QS, at an angle of 50o with PR, in such a way that OQ = OS = 2 cm
• The above steps are shown in fig.12.6(b). We have obtained all the four corners: P, Q, R and S
• These corners are joined together to obtain the required parallelogram PQRS. This is shown in fig.(c)
• Fig.(d) shows a parallelogram with the same diagonals. But the angle between the diagonals is changed to 40o
Solved example 12.5: Construct a parallelogram ABCD with AB = 5 cm, BC = 3 cm, and diagonal AC = 7 cm
Solution: The rough sketch is shown in the fig.12.7(a). We use the properties of a parallelogram to do the construction.
• We know that, for a parallelogram, opposite sides are equal. So AD = 3 cm, and CD = 5 cm
• So the diagonal AC splits the parallelogram into 2 equal triangles: ΔABC and ΔACD
• We can construct ΔABC because, all it's 3 sides are known:
• First draw a horizontal line AB of 5 cm length
• With A as center, draw an arc of 7 cm radius. With B as center, draw an arc with 3 cm radius
• These two arcs will intersect at C. So ABC is complete. We proceed to construct ΔACD:
• With A as center, draw an arc of 3 cm radius. With C as center, draw an arc of 5 cm radius
• These two arcs will intersect at D. Thus we get the parallelogram ABCD. The construction steps and the completed parallelogram are shown in figs.(b) and (c)
Solved example 12.6: Draw a parallelogram PQRS whose diagonals PR = 8 cm, SQ = 6 cm, and one side SR = 5 cm
Solution: First we draw a rough sketch as shown in fig.12.8(a)
• In a parallelogram, the diagonals bisect each other. So PR will be split into PO and RO, each of 4 cm. Similarly, SQ will be split into SO and QO, each of 3 cm. This is shown in the rough sketch. We can now proceed to do the construction.
• Draw a horizontal line PR of 8 cm length, and draw it's bisector P'R'.
• P'R' (shown in blue colour) will intersect PR at O. Note that P'R' is drawn only to obtain the midpoint 'O' of PR. It does not have any other purpose.
• We have obtained P, R and O. Our next step is to obtain 'S'. For that, we use the information: OS = 3 cm and SR = 5 cm
• So with O as center, draw an arc of 3 cm radius. With R as center, draw an arc of 5 cm radius
• These two arcs will intersect at S. So we have obtained the three points: P, R and S. The remaining point is Q
• For this we use a property of parallelograms: Opposite sides are equal. From the rough sketch, it is clear that PQ = 5 cm, since it is the side opposite to SR
• So with O as center, draw an arc of 3 cm radius. With P as center, draw an arc of 5 cm radius
• These two arcs will intersect at Q. Join P, Q, R and S. We will get the required parallelogram
■ We can see an interesting feature in this parallelogram: All the sides are equal. So it is a Rhombus. Note that the bisector P'R' is a perpendicular bisector. This happens only in a Rhombus and a Square. P'R' coincides with one of the diagonals SQ
• After the construction, we can measure and check that all the four sides PQ, QR, RS and SP are equal to 5 cm
• Even without measuring, it can be proved theoretically:
• The two diagonals PR and SQ split the rhombus into four triangles.
• Consider ΔSOR. The lengths of the sides are 3, 4 and 5 cm. By Pythagoras theorem, these are the sides of a right angled triangle. (∵ 32 + 42 = 52).
• The longest side 5 cm is the hypotenuse. So the angle opposite to it (∠SOR) is the right angle
• Now, ∠SOR and ∠SOP form a linear pair. So ∠SOP = 90o
• ∠SOP and ∠QOP form a linear pair. So ∠QOP = 90o
• ∠QOP and ∠QOR form a linear pair. So ∠QOP = 90o
• Thus we see that all the angles at O is 90o. All the triangles have the same two sides (3 cm and 4 cm) and the same included angle 90o.
• So it is a case of SAS congruence. All the four triangles are equal. Thus we get PQ = QR = RS = SP = 5 cm
From the above solved example, we can say that, any Rhombus will satisfy the conditions of a Parallelogram. But any Parallelogram need not satisfy the conditions of a Rhombus. This is similar to the case of Square and Rectangle:
Any Square will satisfy the conditions of a Rectangle. But any Rectangle need not satisfy the conditions of a square.
We have seen three methods for the construction of a Rhombus:
■ When lengths of sides and an angle is given
■ When the lengths of both the diagonals and the angle between them are given
■ When the lengths of both the diagonals and length of one side are given
In the next section, we will learn the construction of Trapeziums.
Construction of a Parallelogram
A Parallelogram is shown in fig.12.4. In the fig., (a), (b) and (c) shows the same Parallelogram. They are shown separately only to avoid congestion of details. |
| Fig.12.4 |
• Opposite sides are equal. In the fig.12.4(a), AB = CD and BC = 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.(b), ∠A + ∠B = 180o, ∠B + ∠C = 180o, ∠C + ∠D = 180o, and ∠D + ∠A = 180o
• Diagonals bisect each other. In the fig.(c), ‘O’ is the midpoint of both AC and BD. Note that, though the diagonals bisect each other, unlike the rhombus, they are not perpendicular to each other.
Solved example 12.3
Construct a parallelogram ABCD, with AB = 6 cm, BC = 3 cm, and ∠DAB = 50o
Solution: First we draw a rough sketch as shown in the fig.12.5(a)
 |
| Fig.12.5 |
• Draw a line AD' at an angle 50o with AB
• With A as center, draw an arc of radius 3 cm, cutting AD' at D
• So we have fixed three corners A, B and D. We have to fix the remaining corner C
• Through D, draw a line DC' parallel to AB
• With D as center, draw an arc of radius 6 cm, cutting DC' at C
• Join B and C. Thus we get the required parallelogram ABCD
• The above steps and the completed parallelogram is shown in fig.(b)
• Fig.(c) shows a parallelogram with the same sides 6 cm and 3 cm. But the angle is changed to 60o. The method of construction is the same
Solved example 12.4
Construct a parallelogram PQRS with diagonals PR = 6 cm, SQ = 4 cm, and the angle between them 50o
Solution: First we draw a rough sketch as shown in the fig.12.6(a)
 |
| Fig.12.6 |
• Draw a horizontal line PR of 6 cm length, and draw it's perpendicular bisector P'R'.
• P'R' (shown in blue colour) will intersect PR at O. Note that P'R' is drawn only to obtain the midpoint of PR. It does not have any other purpose.
• Through O, draw QS, at an angle of 50o with PR, in such a way that OQ = OS = 2 cm
• The above steps are shown in fig.12.6(b). We have obtained all the four corners: P, Q, R and S
• These corners are joined together to obtain the required parallelogram PQRS. This is shown in fig.(c)
• Fig.(d) shows a parallelogram with the same diagonals. But the angle between the diagonals is changed to 40o
Solved example 12.5: Construct a parallelogram ABCD with AB = 5 cm, BC = 3 cm, and diagonal AC = 7 cm
Solution: The rough sketch is shown in the fig.12.7(a). We use the properties of a parallelogram to do the construction.
• We know that, for a parallelogram, opposite sides are equal. So AD = 3 cm, and CD = 5 cm
• So the diagonal AC splits the parallelogram into 2 equal triangles: ΔABC and ΔACD
 |
| Fig.12.7 |
• First draw a horizontal line AB of 5 cm length
• With A as center, draw an arc of 7 cm radius. With B as center, draw an arc with 3 cm radius
• These two arcs will intersect at C. So ABC is complete. We proceed to construct ΔACD:
• With A as center, draw an arc of 3 cm radius. With C as center, draw an arc of 5 cm radius
• These two arcs will intersect at D. Thus we get the parallelogram ABCD. The construction steps and the completed parallelogram are shown in figs.(b) and (c)
Solved example 12.6: Draw a parallelogram PQRS whose diagonals PR = 8 cm, SQ = 6 cm, and one side SR = 5 cm
Solution: First we draw a rough sketch as shown in fig.12.8(a)
 |
| Fig.12.8 |
• Draw a horizontal line PR of 8 cm length, and draw it's bisector P'R'.
• P'R' (shown in blue colour) will intersect PR at O. Note that P'R' is drawn only to obtain the midpoint 'O' of PR. It does not have any other purpose.
• We have obtained P, R and O. Our next step is to obtain 'S'. For that, we use the information: OS = 3 cm and SR = 5 cm
• So with O as center, draw an arc of 3 cm radius. With R as center, draw an arc of 5 cm radius
• These two arcs will intersect at S. So we have obtained the three points: P, R and S. The remaining point is Q
• For this we use a property of parallelograms: Opposite sides are equal. From the rough sketch, it is clear that PQ = 5 cm, since it is the side opposite to SR
• So with O as center, draw an arc of 3 cm radius. With P as center, draw an arc of 5 cm radius
• These two arcs will intersect at Q. Join P, Q, R and S. We will get the required parallelogram
■ We can see an interesting feature in this parallelogram: All the sides are equal. So it is a Rhombus. Note that the bisector P'R' is a perpendicular bisector. This happens only in a Rhombus and a Square. P'R' coincides with one of the diagonals SQ
• After the construction, we can measure and check that all the four sides PQ, QR, RS and SP are equal to 5 cm
• Even without measuring, it can be proved theoretically:
• The two diagonals PR and SQ split the rhombus into four triangles.
• Consider ΔSOR. The lengths of the sides are 3, 4 and 5 cm. By Pythagoras theorem, these are the sides of a right angled triangle. (∵ 32 + 42 = 52).
• The longest side 5 cm is the hypotenuse. So the angle opposite to it (∠SOR) is the right angle
• Now, ∠SOR and ∠SOP form a linear pair. So ∠SOP = 90o
• ∠SOP and ∠QOP form a linear pair. So ∠QOP = 90o
• ∠QOP and ∠QOR form a linear pair. So ∠QOP = 90o
• Thus we see that all the angles at O is 90o. All the triangles have the same two sides (3 cm and 4 cm) and the same included angle 90o.
• So it is a case of SAS congruence. All the four triangles are equal. Thus we get PQ = QR = RS = SP = 5 cm
From the above solved example, we can say that, any Rhombus will satisfy the conditions of a Parallelogram. But any Parallelogram need not satisfy the conditions of a Rhombus. This is similar to the case of Square and Rectangle:
Any Square will satisfy the conditions of a Rectangle. But any Rectangle need not satisfy the conditions of a square.
We have seen three methods for the construction of a Rhombus:
■ When lengths of sides and an angle is given
■ When the lengths of both the diagonals and the angle between them are given
■ When the lengths of both the diagonals and length of one side are given
In the next section, we will learn the construction of Trapeziums.
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