In the previous section we learned the construction of Parallelogram. Next we have to learn the construction of Trapeziums. These shapes are frequently used in Science and Engineering. For example, the road embankments for the transportation of vehicles, and canal sections for conveyance of water, are usually in the shape of Trapeziums. This is because, the sides of the trapezium, which are at an angle, will have more stability, than when they are vertical. So this shape is preferred over the 'rectangular shape'. There are two types of Trapeziums: (i) Isosceles Trapeziums and (ii) Non Isosceles Trapeziums. In this section, we will learn the construction of the Isosceles type.
The properties of this type of Trapezium are:
• Only one pair of opposite sides are equal. In the fig.12.(a), there are two pairs of opposite sides: (i) [AB, CD] and (ii) [AD, BC]. Out of these two pairs, only the sides in the first pair are parallel to each other. The sides AD and BC in the second pair are Non parallel
• Non parallel sides are equal. We have seen above that AD and BC are non parallel. These two sides are equal in length
• Angles on each parallel sides are equal. In the fig.(b), the parallel sides are AB and CD. Consider them separately:
♦ Consider side AB first. The angles on this side are ∠A and ∠B. They are equal in measurement. That is., ∠A = ∠B.
♦ Consider side CD. The angles on this side are ∠C and ∠D. These two angles are equal. That is., ∠C = ∠D
♦ The angles on the bottom side, A and B, are called the Base angles. So we can say: In an Isosceles Trapezium, base angles are equal
• Sum of the angles on each non parallel sides is equal 180o. In the fig., the non parallel sides are AD and BC. Consider them separately:
♦ Consider side AD first. The angles on this side are ∠A and ∠D. Sum of these two angles is 180o. That is., ∠A + ∠D = 180o
♦ Consider side BC. The angles on this side are ∠B and ∠C. Sum of these two angles is 180o. That is., ∠C + ∠D = 180o
• Diagonals are equal. In the fig.12.9(c), AC and BD are equal in length.
♦ Note that, though they are equal, none of them bisect the other, and they are not perpendicular to each other.
Now we will see a special feature of Isosceles Trapezium:
Fig.12.10(a) shows an Isosceles Trapezium ABCD. Consider the non parallel side AD. It makes a certain angle ∠DAB with the base AB.
In the fig.(b), A line DE is drawn in such a way that, length of DE is equal to the length of AD. And also, DE makes the same angle ∠DAB with the base. That is.,
AB = DE - - - (1) and
∠DAB = ∠DEA - - - (2)
(1) and (2) are properties of an Isosceles Triangle:
Two sides equal, and the angles opposite to those equal sides (the base angles) also equal.
So ΔAED is an Isosceles Triangle. If we remove this isosceles triangle from the trapezium, what will remain? We can see that, the remaining portion EBCD is a parallelogram. So we can say that:
■ Every Isosceles Trapezium is made up of an Isosceles Triangle and a parallelogram
This is shown in fig.(c)
So, an Isosceles Triangle is hidden inside an Isosceles Trapezium. We can bring out that hidden triangle by drawing a ‘mirror image’ of the non parallel side.
Now we will see one more feature of Isosceles Trapezium:
In fig.12.11, the Isosceles Trapezium ABCD is already split into an Isosceles Triangle AED and a Parallelogram EBCD.
Now concentrate on the Parallelogram. Length of EB is equal to the length of CD (∵ since they are opposite sides of a parallelogram). This gives us a method to split the 'length of the base' of the Trapezium:
Length of the base AB of the Isosceles Trapezium = Length of the Top parallel side CD + Length of AE, the base of the Isosceles Triangle.
Rearranging the above, we get an equation for calculating the length of base of the hidden Isosceles triangle:
Base of the Isosceles Triangle = Total base of the Isosceles Trapezium – Top parallel side of the Trapezium
The above equation is very useful for the construction of an Isosceles Trapezium. We will see it’s application in a solved example.
Solved example 12.6
Draw an Isosceles Trapezium in which, Base = 6 cm, Top parallel side = 4 cm, and non parallel sides = 3 cm
Solution: First step is to draw a rough sketch as shown in the fig.12.12(a)
Based on the rough sketch, we can proceed to do the construction
• Draw a horizontal line AB, 6 cm in length
• The base of the hidden Isosceles Triangle = AE = AB – CD = 6 – 4 = 2 cm
• With A as center, draw an arc of 2 cm radius (shown in green colour in fig.b), cutting AB at E. So we obtained corners A and E of the Isosceles Triangle.
• Now we want the other corner D. This D is at a distance of 3 cm from both A and E (∵ ΔAED is an isosceles triangle). So draw an arc with center A and radius 3 cm (shown in magenta colour in fig.b). Draw another arc with E as center and radius 3 cm (shown in yellow colour in fig.b).
• These two arcs will intersect at D. So we have obtained the third corner. This is also the third corner of the Trapezium. We must now locate the remaining corner C
• DC is parallel to AB. So, through D, draw a line DC' of any convenient length, parallel to AB
• With D as center, draw an arc (shown in white colour in fig.c) of radius 4 cm, cutting DC' at C. Thus the remaining corner is also obtained. Join B to C. Thus we get the required Trapezium ABCD
Solved example 12.7
Draw an Isosceles Trapezium in which, Base = 7 cm, Top parallel side = 2 cm, and non parallel sides = 4 cm
Solution: First step is to draw a rough sketch as shown in the fig.12.13(a)
Based on the rough sketch, we can proceed to do the construction
• Draw a horizontal line AB, 7 cm in length
• The base of the hidden Isosceles Triangle = AE = AB – CD = 7 – 2 = 5 cm
• With A as center, draw an arc of 5 cm radius (shown in green colour in fig.b), cutting AB at E. So we obtained corners A and E of the Isosceles Triangle.
• Now we want the other corner D. This D is at a distance of 4 cm from both A and E (∵ ΔAED is an isosceles triangle). So draw an arc with center A and radius 4 cm (shown in magenta colour in fig.b). Draw another arc with E as center and radius 4 cm (shown in yellow colour in fig.b).
• These two arcs will intersect at D. So we have obtained the third corner. This is also the third corner of the Trapezium. We must now locate the remaining corner C
• DC is parallel to AB. So, through D, draw a line DC' of any convenient length, parallel to AB
• With D as center, draw an arc (shown in white colour in fig.c) of radius 2 cm, cutting DC' at C. Thus the remaining corner is also obtained. Join B to C. Thus we get the required parallelogram ABCD
Solved example 12.8
Construct an Isosceles Trapezium in which Base = 6 cm, Non parallel sides = 3 cm, and Base angles = 50o
Solution:
In the previous problem, we were given lengths of all the sides. In this problem, length of one side (the top parallel side) is missing. Instead, we are given the Base angles. With those, we can indeed construct the Trapezium. The first step is to draw a rough sketch as shown in the fig. 12.14(a)
Based on the rough sketch, we can proceed to do the construction
• Draw a horizontal line AB, 6 cm in length
• Draw line AD' at an angle 50o with AB. Draw a line BC' at an angle 50o with AB
• With A as center, draw an arc of radius 3 cm, cutting AD' at D. With B as center, draw an arc of radius 3 cm, cutting BC' at C. These arcs are shown in magenta colour in fig.(a)
• Join D and C. Thus we get the Required Trapezium ABCD
■ CD must be parallel to AB. We can prove it in the following way:
• Draw DE Perpendicular to AB. So ∠AED = 90o. In the same way, draw CF perpendicular to AB. So ∠CFB = 90o
• Consider the ΔAED. In that triangle, ∠ADE = 180 - (90 + 50) = 40o (∵ sum of the three interior angles of a triangle = 180o). In the same way, ∠BCF is also equal to 40o
• In ΔAED, we have a side AD of 3 cm length, and with angles 50o and 40o at it's ends
• In ΔBFC, we have a side BC of 3 cm length, and with angles 50o and 40o at it's ends
• This is a case of ASA congruence. That is., ΔAED and ΔBFC are congruent
• We can write the correspondence:
• The sides that we considered are AD and BC. ∠A = ∠B = 50o. So A↔B
• ∠C = ∠D = 40o. So C↔D. The only remaining corners are E and F. So E↔F
• So ΔAED and ΔBFC are congruent under the correspondence: AED↔BFC
• From the above correspondence, we can write the corresponding sides: AE↔BF, ED↔FC, AD↔BC
• Out of the above three, consider ED↔FC. It indicates that, the two perpendiculars shown in yellow colour, are equal in length.
• So the corners D and C are at the same distance from AB. That is., CD is parallel to AB
We have learned to construct an Isosceles Trapezium under two conditions:
■ When the lengths of all the sides are given
■ When the lengths of Base, Non parallel sides, and Base angles are given
In the next section, we will learn the construction of Non Isosceles Trapeziums.
Construction of an Isosceles Trapezium
An Isosceles Trapezium is shown in the fig.12.9. In the fig., (a), (b) and (c) shows the same Isosceles Trapezium. They are shown separately, only to avoid congestion of details. |
| Fig.12.9 |
• Only one pair of opposite sides are equal. In the fig.12.(a), there are two pairs of opposite sides: (i) [AB, CD] and (ii) [AD, BC]. Out of these two pairs, only the sides in the first pair are parallel to each other. The sides AD and BC in the second pair are Non parallel
• Non parallel sides are equal. We have seen above that AD and BC are non parallel. These two sides are equal in length
• Angles on each parallel sides are equal. In the fig.(b), the parallel sides are AB and CD. Consider them separately:
♦ Consider side AB first. The angles on this side are ∠A and ∠B. They are equal in measurement. That is., ∠A = ∠B.
♦ Consider side CD. The angles on this side are ∠C and ∠D. These two angles are equal. That is., ∠C = ∠D
♦ The angles on the bottom side, A and B, are called the Base angles. So we can say: In an Isosceles Trapezium, base angles are equal
• Sum of the angles on each non parallel sides is equal 180o. In the fig., the non parallel sides are AD and BC. Consider them separately:
♦ Consider side AD first. The angles on this side are ∠A and ∠D. Sum of these two angles is 180o. That is., ∠A + ∠D = 180o
♦ Consider side BC. The angles on this side are ∠B and ∠C. Sum of these two angles is 180o. That is., ∠C + ∠D = 180o
• Diagonals are equal. In the fig.12.9(c), AC and BD are equal in length.
♦ Note that, though they are equal, none of them bisect the other, and they are not perpendicular to each other.
Now we will see a special feature of Isosceles Trapezium:
Fig.12.10(a) shows an Isosceles Trapezium ABCD. Consider the non parallel side AD. It makes a certain angle ∠DAB with the base AB.
 |
| Fig.12.10 |
AB = DE - - - (1) and
∠DAB = ∠DEA - - - (2)
(1) and (2) are properties of an Isosceles Triangle:
Two sides equal, and the angles opposite to those equal sides (the base angles) also equal.
So ΔAED is an Isosceles Triangle. If we remove this isosceles triangle from the trapezium, what will remain? We can see that, the remaining portion EBCD is a parallelogram. So we can say that:
■ Every Isosceles Trapezium is made up of an Isosceles Triangle and a parallelogram
This is shown in fig.(c)
So, an Isosceles Triangle is hidden inside an Isosceles Trapezium. We can bring out that hidden triangle by drawing a ‘mirror image’ of the non parallel side.
Now we will see one more feature of Isosceles Trapezium:
In fig.12.11, the Isosceles Trapezium ABCD is already split into an Isosceles Triangle AED and a Parallelogram EBCD.
 |
| Fig.12.11 |
Length of the base AB of the Isosceles Trapezium = Length of the Top parallel side CD + Length of AE, the base of the Isosceles Triangle.
Rearranging the above, we get an equation for calculating the length of base of the hidden Isosceles triangle:
Base of the Isosceles Triangle = Total base of the Isosceles Trapezium – Top parallel side of the Trapezium
The above equation is very useful for the construction of an Isosceles Trapezium. We will see it’s application in a solved example.
Solved example 12.6
Draw an Isosceles Trapezium in which, Base = 6 cm, Top parallel side = 4 cm, and non parallel sides = 3 cm
Solution: First step is to draw a rough sketch as shown in the fig.12.12(a)
 |
| Fig.12.12 |
• Draw a horizontal line AB, 6 cm in length
• The base of the hidden Isosceles Triangle = AE = AB – CD = 6 – 4 = 2 cm
• With A as center, draw an arc of 2 cm radius (shown in green colour in fig.b), cutting AB at E. So we obtained corners A and E of the Isosceles Triangle.
• Now we want the other corner D. This D is at a distance of 3 cm from both A and E (∵ ΔAED is an isosceles triangle). So draw an arc with center A and radius 3 cm (shown in magenta colour in fig.b). Draw another arc with E as center and radius 3 cm (shown in yellow colour in fig.b).
• These two arcs will intersect at D. So we have obtained the third corner. This is also the third corner of the Trapezium. We must now locate the remaining corner C
• DC is parallel to AB. So, through D, draw a line DC' of any convenient length, parallel to AB
• With D as center, draw an arc (shown in white colour in fig.c) of radius 4 cm, cutting DC' at C. Thus the remaining corner is also obtained. Join B to C. Thus we get the required Trapezium ABCD
Solved example 12.7
Draw an Isosceles Trapezium in which, Base = 7 cm, Top parallel side = 2 cm, and non parallel sides = 4 cm
Solution: First step is to draw a rough sketch as shown in the fig.12.13(a)
 |
| Fig.12.13 |
• Draw a horizontal line AB, 7 cm in length
• The base of the hidden Isosceles Triangle = AE = AB – CD = 7 – 2 = 5 cm
• With A as center, draw an arc of 5 cm radius (shown in green colour in fig.b), cutting AB at E. So we obtained corners A and E of the Isosceles Triangle.
• Now we want the other corner D. This D is at a distance of 4 cm from both A and E (∵ ΔAED is an isosceles triangle). So draw an arc with center A and radius 4 cm (shown in magenta colour in fig.b). Draw another arc with E as center and radius 4 cm (shown in yellow colour in fig.b).
• These two arcs will intersect at D. So we have obtained the third corner. This is also the third corner of the Trapezium. We must now locate the remaining corner C
• DC is parallel to AB. So, through D, draw a line DC' of any convenient length, parallel to AB
• With D as center, draw an arc (shown in white colour in fig.c) of radius 2 cm, cutting DC' at C. Thus the remaining corner is also obtained. Join B to C. Thus we get the required parallelogram ABCD
Solved example 12.8
Construct an Isosceles Trapezium in which Base = 6 cm, Non parallel sides = 3 cm, and Base angles = 50o
Solution:
In the previous problem, we were given lengths of all the sides. In this problem, length of one side (the top parallel side) is missing. Instead, we are given the Base angles. With those, we can indeed construct the Trapezium. The first step is to draw a rough sketch as shown in the fig. 12.14(a)
• Draw a horizontal line AB, 6 cm in length
• Draw line AD' at an angle 50o with AB. Draw a line BC' at an angle 50o with AB
• With A as center, draw an arc of radius 3 cm, cutting AD' at D. With B as center, draw an arc of radius 3 cm, cutting BC' at C. These arcs are shown in magenta colour in fig.(a)
• Join D and C. Thus we get the Required Trapezium ABCD
■ CD must be parallel to AB. We can prove it in the following way:
• Draw DE Perpendicular to AB. So ∠AED = 90o. In the same way, draw CF perpendicular to AB. So ∠CFB = 90o
• Consider the ΔAED. In that triangle, ∠ADE = 180 - (90 + 50) = 40o (∵ sum of the three interior angles of a triangle = 180o). In the same way, ∠BCF is also equal to 40o
• In ΔAED, we have a side AD of 3 cm length, and with angles 50o and 40o at it's ends
• In ΔBFC, we have a side BC of 3 cm length, and with angles 50o and 40o at it's ends
• This is a case of ASA congruence. That is., ΔAED and ΔBFC are congruent
• We can write the correspondence:
• The sides that we considered are AD and BC. ∠A = ∠B = 50o. So A↔B
• ∠C = ∠D = 40o. So C↔D. The only remaining corners are E and F. So E↔F
• So ΔAED and ΔBFC are congruent under the correspondence: AED↔BFC
• From the above correspondence, we can write the corresponding sides: AE↔BF, ED↔FC, AD↔BC
• Out of the above three, consider ED↔FC. It indicates that, the two perpendiculars shown in yellow colour, are equal in length.
• So the corners D and C are at the same distance from AB. That is., CD is parallel to AB
We have learned to construct an Isosceles Trapezium under two conditions:
■ When the lengths of all the sides are given
■ When the lengths of Base, Non parallel sides, and Base angles are given
In the next section, we will learn the construction of Non Isosceles Trapeziums.
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