Chapter 13.3 - Area of Isosceles Trapezium

In the previous section we completed the discussion on Area of Rhombus. In this chapter, we will discuss about Area of Isosceles Trapezium.

Area of Isosceles Trapezium

In a previous section, we saw how a rectangle is modified to make a parallelogram. Now we will see how the rectangle can be modified to get an isosceles trapezium. Consider the rectangle ABCD shown in fig.13.14(a). It is a cardboard cut out, having a length of 6 cm and height of 3 cm. So it’s area is 6 × 3 = 18 cm².

Fig.13.14

We are going to make some modifications to this rectangle: Mark a point E, 2 cm from D as shown in fig.13.14(b). Join A to E. We get a triangle AED. Cut this triangle out, and move it to the right side. Attach it to the original piece in such a way that the vertical edge of the triangle coincides with the vertical edge BC. By doing this, we get a parallelogram. This much, we have seen earlier in fig.13.1. Now flip the triangle in a vertical direction. 

How do we do the vertical flipping? Initially, the base DE of the ΔAED is in-line with the top side CD of the rectangle. And the vertex A is in-line with the bottom side AB. But after the ‘vertical flip’, this is reversed. The vertex A comes to the top and base DE goes to the bottom. That is., the top and bottom are interchanged. This is shown in the animation in fig.(c). Note that there is no flipping in the horizontal direction. That is., there is no interchanging of left and right sides.

As a result of the vertical flip, we get an isosceles trapezium. We have not added any thing new. Neither have we taken away any thing. So:
■ Area of the newly obtained trapezium is same as the area of the original rectangle. 

Another point can also be noted: The top side CD is reduced by 2 cm, and the bottom side AB is increasd by 2 cm. So there is no change in the sum of top length and bottom lengths. Thus:
■ Sum of the top and bottom parallel sides of the isosceles trapezium is equal to the sum of the lengths of the original rectangle.

In the above experiment, we marked E at a distance of 2 cm from D. Instead of 2 cm, we can use different values like, 3 cm, 1 cm, 4 cm etc., In all cases, we will find that, the area of the final isosceles trapezium will be equal to 18 cm². This is shown in the fig.13.15 below:

Fig.13.15

The following two points have to be specially noted:
■ The base AB is increased by the length DE
■ The top side CD is decreased by the length DE
■ The height AD is same for the original rectangle and the newly obtained trapezium
■ In all the three cases, the area of the newly obtained trapezium is equal to the area of the original rectangle because, nothing is taken away and, nothing new is added.

Consider fig.13.15(b):
• The base of the newly formed trapezium = AB + BG = 6 + 3 = 9     
• The top side of the newly formed trapezium = CE = CD – CE = 6 – 3 = 3
• Sum of the above = 9 + 3 = 12
• Sum of the lengths of the original rectangle in fig.a = AB + CD = 6 + 6 = 12 (same as above)

Let us see if this is true for the other two cases as well:
Consider fig.13.15(c):
• The base of the newly formed trapezium = AB + BG = 6 + 1 = 7     
• The top side of the newly formed trapezium = CE = CD – CE = 6 – 1 = 5
• Sum of the above = 7 + 5 = 12
• Sum of the lengths of the original rectangle in fig.a = AB + CD = 6 + 6 = 12 (same as above)

Consider fig.13.15(d):
• The base of the newly formed trapezium = AB + BG = 6 + 4 = 10     
• The top side of the newly formed trapezium = CE = CD – CE = 6 – 4 = 2
• Sum of the above = 10 + 2 = 12
• Sum of the lengths of the original rectangle in fig.a = AB + CD = 6 + 6 = 12 (same as above)

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• From this we can write: Sum of bottom and top sides of the trapezium = Sum of lengths of the original rectangle
• But the ‘sum of lengths’ of original rectangle is 'two times the length' of that rectangle. Because, the two lengths are equal. This is shown in the fig.13.16 below:

Fig.13.16

• From this we can write: Length of original rectangle 2l = (a + b)/2 ⇒ l = (a + b)/2
• When we did the modifications to the rectangle, height was not changed. So the height d of the trapezium is same as that of the original rectangle. 
• So the area of the original rectangle = l × d = (a + b)/2 × d. 
• This is the area of newly formed trapezium also. It is generally written as:

So we can say:
■ The area of a trapezium is equal to the product: [half the sum of parallel sides] × [Perpendicular distance between parallel sides]

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We will now see some solved examples:
Solved example 13.10
Calculate the area of the isosceles trapezium shown in fig.13.17(a)

Fig.13.17

Solution: We can calculate the area of the given trapezium by using 3 methods
Method 1, From rectangle:
• The given trapezium in fig.13.17(a) is formed from a rectangle of equal area
• Sum of the lengths of that rectangle = Sum of the parallel sides of the trapezium = 3 + 7 = 10
∴ length of that rectangle = 10/2 = 5 cm
• Height of that rectangle = height of trapezium = 4 cm
∴ area of rectangle = Area of trapezium = 5 × 4 = 20 cm²

Method 2, Using formula:

Method 3, By splitting into simple areas:
• The given trapezium in fig.13.17(a) can be split into three parts as shown in fig(b): 2 triangles (marked as (i)) and one rectangle (marked as (ii)). This splitting is done as follows:
    ♦ Two perpendicular lines DE and CF are drawn to AB. So EF = CD = 3 cm
    ♦ The remaining length in the base is AB – 3 cm = 7 – 3 = 4 cm
    ♦ This must be equal to AE + BF. That is., AE + BF = 4 cm
    ♦ But since it is an isosceles trapezium, AE = BF
         ∴  AE = BF = 4/2 = 2 cm
    ♦ So we get two triangles (with base 2 cm and height 4 cm) and one rectangle (with width 3 cm and height 4 cm)
• Area of each triangle =  ¹⁄₂ × b × h = ¹⁄₂ × 2 × 4 = 4 cm² 
• Area of rectangle =  width × height = 3 × 4 = 12 cm²
∴ Total area = 4 + 12 + 4 = 20 cm²


Solved example 13.11
In an Isosceles Trapezium, the parallel sides are 8 cm and 14 cm long. Each of the non parallel sides are 5 cm long. What is the area?
Solution:
Fig. 13.18(a) below is a rough sketch showing all the given details

Fig.13.18

• The height of the trapezium is not given. We must find it first. For that, we must split the trapezium into two triangles and one rectangle. This splitting is done exactly as we did in the previous example 13.10. This is shown in fig.(b)
• We need only one triangle for our calculations. Let us take ΔAED

• It is a right triangle. The hypotenuse AD and one leg AE are known. To find the other leg DE, we will apply the Pythagoras theorem:
• AE² + DE² = AD² 
• So 3² + DE² = 5² ⇒ 9 + DE² = 25 ⇒ DE² = 25 - 9 = 16 ∴ DE = √16 = 4
• To find the area, we can use any of the three methods that we used in the previous example. Here, we will use the formula:

Solved example 13.12
Construct a rectangle of sides 7 cm and 4 cm. Construct an Isosceles trapezium of the same area, with the following specifications:
(i) Lengths of parallel sides 9 cm and 5 cm
(ii) Lengths of non parallel sides 5 cm
Solution:
(i) Fig.13.19(a) below shows the rectangle.

Fig.13.19

• We have to construct an isosceles trapezium with the same area. It should have parallel sides 9 cm and 5 cm. 
• The numbers 9 and 5 are closely related to the given length 7 of the rectangle:
• 7 + 2 = 9, and 7 – 2 = 5
• Mark a point E on CD, 2 cm away from D. Join AE. We get ΔAED
• If we remove ΔAED from the left side, attach it to the right side, and then flip it vertically, We will get the required trapezium. This is shown in fig.b
• Extend the line AB horizontally to AF' . With B as center, draw an arc (shown in yellow colour in fig.b) of radius 2 cm
• This arc will cut AF' at F. Join F to C. AFCE is the required trapezium
(ii) In the original rectangle, with A as center, draw an arc (shown in green colour in fig.c) of radius 5 cm, cutting CD at E. Join A to E. We get ΔAED
• If we remove ΔAED from the left side, attach it to the right side, and then flip it vertically, We will get the required trapezium. This is shown in fig.c
• Extend the line AB horizontally to AF' . With C as center, draw an arc (shown in magenta colour in fig.c) of radius 5 cm. This arc will cut AF' at F. 
• Take the distance DE in the compass. With that radius, draw an arc (shown in yellow colour in fig.c) with B as center. This arc should cut AF' at the same point F. Join F to C. AFCE is the required trapezium

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In the next section, we will see the area of a Non Isosceles Trapezium.

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