In the previous section we completed the discussion on Isosceles Trapezium. In this chapter, we will discuss about Area of Non Isosceles Trapezium.
Lengths of it's parallel sides are a and b, and it's height is d. In fig.b, a diagonal BD is drawn for the same trapezium. This diagonal splits the trapezium into two triangles: ΔABD and ΔBCD.
• Consider ΔABD. It has a base a and height d. So it's area = 1⁄2 × a × d = 1⁄2 × ad
• Consider ΔBCD. It has a base b and height d. So it's area = 1⁄2 × b × d = 1⁄2 × bd
• Total area of the two triangles = Area of the trapezium = 1⁄2 × ad + 1⁄2 × bd = 1⁄2 × (a + b)d
This is the same expression that we derived for an Isosceles trapezium in the previous section. So, here also we can say:
■ The area of a trapezium is equal to the product: [half the sum of parallel sides] × [Perpendicular distance between parallel sides]
Solved example 13.13
The lengths of the parallel sides of a trapezium are 30 cm and 10 cm. The distance between the parallel sides is 20 cm. Calculate the area of the trapezium
Solution:
Given: a = 30, b = 10, and d = 20 cm
Area = 1⁄2 × (a + b)d = 1⁄2 × (30 + 10) 20 = 1⁄2 × 40 × 20 = 400 cm2
Solved example 13.14
Area of a trapezium is 126 cm2 . Distance between it's parallel sides is 6 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Solution:
• Given: Area = 126 cm2, a = 28 cm, d = 6 cm. We have to find b
• Area = 1⁄2 × (a + b)d ⇒ 126 = 1⁄2 × (28 + b) 6 ⇒ 126 = (28 + b) × 3 ⇒ 126 = 84 + 3b ⇒ 3b = 42
∴ b = 42/3 = 14 cm
Solved example 13.15
Compute the area of the trapezium ABCD shown in fig.13.21 below:
Solution:
• In the fig., we are given the lengths of the parallel sides. But the height is not given. Instead, we are given the length of a diagonal BD.
• This diagonal, together with sides AD and AB forms a right triangle ABD. So we can apply the Pythagoras theorem:
• AB2 + AD2 = BD2
• So 122 + AD2 = 132 ⇒ 144 + AD2 = 169 ⇒ AD2 = 169 - 144 = 25 ∴ AD = √25 = 5
Now we have the height. We can write: a = 12, b = 4, and d = 5
Area = 1⁄2 × (a + b)d = 1⁄2 × (12 + 4) 5 = 1⁄2 × 16 × 5 = 40 cm2
Solved example 13.16
Compute the area of the hexagon shown in fig.13.22 below:
Solution:
• The hexagon ABCDEF is symmetric about the diagonal FC. This diagonal splits the hexagon into two equal trapeziums: ABCF and FCDE. We need to find the area of any one trapezium only. Twice that area will give the area of the hexagon.
• In fig.b, the hexagon has been split into the two trapeziums. Consider the upper trapezium FCDE.
• The length of the shorter parallel side is 7 cm. The height is 24/2 = 12 cm. We need to find the length of the longer parallel side.
• For that, we drop two perpendicular lines EG and DH to FC. We get two right triangles: ΔFGE and ΔCHD. These two triangles are equal. We want the length of FG. For that, we can apply the Pythagoras theorem:
• GE2 + FG2 = FE2
• So 122 + FG2 = 152 ⇒ 144 + FG2 = 225 ⇒ FG2 = 225 - 144 = 81 ∴ FG = √81 = 9
• FG = HC. So diagonal FC = 9 + 7 + 9 = 25. This is the bottom parallel side of the trapezium
• Now we can use the formula:
• Area = 1⁄2 × (a + b)d = 1⁄2 × (7 + 25) 12 = 1⁄2 × 32 × 12 = 192 cm2
∴ Total area = 192 × 2 = 384 cm2
In the next section, we will see the area of Quadrilaterals in general.
Area of Non Isosceles Trapezium
Non isosceles trapezium is also a quadrilateral. Two of it's opposite sides will be parallel. The other two opposite sides need not be parallel. Consider the Non isosceles trapezium shown in fig.13.20.(a) |
| Fig.13.20 |
• Consider ΔABD. It has a base a and height d. So it's area = 1⁄2 × a × d = 1⁄2 × ad
• Consider ΔBCD. It has a base b and height d. So it's area = 1⁄2 × b × d = 1⁄2 × bd
• Total area of the two triangles = Area of the trapezium = 1⁄2 × ad + 1⁄2 × bd = 1⁄2 × (a + b)d
This is the same expression that we derived for an Isosceles trapezium in the previous section. So, here also we can say:
■ The area of a trapezium is equal to the product: [half the sum of parallel sides] × [Perpendicular distance between parallel sides]
Solved example 13.13
The lengths of the parallel sides of a trapezium are 30 cm and 10 cm. The distance between the parallel sides is 20 cm. Calculate the area of the trapezium
Solution:
Given: a = 30, b = 10, and d = 20 cm
Area = 1⁄2 × (a + b)d = 1⁄2 × (30 + 10) 20 = 1⁄2 × 40 × 20 = 400 cm2
Solved example 13.14
Area of a trapezium is 126 cm2 . Distance between it's parallel sides is 6 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Solution:
• Given: Area = 126 cm2, a = 28 cm, d = 6 cm. We have to find b
• Area = 1⁄2 × (a + b)d ⇒ 126 = 1⁄2 × (28 + b) 6 ⇒ 126 = (28 + b) × 3 ⇒ 126 = 84 + 3b ⇒ 3b = 42
∴ b = 42/3 = 14 cm
Solved example 13.15
Compute the area of the trapezium ABCD shown in fig.13.21 below:
 |
| Fig.13.21 |
• In the fig., we are given the lengths of the parallel sides. But the height is not given. Instead, we are given the length of a diagonal BD.
• This diagonal, together with sides AD and AB forms a right triangle ABD. So we can apply the Pythagoras theorem:
• AB2 + AD2 = BD2
• So 122 + AD2 = 132 ⇒ 144 + AD2 = 169 ⇒ AD2 = 169 - 144 = 25 ∴ AD = √25 = 5
Now we have the height. We can write: a = 12, b = 4, and d = 5
Area = 1⁄2 × (a + b)d = 1⁄2 × (12 + 4) 5 = 1⁄2 × 16 × 5 = 40 cm2
Solved example 13.16
Compute the area of the hexagon shown in fig.13.22 below:
 |
| Fig.13.22 |
• The hexagon ABCDEF is symmetric about the diagonal FC. This diagonal splits the hexagon into two equal trapeziums: ABCF and FCDE. We need to find the area of any one trapezium only. Twice that area will give the area of the hexagon.
• In fig.b, the hexagon has been split into the two trapeziums. Consider the upper trapezium FCDE.
• The length of the shorter parallel side is 7 cm. The height is 24/2 = 12 cm. We need to find the length of the longer parallel side.
• For that, we drop two perpendicular lines EG and DH to FC. We get two right triangles: ΔFGE and ΔCHD. These two triangles are equal. We want the length of FG. For that, we can apply the Pythagoras theorem:
• GE2 + FG2 = FE2
• So 122 + FG2 = 152 ⇒ 144 + FG2 = 225 ⇒ FG2 = 225 - 144 = 81 ∴ FG = √81 = 9
• FG = HC. So diagonal FC = 9 + 7 + 9 = 25. This is the bottom parallel side of the trapezium
• Now we can use the formula:
• Area = 1⁄2 × (a + b)d = 1⁄2 × (7 + 25) 12 = 1⁄2 × 32 × 12 = 192 cm2
∴ Total area = 192 × 2 = 384 cm2
In the next section, we will see the area of Quadrilaterals in general.
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