In the previous section we completed the discussion on the area of Non Isosceles Trapezium. In this section, we will discuss about Area of Quadrilaterals in general.
The splitting is done by drawing a diagonal BD. The length of this diagonal is denoted as 'd'. We get two triangles ΔABD and ΔBCD. We can find the areas of these triangles. The total area of the two triangles will be the area of the quadrilateral ABCD.
But to find the area of a triangle, we need it's base and height. For both the triangles, base is equal to d. To get the height, perpendicular lines AE and CF are dropped from A and C. The length of AE is h1 and length of CF is h2. Now we can calculate the areas of each triangle:
• Area of ΔABD = 1⁄2 × dh1
• Area of ΔBCD = 1⁄2 × dh2
• Total area of the two triangles = Area of the trapezium = 1⁄2 × dh1 + 1⁄2 × dh2 = 1⁄2 × d(h1 + h2)
So we can write:
■ Area of a quadrilateral is half of the product: [Length of a diagonal] × [Sum of the perpendicular distances from opposite vertices to the diagonal]
Solved example 13.17
Compute the area of the quadrilateral ABCD shown in the fig.13.24 below
Solution:
• We have, Area = 1⁄2 × d(h1 + h2)
• Given; d = 8 cm, h1 = 4 cm and h2 = 6 cm
• Substituting the values, we get Area = 1⁄2 × 8(4 + 6) = 1⁄2 × 80 = 40 cm2
Solved example 13.18
Compute the area of the quadrilateral ABCD shown in the fig.13.25 below
Solution:
In the fig., only the lengths of four sides are given. We need one diagonal and the perpendicular distances. We must calculate them ourselves. There are two clues:
■ Angle at C is 90o
■ Sides AB and AD are equal in length
We can use these clues to find the area:
• Draw the diagonal BD. (shown in fig.b) The quadrilateral is then split into ΔABD and ΔBCD
• ΔBCD is a right triangle. We can find it's area easily because base BC and height CD are known. Thus:
■ Area of ΔBCD = 1⁄2 × b × d = 1⁄2 × 16 × 12 = 96 cm2
• While we are working on this ΔBCD, let us calculate the hypotenuse BD also. Because, this hypotenuse will be required at a later stage in this problem.
• We can apply the Pythagoras theorem:
• BC2 + CD2 = BD2
• So 162 + 122 = BD2 ⇒ 256 + 144 = BD2 ⇒ 400 = BD2 ∴ BD = √400 = 20
• Now we have to calculate the area of ΔABD. We have the base BD. We want the height to calculate the area. We can determine the height as follows:
• ΔABD is an isosceles triangle, with equal sides AB and AD, and the base BD. So, if we drop a perpendicular AE (as shown in fig.c) from the vertex A to the base BD, that perpendicular will bisect the base. That means, BE = DE
• We get two right triangles ΔAEB and ΔAED. We need only one of them for our calculations, because, they are equal.
• Consider ΔAEB: AB = 26 cm, BE = BD/2 = 20/2 = 10
Apply Pythagoras theorem:
• AB2 - BE2 = AE2
• So 262 - 102 = AE2 ⇒ 676 - 100 = AE2 ⇒ AE2 = 576 ∴ AE = √576 = 24
• Now ΔABD is complete. We have the base and height. So area = 1⁄2 × BD × AE = 1⁄2 × 20 × 24 = 240 cm2
• So area of quadrilateral ABCD = area of ΔBCD + area of ΔABD = 96 + 240 = 336 cm2
Solved example 13.19
The three blue lines in the fig.13.26(a) are parallel.
(i) Prove that: [Area of ABCD ÷ Area of PQRS] = [Length of diagonal AC ÷ Length of diagonal PR]
(ii) How should the diagonals be related, if ABCD and PQRS is to have the same area?
(iii) Draw two quadrilaterals, neither parallelograms, nor trapeziums, each with area 15cm2
Solution:
• We are given two quadrilaterals and three parallel lines in fig.a. The diagonals AC and PR of the quadrilaterals are shown in yellow colour in fig.b.
• These diagonals split each of the quadrilaterals into two triangles. The heights of the triangles are also marked: h1 and h2
• Area of ΔACD = 1⁄2 × AC × h1 • Area of ΔACB = 1⁄2 × AC × h2
■ Area of quadrilateral ABCD = sum of above two areas = [1⁄2 × AC × h1 + 1⁄2 × AC × h2] = 1⁄2 × AC × (h1 + h2)
• Area of ΔPRS = 1⁄2 × PR × h1 • Area of ΔPQR = 1⁄2 × PR × h2
■ Area of quadrilateral PQRS =[1⁄2 × PR × h1 + 1⁄2 × PR × h2] = 1⁄2 × PR × (h1 + h2)
∴ [Area of ABCD ÷ Area of PQRS] = {[1⁄2 × AC × (h1 + h2)] ÷ [1⁄2 × PR × (h1 + h2)]} = AC ÷ PR - - - (1)
■ The above result can be written in another form:
The ratio of 'Area of ABCD' to 'Area of PQRS' is same as the ratio of 'Length of AC' to 'Length of PR'
(ii) • If ABCD and PQRS have the same area, [Area of ABCD ÷ Area of PQRS] will be equal to 1
• But from (1), [Area of ABCD ÷ Area of PQRS] = AC ÷ PR
• So, if the areas are equal, AC ÷ PR will also become equal to 1
• If AC ÷ PR is equal to 1, AC = PR
■ So we can write: If the areas are equal, their diagonals will also be equal
(iii) Consider the two quadrilaterals ABCD and PQRS in fig.13.27 below:
• The three blue lines are parallel
• Both the quadrilaterals must satisfy the following conditions:
♦ AC and PR must always be 6 cm
♦ AC and PR must always lie on the middle blue line
♦ Vertices 'S' and 'D' must always lie on the top most blue line
♦ Vertices 'B' and 'Q' must always lie on the bottom most blue line
• If the above conditions are satisfied, both ABCD and PQRS will have an area of 15 cm2, what ever be the shapes.
Proof:
• Area of ΔACD = 1⁄2 × 6 × 3 = 9.
• Δ PRS has the same base and height. So it also has an area of 9
■ This area will not change with shape. As long as the base is 6 cm in length, it lies on the middle blue line, and the top vertex lies on the top blue line
• Area of ΔABC = 1⁄2 × 6 × 2 = 6
• Δ PQR has the same base and height. So it also has an area of 6
■ This area will not change with shape. As long as the base is 6 cm in length, it lies on the middle blue line, and the bottom vertex lies on the bottom blue line
■ So the total area of each quadrilateral = 9 + 6 = 15 cm2
In the next section, we will discuss about 'Unchanging areas of triangles'.
Area of Quadrilaterals
Fig.13.23(a) shows a quadrilateral ABCD with no specified shape. To find it's area, we can split it into two triangles as shown in fig.b. |
| Fig.13.23 |
But to find the area of a triangle, we need it's base and height. For both the triangles, base is equal to d. To get the height, perpendicular lines AE and CF are dropped from A and C. The length of AE is h1 and length of CF is h2. Now we can calculate the areas of each triangle:
• Area of ΔABD = 1⁄2 × dh1
• Area of ΔBCD = 1⁄2 × dh2
• Total area of the two triangles = Area of the trapezium = 1⁄2 × dh1 + 1⁄2 × dh2 = 1⁄2 × d(h1 + h2)
So we can write:
■ Area of a quadrilateral is half of the product: [Length of a diagonal] × [Sum of the perpendicular distances from opposite vertices to the diagonal]
Solved example 13.17
Compute the area of the quadrilateral ABCD shown in the fig.13.24 below
 |
| Fig.13.24 |
• We have, Area = 1⁄2 × d(h1 + h2)
• Given; d = 8 cm, h1 = 4 cm and h2 = 6 cm
• Substituting the values, we get Area = 1⁄2 × 8(4 + 6) = 1⁄2 × 80 = 40 cm2
Solved example 13.18
Compute the area of the quadrilateral ABCD shown in the fig.13.25 below
 |
| Fig.13.25 |
In the fig., only the lengths of four sides are given. We need one diagonal and the perpendicular distances. We must calculate them ourselves. There are two clues:
■ Angle at C is 90o
■ Sides AB and AD are equal in length
We can use these clues to find the area:
• Draw the diagonal BD. (shown in fig.b) The quadrilateral is then split into ΔABD and ΔBCD
• ΔBCD is a right triangle. We can find it's area easily because base BC and height CD are known. Thus:
■ Area of ΔBCD = 1⁄2 × b × d = 1⁄2 × 16 × 12 = 96 cm2
• While we are working on this ΔBCD, let us calculate the hypotenuse BD also. Because, this hypotenuse will be required at a later stage in this problem.
• We can apply the Pythagoras theorem:
• BC2 + CD2 = BD2
• So 162 + 122 = BD2 ⇒ 256 + 144 = BD2 ⇒ 400 = BD2 ∴ BD = √400 = 20
• Now we have to calculate the area of ΔABD. We have the base BD. We want the height to calculate the area. We can determine the height as follows:
• ΔABD is an isosceles triangle, with equal sides AB and AD, and the base BD. So, if we drop a perpendicular AE (as shown in fig.c) from the vertex A to the base BD, that perpendicular will bisect the base. That means, BE = DE
• We get two right triangles ΔAEB and ΔAED. We need only one of them for our calculations, because, they are equal.
• Consider ΔAEB: AB = 26 cm, BE = BD/2 = 20/2 = 10
Apply Pythagoras theorem:
• AB2 - BE2 = AE2
• So 262 - 102 = AE2 ⇒ 676 - 100 = AE2 ⇒ AE2 = 576 ∴ AE = √576 = 24
• Now ΔABD is complete. We have the base and height. So area = 1⁄2 × BD × AE = 1⁄2 × 20 × 24 = 240 cm2
• So area of quadrilateral ABCD = area of ΔBCD + area of ΔABD = 96 + 240 = 336 cm2
Solved example 13.19
The three blue lines in the fig.13.26(a) are parallel.
(i) Prove that: [Area of ABCD ÷ Area of PQRS] = [Length of diagonal AC ÷ Length of diagonal PR]
(ii) How should the diagonals be related, if ABCD and PQRS is to have the same area?
(iii) Draw two quadrilaterals, neither parallelograms, nor trapeziums, each with area 15cm2
 |
| Fig.13.26 |
• We are given two quadrilaterals and three parallel lines in fig.a. The diagonals AC and PR of the quadrilaterals are shown in yellow colour in fig.b.
• These diagonals split each of the quadrilaterals into two triangles. The heights of the triangles are also marked: h1 and h2
• Area of ΔACD = 1⁄2 × AC × h1 • Area of ΔACB = 1⁄2 × AC × h2
■ Area of quadrilateral ABCD = sum of above two areas = [1⁄2 × AC × h1 + 1⁄2 × AC × h2] = 1⁄2 × AC × (h1 + h2)
• Area of ΔPRS = 1⁄2 × PR × h1 • Area of ΔPQR = 1⁄2 × PR × h2
■ Area of quadrilateral PQRS =[1⁄2 × PR × h1 + 1⁄2 × PR × h2] = 1⁄2 × PR × (h1 + h2)
∴ [Area of ABCD ÷ Area of PQRS] = {[1⁄2 × AC × (h1 + h2)] ÷ [1⁄2 × PR × (h1 + h2)]} = AC ÷ PR - - - (1)
■ The above result can be written in another form:
The ratio of 'Area of ABCD' to 'Area of PQRS' is same as the ratio of 'Length of AC' to 'Length of PR'
(ii) • If ABCD and PQRS have the same area, [Area of ABCD ÷ Area of PQRS] will be equal to 1
• But from (1), [Area of ABCD ÷ Area of PQRS] = AC ÷ PR
• So, if the areas are equal, AC ÷ PR will also become equal to 1
• If AC ÷ PR is equal to 1, AC = PR
■ So we can write: If the areas are equal, their diagonals will also be equal
(iii) Consider the two quadrilaterals ABCD and PQRS in fig.13.27 below:
 |
| Fig.13.27 |
• Both the quadrilaterals must satisfy the following conditions:
♦ AC and PR must always be 6 cm
♦ AC and PR must always lie on the middle blue line
♦ Vertices 'S' and 'D' must always lie on the top most blue line
♦ Vertices 'B' and 'Q' must always lie on the bottom most blue line
• If the above conditions are satisfied, both ABCD and PQRS will have an area of 15 cm2, what ever be the shapes.
Proof:
• Area of ΔACD = 1⁄2 × 6 × 3 = 9.
• Δ PRS has the same base and height. So it also has an area of 9
■ This area will not change with shape. As long as the base is 6 cm in length, it lies on the middle blue line, and the top vertex lies on the top blue line
• Area of ΔABC = 1⁄2 × 6 × 2 = 6
• Δ PQR has the same base and height. So it also has an area of 6
■ This area will not change with shape. As long as the base is 6 cm in length, it lies on the middle blue line, and the bottom vertex lies on the bottom blue line
■ So the total area of each quadrilateral = 9 + 6 = 15 cm2
In the next section, we will discuss about 'Unchanging areas of triangles'.
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