In the previous section we completed the discussion on the calculation of area of a parallelogram. We also saw a number of solved examples. In this section, we will see the area of a Rhombus.
When we are given the length of side of a square, we can easily calculate the area by using the formula:
Area of a square of side l = l2
But how do we calculate the area when diagonal is given? For that, we use a 'special property of diagonals of a square':
■ The diagonals of a square bisect each other, and are perpendicular to each other.
So the area consists of 4 equal triangles. This is shown in fig.b. They are: OAB, OBC, OCD and OAD. Not just triangles, they are isosceles triangles. Because two sides are equal (3 cm).
• In each triangle, the base is 3 cm, and height is 3 cm. So area of one triangle = 1⁄2 × b × h = 1⁄2 × 3 × 3 = 41⁄2
• ∴ Total area = 4 × 41⁄2 = 18 cm2
• Fig.c shows a square PQRS whose diagonal is 5 cm. Area of PQRS will be equal to 4 × 1⁄2 × 2.5 × 2.5 = 12.5 cm2
We can use algebra to get a general form:
From the above we can write:
■ Area of a square is half the square of it's diagonal
• Now consider a Rhombus ABCD in fig.13.11(a). Here also diagonals are given instead of sides. Diagonals of rhombus also have a special property: They bisect each other, and are perpendicular to each other.
• But unlike a square, the diagonals of a rhombus are not of equal length. So the four segments of the diagonals will not be of equal length.
• Consequently, the triangles formed by the diagonals are not isosceles.
• Even if they are not isosceles, we can calculate the area of each triangle easily. Because, we know the base and height. In fig.13.11(a), the area of each triangle is 1⁄2 × b × h = 1⁄2 × 3 × 2 = 3
• So the total area of the Rhombus ABCD = Total area of 4 squares = 4 × 3 = 12 cm2
• Similarly, the area of the rhombus PQRS in fig will be equal to 4 × 1⁄2 × 2.5 × 2 = 10 cm2
• We can use algebra to get a general form:
From the above result, we can write:
■ Area of a Rhombus is half the product of it's diagonals
Solved example 13.7
A square has an area of 4.5 cm2. What is the length of it’s diagonal?
Solution:
Solved example 13.8
The area of a non square rhombus is 216 cm2. The length of one of it’s diagonal is 24 cm. Compute the following:
(1) Length of the other diagonal (2) Length of a side (3) Perimeter (4) Distance between sides
Solution:
(1) Length of the other diagonal:
(2) Length of a side:
Let us draw a rough sketch (shown in fig.13.12.a), which shows all the available details:
• We have two diagonals BD = 24 cm and AC = 18 cm.
• The diagonals of a rhombus bisects each other. So AC will be split into AO and CO, each of 9 cm. Similarly, BD will be split into BO and DO, each of 12 cm.
• Consider any triangle, say ΔOBC. It is a right triangle, right angled at O. (∵ diagonals of a rhombus are perpendicular to each other)
• We can apply the Pythagoras theorem to this triangle: OB2 + OC2 = BC2
So 122 + 92 = BC2 ⇒ 144 + 81 = BC2 ⇒ 225 = BC2 ∴ BC = √225 = 15
(3) Perimeter: We have obtained the length of one side. For a rhombus, all four sides are equal. So Perimeter = 4 × 15 = 60 cm
(4) Distance between sides:
• Consider fig 13.12.b above. CE is drawn perpendicular to AB. So the length of CE is the distance between sides AB and CD. This will be the same distance between the other two sides AD and CB. So our aim is to find the length of CE.
• For that, we consider the rhombus as a parallelogram. A rhombus is actually a special parallelogram. So the method that we use to calculate the area of a parallelogram can be use to calculate the area of a rhombus also
• The method for parallelogram was explained here. We have to multiply the base by the height
• In our present problem, AB is the base, and CE is the height. So area of the rhombus ABCD = AB×CE
• The area is given to us as 216 cm2. earlier, we calculated AB (length of one side) as 15 cm. So, in the above equation, CE is the only unknown
• We can write: 216 = 15×CE. ∴ CE = 216/15 = 14.4 cm
Solved example 13.9
A 68 cm long rope is used to make a rhombus. The distance between two opposite corners is 16 cm. (1) What is the distance between the other two opposite corners? (2) What is the area of the rhombus?
Solution:
A rough sketch is shown in fig.13.13 below:
• The perimeter is given as 68 cm. We know that all the four sides of a rhombus are equal. So length of one side = 68/4 = 17 cm.
• Distance between one pair of opposite corners is given as 16 cm. This is the length d1 of one diagonal
• Let DB = 16 cm. Then DO = OB = 8 cm (∵ diagonals of a rhombus are perpendicular bisectors of each other)
• We can apply the Pythagoras theorem to the ΔOBC:
• OC2 + 82 = 172 ⇒ OC2 = 172 – 82 = 289 – 64 = 225 ∴ OC = √225 = 15
• Now, OA will also be 15 cm. So distance between the other pair of opposite corners = d2 = AC = 15 + 15 = 30 cm
• To find the area:
In the next section, we will learn about the Area of Isosceles Trapezium.
Area of a Rhombus
Consider a square shown in fig.13.10(a). Usually, a square is specified by it's side. But here, side is not given. Instead, the length of diagonal is given as 6 cm as in fig.b (for a square, both diagonals are equal). |
| Fig.13.10 |
Area of a square of side l = l2
But how do we calculate the area when diagonal is given? For that, we use a 'special property of diagonals of a square':
■ The diagonals of a square bisect each other, and are perpendicular to each other.
So the area consists of 4 equal triangles. This is shown in fig.b. They are: OAB, OBC, OCD and OAD. Not just triangles, they are isosceles triangles. Because two sides are equal (3 cm).
• In each triangle, the base is 3 cm, and height is 3 cm. So area of one triangle = 1⁄2 × b × h = 1⁄2 × 3 × 3 = 41⁄2
• ∴ Total area = 4 × 41⁄2 = 18 cm2
• Fig.c shows a square PQRS whose diagonal is 5 cm. Area of PQRS will be equal to 4 × 1⁄2 × 2.5 × 2.5 = 12.5 cm2
We can use algebra to get a general form:
■ Area of a square is half the square of it's diagonal
• Now consider a Rhombus ABCD in fig.13.11(a). Here also diagonals are given instead of sides. Diagonals of rhombus also have a special property: They bisect each other, and are perpendicular to each other.
 |
| Fig.13.11 |
• Consequently, the triangles formed by the diagonals are not isosceles.
• Even if they are not isosceles, we can calculate the area of each triangle easily. Because, we know the base and height. In fig.13.11(a), the area of each triangle is 1⁄2 × b × h = 1⁄2 × 3 × 2 = 3
• So the total area of the Rhombus ABCD = Total area of 4 squares = 4 × 3 = 12 cm2
• Similarly, the area of the rhombus PQRS in fig will be equal to 4 × 1⁄2 × 2.5 × 2 = 10 cm2
• We can use algebra to get a general form:
■ Area of a Rhombus is half the product of it's diagonals
Solved example 13.7
A square has an area of 4.5 cm2. What is the length of it’s diagonal?
Solution:
The area of a non square rhombus is 216 cm2. The length of one of it’s diagonal is 24 cm. Compute the following:
(1) Length of the other diagonal (2) Length of a side (3) Perimeter (4) Distance between sides
Solution:
(1) Length of the other diagonal:
Let us draw a rough sketch (shown in fig.13.12.a), which shows all the available details:
 |
| Fig.13.12 |
• The diagonals of a rhombus bisects each other. So AC will be split into AO and CO, each of 9 cm. Similarly, BD will be split into BO and DO, each of 12 cm.
• Consider any triangle, say ΔOBC. It is a right triangle, right angled at O. (∵ diagonals of a rhombus are perpendicular to each other)
• We can apply the Pythagoras theorem to this triangle: OB2 + OC2 = BC2
So 122 + 92 = BC2 ⇒ 144 + 81 = BC2 ⇒ 225 = BC2 ∴ BC = √225 = 15
(3) Perimeter: We have obtained the length of one side. For a rhombus, all four sides are equal. So Perimeter = 4 × 15 = 60 cm
(4) Distance between sides:
• Consider fig 13.12.b above. CE is drawn perpendicular to AB. So the length of CE is the distance between sides AB and CD. This will be the same distance between the other two sides AD and CB. So our aim is to find the length of CE.
• For that, we consider the rhombus as a parallelogram. A rhombus is actually a special parallelogram. So the method that we use to calculate the area of a parallelogram can be use to calculate the area of a rhombus also
• The method for parallelogram was explained here. We have to multiply the base by the height
• In our present problem, AB is the base, and CE is the height. So area of the rhombus ABCD = AB×CE
• The area is given to us as 216 cm2. earlier, we calculated AB (length of one side) as 15 cm. So, in the above equation, CE is the only unknown
• We can write: 216 = 15×CE. ∴ CE = 216/15 = 14.4 cm
Solved example 13.9
A 68 cm long rope is used to make a rhombus. The distance between two opposite corners is 16 cm. (1) What is the distance between the other two opposite corners? (2) What is the area of the rhombus?
Solution:
A rough sketch is shown in fig.13.13 below:
 |
| Fig.13.13 |
• Distance between one pair of opposite corners is given as 16 cm. This is the length d1 of one diagonal
• Let DB = 16 cm. Then DO = OB = 8 cm (∵ diagonals of a rhombus are perpendicular bisectors of each other)
• We can apply the Pythagoras theorem to the ΔOBC:
• OC2 + 82 = 172 ⇒ OC2 = 172 – 82 = 289 – 64 = 225 ∴ OC = √225 = 15
• Now, OA will also be 15 cm. So distance between the other pair of opposite corners = d2 = AC = 15 + 15 = 30 cm
• To find the area:
In the next section, we will learn about the Area of Isosceles Trapezium.
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