In the previous section we completed the discussion on the calculation of area of a parallelogram. We also saw a solved example. In this section, we will see more solved examples.
Solved example 13.3
Draw a parallelogram of area 25 and perimeter 24 cm
Solution:
The parallelogram has to satisfy two conditions:
■ The area must be 25 cm2
■ The perimeter must be 24 cm
• We can first make a rectangle of area 25, and then modify it to make a parallelogram of the same area. For that, we remove a triangle from the left side of the rectangle, and attach it to the right side.
• The perimeter of the parallelogram has to be 24 cm. That means, the sum of two adjacent sides must be 12 cm. The possible combinations which give 12 cm are:
• 1 +11, 2 +10, 3 +9, 4 +8, 5 +7 and 6 +6. Out of these combinations, only 5 is available as a factor of 25. So, we have to choose the combination 5 +7. That is., One side of the parallelogram is 5 cm, and it's adjacent side is 7 cm.
• The area of this rectangle is to be 25. So we must select 5 cm as the base. If we choose 7 cm, the other side will not be a whole number. (∵ 25 ÷ 7 gives a decimal value, not a whole number)
• When we choose 5 cm. the adjacent side will also be 5 cm (∵5 × 5 = 25)
• So we get the final fix: A square of side 5 cm. This square can be modified into a parallelogram as shown in fig below:
• The side of the parallelogram is to be 7 cm. So, with A as center, draw an arc (shown in green colour in fig.b) of radius 7 cm, cutting CD at E. Join AE. The length of AE = 7 cm.
• Through B, draw a line BF', parallel to AE.
• Extend the line CD towards the right until it meets BF' at F.
• Change the firm lines AD, BC and ED into dashed lines. Then ABFE is the required Parallelogram.
Proof:
• ABFE is a parallelogram. It is evident because:
♦ Opposite sides AB and EF are already parallel as they are opposite sides of the original square ABCD
♦ Opposite sides AE and BF are parallel because BF is drawn parallel to AB
• In fig.c , we can see that, from the square ABCD, a triangle AED is removed from the left side, and another triangle BFC is attached at the right side. We have to prove that these two triangles are equal. For this, we can use SAS congruence:
• AD = BC = 5 cm (since opposite sides of a square)
• AE = BF = 6 cm (since opposite sides of a parallelogram)
• ∠DAE = ∠CBF (∵ AD∥BC and AE∥BF)
• So the two triangles are equal.
Solved example 13.4
In the fig.13.7(a), PQRS is a parallelogram. ΔPQT is drawn within this parallelogram. If the area of PQT is 18 cm2, what is the area of the parallelogram PQRS?
Solution:
• To find the area of the parallelogram, we need it’s height h and base b. These are marked in fig.b.
• The same h and b can be used to calculate the area of the triangle also. This is because, area of any triangle = 1⁄2 × base × height
• But the area of the triangle is given as 18 cm2
• So 1⁄2 × b × h = 18. ∴ b × h = 36
• But b × h is the area of the parallelogram. So we can write: Area of parallelogram PQRS = 36 cm2
Solved example 13.5
Two pairs of parallel lines intersect to form a parallelogram as shown in the fig.13.8.(a) Calculate the area and perimeter of the parallelogram.
Solution:
• The given fig(a), shows the distances between each set of the parallel lines. These distances are 'important clues' which help us to determine various measurements related to the parallelogram. This is shown in fig(b)
■ The 4 cm is the height h of the parallelogram ABCD because, it is measured perpendicular to the side AB
■ The 3 cm is the height of the parallelogram ABCD in the ‘other direction’ because it is measured perpendicular to BC
■ We are also given the length of AB as 5 cm. This can be taken as the base b
With this information, we can proceed to solve the problem:
• Area of the parallelogram ABCD = b × h = 5 × 4 = 20 cm2
• We calculated this area by working in the horizontal direction. We must get the same area if we work in the ‘other direction’ also. Let us do such a calculation:
• Area = b × h = BC × 3
• In the above equation, Area is known as 20 cm2. The only unknown is BC. So we can write:
BC = Area/3 = 20/3 = 6.67 cm
• Now we have lengths of two adjacent sides: AB and BC. So perimeter = 2 × (5 + 6.67) = 2 × 11.67 = 23.34 cm
Solved example 13.6
Find the area of the Parallelogram ABCD shown in fig.13.9 below:
Solution:
• In the fig., it is given that, the length of diagonal AC = 6 cm, and the perpendicular DE to AC has a length of 3.5 cm.
• These details enable us to calculate the area of ΔACD (∵ AC is the base and DE is the height of ΔACD)
• So we get: Area of ΔACD = 1⁄2 × 6 × 3.5
• In a parallelogram, any one of it's diagonal will split the parallelogram into two equal triangles. So area of ΔACD = area of ΔABC. ∴ two times the area of ΔACD will be equal to the area of the parallelogram ABCD
• So area of parallelogram ABCD = 2 × 1⁄2 × 6 × 3.5 = 6 × 3.5 = 21 cm2
In the next section, we will learn about the Area of a Rhombus.
Solved example 13.3
Draw a parallelogram of area 25 and perimeter 24 cm
Solution:
The parallelogram has to satisfy two conditions:
■ The area must be 25 cm2
■ The perimeter must be 24 cm
• We can first make a rectangle of area 25, and then modify it to make a parallelogram of the same area. For that, we remove a triangle from the left side of the rectangle, and attach it to the right side.
• The perimeter of the parallelogram has to be 24 cm. That means, the sum of two adjacent sides must be 12 cm. The possible combinations which give 12 cm are:
• 1 +11, 2 +10, 3 +9, 4 +8, 5 +7 and 6 +6. Out of these combinations, only 5 is available as a factor of 25. So, we have to choose the combination 5 +7. That is., One side of the parallelogram is 5 cm, and it's adjacent side is 7 cm.
• The area of this rectangle is to be 25. So we must select 5 cm as the base. If we choose 7 cm, the other side will not be a whole number. (∵ 25 ÷ 7 gives a decimal value, not a whole number)
• When we choose 5 cm. the adjacent side will also be 5 cm (∵5 × 5 = 25)
• So we get the final fix: A square of side 5 cm. This square can be modified into a parallelogram as shown in fig below:
 |
| Fig.13.6 |
• Through B, draw a line BF', parallel to AE.
• Extend the line CD towards the right until it meets BF' at F.
• Change the firm lines AD, BC and ED into dashed lines. Then ABFE is the required Parallelogram.
Proof:
• ABFE is a parallelogram. It is evident because:
♦ Opposite sides AB and EF are already parallel as they are opposite sides of the original square ABCD
♦ Opposite sides AE and BF are parallel because BF is drawn parallel to AB
• In fig.c , we can see that, from the square ABCD, a triangle AED is removed from the left side, and another triangle BFC is attached at the right side. We have to prove that these two triangles are equal. For this, we can use SAS congruence:
• AD = BC = 5 cm (since opposite sides of a square)
• AE = BF = 6 cm (since opposite sides of a parallelogram)
• ∠DAE = ∠CBF (∵ AD∥BC and AE∥BF)
• So the two triangles are equal.
Solved example 13.4
In the fig.13.7(a), PQRS is a parallelogram. ΔPQT is drawn within this parallelogram. If the area of PQT is 18 cm2, what is the area of the parallelogram PQRS?
 |
| Fig.13.7 |
• To find the area of the parallelogram, we need it’s height h and base b. These are marked in fig.b.
• The same h and b can be used to calculate the area of the triangle also. This is because, area of any triangle = 1⁄2 × base × height
• But the area of the triangle is given as 18 cm2
• So 1⁄2 × b × h = 18. ∴ b × h = 36
• But b × h is the area of the parallelogram. So we can write: Area of parallelogram PQRS = 36 cm2
Solved example 13.5
Two pairs of parallel lines intersect to form a parallelogram as shown in the fig.13.8.(a) Calculate the area and perimeter of the parallelogram.
 |
| Fig.13.8 |
• The given fig(a), shows the distances between each set of the parallel lines. These distances are 'important clues' which help us to determine various measurements related to the parallelogram. This is shown in fig(b)
■ The 4 cm is the height h of the parallelogram ABCD because, it is measured perpendicular to the side AB
■ The 3 cm is the height of the parallelogram ABCD in the ‘other direction’ because it is measured perpendicular to BC
■ We are also given the length of AB as 5 cm. This can be taken as the base b
With this information, we can proceed to solve the problem:
• Area of the parallelogram ABCD = b × h = 5 × 4 = 20 cm2
• We calculated this area by working in the horizontal direction. We must get the same area if we work in the ‘other direction’ also. Let us do such a calculation:
• Area = b × h = BC × 3
• In the above equation, Area is known as 20 cm2. The only unknown is BC. So we can write:
BC = Area/3 = 20/3 = 6.67 cm
• Now we have lengths of two adjacent sides: AB and BC. So perimeter = 2 × (5 + 6.67) = 2 × 11.67 = 23.34 cm
Solved example 13.6
Find the area of the Parallelogram ABCD shown in fig.13.9 below:
 |
| Fig.13.9 |
• In the fig., it is given that, the length of diagonal AC = 6 cm, and the perpendicular DE to AC has a length of 3.5 cm.
• These details enable us to calculate the area of ΔACD (∵ AC is the base and DE is the height of ΔACD)
• So we get: Area of ΔACD = 1⁄2 × 6 × 3.5
• In a parallelogram, any one of it's diagonal will split the parallelogram into two equal triangles. So area of ΔACD = area of ΔABC. ∴ two times the area of ΔACD will be equal to the area of the parallelogram ABCD
• So area of parallelogram ABCD = 2 × 1⁄2 × 6 × 3.5 = 6 × 3.5 = 21 cm2
In the next section, we will learn about the Area of a Rhombus.
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