In the previous section we completed the discussion on triangles between parallel lines. In this section, we will see parallelograms between parallel lines. Consider the parallelogram ABCD (shown in yellow colour) in fig.14.20.
We know how to find it's area. It is equal to b × h = AB × h. Another parallelogram ABPQ (shown in green colour) is drawn with the same base AB, and height h. So it's area will also be equal to AB × h. So we can write:
Theorem 14.3:
■ Two parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
The converse of this theorem can also be written:
Theorem 14.4:
■ Two parallelograms having the same base (or equal bases) and equal areas lie between the same parallels
Another important point can also be noted:
• The base AB remains the same, both in length and position.
• The top parallel side change position. CD becomes PQ. But it does not change length.
• That is., length of CD = Length of PQ.
• This is because, if the new shape ABPQ is to be a parallelogram, opposite sides AB and PQ must be equal.
• So in the fig.14.20, Three entities: AB, CD, and PQ are equal in length. In the fig., all three of them are 4 units in length.
We will now see some solved examples:
Solved example 14.12
In fig.14.21(a), ABCD is a parallelogram. P is any point on CD. Q is any point on AD.
Prove that ΔAPB and ΔBQC are equal in area.
Solution:
• We have to prove that ar (APB) = ar (BQC)
• Let the parallel distance between AB and CD be h1, and that between BC and AD be h2. This is shown in fig.b. Note that the measurements should be made in an exact perpendicular direction as shown in fig.14.3
1. ar (ABCD) = b × h = AB × h1
2. ar (ABCD) = b × h = BC × h2
3. ar (APB) = 1⁄2 × b × h = 1⁄2 × AB × h1
4. ar (APB) = 1⁄2 × b × h = 1⁄2 × BC × h2
• Steps (1) and (2) are two different 'directions' (details here) to calculate the area of the parallelogram ABCD
• Though the directions are different, the results from the two methods must be the same because, it is the area of the same parallelogram
So we can write:
5. AB × h1 = BC × h2 dividing both sides by 2, we get:
6. 1⁄2 × AB × h1 = 1⁄2 × BC × h2
Comparing step (6) with (3) and (4) we get: ar (APB) = ar (BQC)
Solved example 14.13
ABCD is a parallelogram. E, F, G and H are the midpoints of the sides. Show that ar (EFGH) = 1⁄2 ar (ABCD)
Solution:
The rough sketch is shown in fig.14.22(a)
• In fig.b, a line is drawn through H and F. This will be parallel to both AB and CD
• Let the parallel distance between HF and CD be x, and that between HF and AB be y
• Let the parallel distance between AB and CD be h (the height of the original parallelogram ABCD)
1. ar (ABCD) = b × h = AB × h
2. ar (GFH) = 1⁄2 × b × h = 1⁄2 × HF × x
3. ar (EFH) = 1⁄2 × b × h = 1⁄2 × HF × y
4. ar (EFGH) = ar (GFH) + ar (EFH)
⇒ ar (EFGH) = 1⁄2 × HF × x + 1⁄2 × HF × y = 1⁄2 × HF × (x + y)
⇒ ar (EFGH) = 1⁄2 × AB × (h) [∵ HF = AB and (x + y) = h]
Comparing this result with step (1) we get: ar (EFGH) = 1⁄2 ar (ABCD)
Solved example 14.14
ABCD is a parallelogram. P is any point inside ABCD. Prove that:
(i) ar (APB) + ar (PCD) = 1⁄2 ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
Solution:
The rough sketch is shown in fig.14.23(a)
• In fig.b, an yellow dashed line is drawn through P, parallel to AB. This will be parallel to CD also
• Let the parallel distance between the dashed line and CD be x, and that between the dashed line and AB be y
• Let the parallel distance between AB and CD be h (the height of the original parallelogram ABCD)
1. ar (ABCD) = b × h = AB × h
2. ar (PCD) = 1⁄2 × b × h = 1⁄2 × CD × x
3. ar (APB) = 1⁄2 × b × h = 1⁄2 × AB × y
4. ar (PCD) + ar (APB) = 1⁄2 × CD × x + 1⁄2 × AB × y = 1⁄2 × AB × (x + y) [∵ CD = AB]
⇒ ar (PCD) + ar (APB) = 1⁄2 × AB × (h) [∵ (x + y) = h]
Comparing this result with step (1) we get: ar (PCD) + ar (APB) = 1⁄2 ar (ABCD) - - - Answer 1
• Sum of all the four small triangles = ar (ABCD)
⇒ ar (APB) + ar (PCD) + ar (APD) + ar (PBC) = ar (ABCD)
• Let us group the above:
[ar (APB) + ar (PCD)] + [ar (APD) + ar (PBC)] = ar (ABCD)
• From Answer 1, we have, the first group = 1⁄2 ar (ABCD)
• So we can write: 1⁄2 ar (ABCD) + [ar (APD) + ar (PBC)] = ar (ABCD)
• So the second group is also equal to 1⁄2 ar (ABCD)
• So we can write: [ar (APB) + ar (PCD)] = [ar (APD) + ar (PBC)] - - - Answer 2
Solved example 14.15
ABCD and ABPQ are parallelograms. R is any point on PB. Prove that, ar (ARQ) = 1⁄2 ar (ABCD)
Solution:
The rough sketch is shown in fig.14.24(a)
• Let the parallel distance between AQ and BP be h
1. ar (ABPQ) = b × h = AQ × h
2. ar (ARQ) = 1⁄2 × b × h = 1⁄2 × AQ × h
• Comparing the above two results we get: ar (ARQ) = 1⁄2 ar (ABPQ)
• But ar (ABPQ) = ar (ABCD) [∵ Parallelograms between two parallel lines and with same base]
So we can write: ar (ARQ) = 1⁄2 ar (ABCD)
In the next section we will see Division of Triangles.
 |
| Fig.14.20 |
Theorem 14.3:
■ Two parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
The converse of this theorem can also be written:
Theorem 14.4:
■ Two parallelograms having the same base (or equal bases) and equal areas lie between the same parallels
Another important point can also be noted:
• The base AB remains the same, both in length and position.
• The top parallel side change position. CD becomes PQ. But it does not change length.
• That is., length of CD = Length of PQ.
• This is because, if the new shape ABPQ is to be a parallelogram, opposite sides AB and PQ must be equal.
• So in the fig.14.20, Three entities: AB, CD, and PQ are equal in length. In the fig., all three of them are 4 units in length.
We will now see some solved examples:
Solved example 14.12
In fig.14.21(a), ABCD is a parallelogram. P is any point on CD. Q is any point on AD.
 |
| Fig.14.21 |
Solution:
• We have to prove that ar (APB) = ar (BQC)
• Let the parallel distance between AB and CD be h1, and that between BC and AD be h2. This is shown in fig.b. Note that the measurements should be made in an exact perpendicular direction as shown in fig.14.3
1. ar (ABCD) = b × h = AB × h1
2. ar (ABCD) = b × h = BC × h2
3. ar (APB) = 1⁄2 × b × h = 1⁄2 × AB × h1
4. ar (APB) = 1⁄2 × b × h = 1⁄2 × BC × h2
• Steps (1) and (2) are two different 'directions' (details here) to calculate the area of the parallelogram ABCD
• Though the directions are different, the results from the two methods must be the same because, it is the area of the same parallelogram
So we can write:
5. AB × h1 = BC × h2 dividing both sides by 2, we get:
6. 1⁄2 × AB × h1 = 1⁄2 × BC × h2
Comparing step (6) with (3) and (4) we get: ar (APB) = ar (BQC)
Solved example 14.13
ABCD is a parallelogram. E, F, G and H are the midpoints of the sides. Show that ar (EFGH) = 1⁄2 ar (ABCD)
Solution:
The rough sketch is shown in fig.14.22(a)
 |
| Fig.14.22 |
• Let the parallel distance between HF and CD be x, and that between HF and AB be y
• Let the parallel distance between AB and CD be h (the height of the original parallelogram ABCD)
1. ar (ABCD) = b × h = AB × h
2. ar (GFH) = 1⁄2 × b × h = 1⁄2 × HF × x
3. ar (EFH) = 1⁄2 × b × h = 1⁄2 × HF × y
4. ar (EFGH) = ar (GFH) + ar (EFH)
⇒ ar (EFGH) = 1⁄2 × HF × x + 1⁄2 × HF × y = 1⁄2 × HF × (x + y)
⇒ ar (EFGH) = 1⁄2 × AB × (h) [∵ HF = AB and (x + y) = h]
Comparing this result with step (1) we get: ar (EFGH) = 1⁄2 ar (ABCD)
Solved example 14.14
ABCD is a parallelogram. P is any point inside ABCD. Prove that:
(i) ar (APB) + ar (PCD) = 1⁄2 ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
Solution:
The rough sketch is shown in fig.14.23(a)
 |
| Fig.14.23 |
• Let the parallel distance between the dashed line and CD be x, and that between the dashed line and AB be y
• Let the parallel distance between AB and CD be h (the height of the original parallelogram ABCD)
1. ar (ABCD) = b × h = AB × h
2. ar (PCD) = 1⁄2 × b × h = 1⁄2 × CD × x
3. ar (APB) = 1⁄2 × b × h = 1⁄2 × AB × y
4. ar (PCD) + ar (APB) = 1⁄2 × CD × x + 1⁄2 × AB × y = 1⁄2 × AB × (x + y) [∵ CD = AB]
⇒ ar (PCD) + ar (APB) = 1⁄2 × AB × (h) [∵ (x + y) = h]
Comparing this result with step (1) we get: ar (PCD) + ar (APB) = 1⁄2 ar (ABCD) - - - Answer 1
• Sum of all the four small triangles = ar (ABCD)
⇒ ar (APB) + ar (PCD) + ar (APD) + ar (PBC) = ar (ABCD)
• Let us group the above:
[ar (APB) + ar (PCD)] + [ar (APD) + ar (PBC)] = ar (ABCD)
• From Answer 1, we have, the first group = 1⁄2 ar (ABCD)
• So we can write: 1⁄2 ar (ABCD) + [ar (APD) + ar (PBC)] = ar (ABCD)
• So the second group is also equal to 1⁄2 ar (ABCD)
• So we can write: [ar (APB) + ar (PCD)] = [ar (APD) + ar (PBC)] - - - Answer 2
Solved example 14.15
ABCD and ABPQ are parallelograms. R is any point on PB. Prove that, ar (ARQ) = 1⁄2 ar (ABCD)
Solution:
The rough sketch is shown in fig.14.24(a)
 |
| Fig.14.24 |
1. ar (ABPQ) = b × h = AQ × h
2. ar (ARQ) = 1⁄2 × b × h = 1⁄2 × AQ × h
• Comparing the above two results we get: ar (ARQ) = 1⁄2 ar (ABPQ)
• But ar (ABPQ) = ar (ABCD) [∵ Parallelograms between two parallel lines and with same base]
So we can write: ar (ARQ) = 1⁄2 ar (ABCD)
In the next section we will see Division of Triangles.
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