Areas of Parallelograms and Triangle: A Deep Dive for Class 9

Understanding the Core Concepts: Same Base and Same Parallels

The entire chapter hinges on two simple-sounding but very important ideas: figures on the 'same base' and figures 'between the same parallels'. Let's break them down.

1. Figures on the Same Base:
Two geometric figures are said to be on the same base if they have a common side. For instance, imagine a parallelogram ABCD and a triangle ABE. Both figures share the side AB. In this case, we say that parallelogram ABCD and triangle ABE are on the same base AB. It's that simple! The common side acts as the foundation for both shapes.

2. Figures Between the Same Parallels:
This condition is a bit more specific. For two figures on the same base to be 'between the same parallels', the vertices (or the parts of the figures) opposite to the common base must lie on a single straight line that is parallel to the base. In our example of parallelogram ABCD and triangle ABE on base AB, if the vertex E of the triangle and the vertices D and C of the parallelogram all lie on a line L which is parallel to AB, then we can say the figures are between the same parallels AB and L. If E was not on the line passing through D and C, they would not be between the same parallels, even if they share the same base.

Key Theorems on Areas

Theorem 1: Parallelograms on Same Base

Parallelograms on the same base and between the same parallels are equal in area. If parallelogram ABCD and ABEF are on base AB and between parallels AB and CF, then ar(ABCD) = ar(ABEF).

Theorem 2: Triangle and Parallelogram on Same Base

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram. If △ABE and parallelogram ABCD are on base AB and between parallels AB and CD, then ar(△ABE) = 1/2 * ar(ABCD).

Theorem 3: Triangles on Same Base

Two triangles on the same base (or equal bases) and between the same parallels are equal in area. If △ABC and △ABD are on base AB and between parallels AB and CD, then ar(△ABC) = ar(△ABD).

Converse of Theorem 3

Two triangles having the same base (or equal bases) and equal areas lie between the same parallels. This is useful for proving that two lines are parallel.

Worked Examples: Applying the Theorems

• Problem: In parallelogram ABCD, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find the length of AD. Solution: Step 1: Identify the properties of the parallelogram. Since ABCD is a parallelogram, opposite sides are equal. Therefore, CD = AB = 16 cm. Step 2: Calculate the area of the parallelogram using the first set of base and height. The base is DC and the corresponding perpendicular height is AE. Area(ABCD) = Base × Height = DC × AE Area(ABCD) = 16 cm × 8 cm = 128 cm². Step 3: Use the calculated area with the second set of base and height. We can also calculate the area using base AD and its corresponding height CF. Area(ABCD) = AD × CF Step 4: Equate the area and solve for the unknown side AD. 128 cm² = AD × 10 cm AD = 128 / 10 = 12.8 cm. Final Answer: The length of AD is 12.8 cm.
• Problem: E is any point on the median AD of a △ABC. Show that ar(ABE) = ar(ACE). Solution: Step 1: Understand the property of a median. AD is the median of △ABC. A median divides a triangle into two triangles of equal area. Therefore, ar(ABD) = ar(ACD). --- (1) Step 2: Consider the smaller triangle △EBC. Since E is on AD, ED is the median of △EBC (as D is the midpoint of BC). Therefore, ED divides △EBC into two triangles of equal areas. ar(EBD) = ar(ECD). --- (2) Step 3: Subtract the area of the smaller triangles from the larger ones. We subtract equation (2) from equation (1). ar(ABD) - ar(EBD) = ar(ACD) - ar(ECD). Step 4: Observe the resulting figures. From the figure, ar(ABD) - ar(EBD) gives ar(ABE), and ar(ACD) - ar(ECD) gives ar(ACE). Final Answer: ar(ABE) = ar(ACE). (Hence proved)

Avoid These Common Mistakes!

Pay close attention to these points to avoid losing marks in your exams:

• Confusing Congruence and Equal Area: Remember, if two figures are congruent, they must have equal areas. However, the reverse is not always true! Two triangles can have the same area without being congruent (e.g., one might be tall and narrow, the other short and wide). Don't assume congruence just from equal areas.
• Incorrectly Identifying Height: The height (or altitude) of a parallelogram or triangle is always the perpendicular distance between the base and the opposite side/vertex. A slanted side is never the height unless it's a rectangle or a right-angled triangle.
• Assuming Parallel Lines: Do not assume that two lines are parallel just because they look parallel in a diagram. You can only use the 'between same parallels' theorems if the problem explicitly states the lines are parallel or if you can prove it using other geometric properties.

Practice Questions with Solutions

• Q: The median of a triangle divides it into two: A: Step 1: Recall the definition of a median. A median connects a vertex to the midpoint of the opposite side. Step 2: Recall the theorem related to medians and area. A median of a triangle divides it into two triangles of equal areas. Final answer: The median of a triangle divides it into two triangles of equal areas.
• Q: In the given figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that ar(ABCD) = ar(EFCD) and ar(ABCD) = DC x AL. A: Step 1: To prove ar(ABCD) = ar(EFCD). A rectangle is a special type of parallelogram. Both parallelogram ABCD and rectangle EFCD are on the same base DC and are between the same parallels DC and AF (since AB || DC and EF || DC). By the theorem, parallelograms on the same base and between the same parallels are equal in area. Thus, ar(ABCD) = ar(EFCD). Step 2: To prove ar(ABCD) = DC x AL. We know the area of a parallelogram is the product of its base and corresponding altitude. For parallelogram ABCD, DC is the base and AL is the corresponding altitude. Therefore, ar(ABCD) = DC x AL. Final answer: Both statements are proved based on fundamental area theorems.
• Q: Let P and Q be any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC). A: Step 1: Consider △APB and parallelogram ABCD. They lie on the same base AB and are between the same parallel lines AB and DC. According to the theorem, the area of the triangle is half the area of the parallelogram. So, ar(APB) = 1/2 ar(ABCD). --- (1) Step 2: Now, consider △BQC and parallelogram ABCD. They lie on the same base BC and are between the same parallel lines BC and AD. Therefore, ar(BQC) = 1/2 ar(ABCD). --- (2) Step 3: Compare the results from Step 1 and Step 2. From equations (1) and (2), we can see that both areas are equal to half the area of the same parallelogram. Final answer: ar(APB) = ar(BQC).
• Q: A farmer has a field in the shape of a parallelogram PQRS. She takes a point A on RS and joins it to points P and Q. In how many parts is the field divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it? A: Step 1: Visualize the situation. Joining point A on RS to P and Q divides the field into three parts. Step 2: Identify the shapes. The three parts are △PSA, △PAQ, and △QRA. They are all triangles. Step 3: Apply the area theorem. △PAQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS. Therefore, ar(PAQ) = 1/2 ar(PQRS). Step 4: Determine the area of the remaining parts. The remaining area is ar(PSA) + ar(QRA) = ar(PQRS) - ar(PAQ) = ar(PQRS) - 1/2 ar(PQRS) = 1/2 ar(PQRS). Step 5: Formulate the advice for the farmer. The farmer can sow wheat in △PAQ and pulses in the other two triangles (△PSA and △QRA) combined. Or, she can sow pulses in △PAQ and wheat in the other two combined. This way, both crops are sown in equal portions (half the area of the field each). Final answer: The field is divided into 3 triangles. She should plant one crop in the middle triangle △PAQ and the other crop in the remaining two triangles △PSA and △QRA.

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