Areas of Parallelograms and Triangle Ex 9.4 NCERT Solutions

Conceptual Foundations of Exercise 9.4

To successfully tackle the challenges of areas of parallelograms and triangle ex 9 4 class 9 ncert, we must arm ourselves with three foundational properties. First, parallelograms on the same base and between the same parallels are equal in area. If parallelogram ABCD and EFCD share base CD and lie between parallel lines AB and CD, then Area(ABCD) = Area(EFCD). Second, two triangles on the same base (or equal bases) and between the same parallels are equal in area. This is crucial when simplifying complex quadrilateral regions. Third, the median of a triangle divides it into two triangles of equal areas. If AD is a median of triangle ABC, then Area(ABD) = Area(ACD). Exercise 9.4 builds on these core postulates by combining them. For instance, if you join the midpoints of the sides of any triangle, you form four smaller triangles, each having exactly one-fourth the area of the original triangle. We solve these problems not by measuring, but by logically linking bases and heights through parallel lines.

Step-by-Step Proof Techniques

1. Identify the Base and Parallel Lines — Scan the geometric figure to find triangles or parallelograms sharing a common side (base) and located between parallel lines to establish initial area equality.
2. Locate Medians and Midpoints — Identify if any line segment joins a vertex to the midpoint of the opposite side. Apply the median property: ar(ABD) = ar(ACD) = 0.5 * ar(ABC).
3. Set Up Area Equations — Express the larger geometric figure as a sum of its non-overlapping constituent parts (e.g., ar(ABCD) = ar(ADE) + ar(BCDE)) to break down complex polygons.
4. Substitute and Simplify — Use algebraic substitution of equal areas to eliminate unwanted terms and isolate the required expression to be proved.

Key Theoretical Proofs Explained

• Theorem: Prove that the median of a triangle divides it into two triangles of equal area. Step 1: Let ABC be a triangle and AD be its median. Draw altitude AM perpendicular to BC. Step 2: Area(ABD) = 0.5 base height = 0.5 BD AM. Step 3: Area(ACD) = 0.5 base height = 0.5 CD AM. Step 4: Since AD is the median, BD = CD. Thus, Area(ABD) = Area(ACD).
• Problem: In triangle ABC, D and E are midpoints of BC and AD respectively. Prove that ar(BED) = 1/4 ar(ABC). Step 1: AD is a median of triangle ABC. Therefore, ar(ABD) = 0.5 ar(ABC). Step 2: In triangle ABD, BE is the median on AD (since E is the midpoint of AD). Step 3: Therefore, ar(BED) = 0.5 ar(ABD). Step 4: Substituting step 1 into step 3: ar(BED) = 0.5 (0.5 ar(ABC)) = 0.25 ar(ABC).

Common Mistakes to Avoid in Exams

Don't skip the Parallel Postulate proof! A common mistake in the CBSE board exam is assuming two figures are between the same parallels without explicitly stating why the lines are parallel. Always state: 'Since line L1 is parallel to line L2, and both figures lie on the same base...'. If lines are not given parallel, find alternate interior angles or opposite parallel sides first. Another major error is confusing perimeter with area. Two triangles with equal areas do not necessarily have equal perimeters!

Practice Questions with Solutions

• Q: In triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC). A: Step 1: AD is a median of triangle ABC. Therefore, ar(ABD) = ar(ACD) = 1/2 ar(ABC). Step 2: In triangle ABD, BE is the median on AD (since E is the midpoint of AD). Step 3: Therefore, ar(BED) = 1/2 ar(ABD). Step 4: Substitute equation 1 into equation 2: ar(BED) = 1/2 [1/2 ar(ABC)] = 1/4 ar(ABC). Final answer: ar(BED) = 1/4 * ar(ABC) is proved.
• Q: Show that the diagonals of a parallelogram divide it into four triangles of equal area. A: Step 1: Let ABCD be a parallelogram with diagonals AC and BD intersecting at O. Since the diagonals of a parallelogram bisect each other, O is the midpoint of AC and BD. Step 2: In triangle ABC, BO is the median. Thus, ar(AOB) = ar(BOC). Step 3: In triangle BCD, CO is the median. Thus, ar(BOC) = ar(COD). Step 4: Similarly, in triangle ACD, DO is the median, so ar(COD) = ar(AOD). Step 5: Equating all relations: ar(AOB) = ar(BOC) = ar(COD) = ar(AOD). Final answer: The four triangles formed by the intersecting diagonals have equal areas.
• Q: D, E, and F are respectively the mid-points of the sides BC, CA, and AB of a triangle ABC. Show that BDEF is a parallelogram. A: Step 1: In triangle ABC, E and F are the mid-points of AC and AB. By Mid-point Theorem, FE is parallel to BC, which means FE is parallel to BD. Step 2: Similarly, D and E are the mid-points of BC and AC. Thus, DE is parallel to AB, which means DE is parallel to FB. Step 3: In quadrilateral BDEF, opposite sides are parallel (FE || BD and DE || FB). Final answer: Since both pairs of opposite sides are parallel, BDEF is a parallelogram.
• Q: If the medians of a triangle ABC intersect at G, show that ar(AGC) = 1/3 ar(ABC). A: Step 1: Let AD, BE, and CF be the medians intersecting at centroid G. Median AD divides triangle ABC into two equal areas: ar(ABD) = ar(ACD). Step 2: GD is the median of triangle GBC, so ar(GBD) = ar(GCD). Step 3: Subtracting step 2 from step 1: ar(ABD) - ar(GBD) = ar(ACD) - ar(GCD) which simplifies to ar(ABG) = ar(ACG). Step 4: Similarly, we can show ar(ABG) = ar(BCG). Therefore, ar(ABG) = ar(BCG) = ar(ACG). Step 5: Since the total area is ar(ABC) = ar(ABG) + ar(BCG) + ar(ACG) = 3 ar(ACG), we get ar(ACG) = 1/3 ar(ABC). Final answer: ar(AGC) = 1/3 ar(ABC) is proved.

Frequently Asked Questions