Areas of Parallelograms and Triangle Exercise 9.3 - CBSE Class 9

Key Geometric Theorems of Exercise 9.3

Exercise 9.3 of CBSE Class 9 Mathematics is built upon two core geometrical truths. The first is that triangles on the same base (or equal bases) and between the same parallels are equal in area. To understand why, recall that the area of a triangle is given by the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. When two triangles share the same base and lie between the same parallel lines, their perpendicular heights (altitudes) are identical. Consequently, their calculated areas must be equal.

The second core concept is the median property: a median of a triangle divides it into two triangles of equal areas. A median connects a vertex to the midpoint of the opposite side. Since the midpoint divides the base into two equal segments, and both triangles share the same vertex (meaning they share the same perpendicular height), their areas are exactly equal. Mastering these properties allows you to solve complex multi-step geometric proofs step-by-step.

Step-by-Step: Proving the Median Theorem

1. Construct the Triangle and Median — Draw any triangle ABC. Construct a median AD from vertex A to the base BC. By definition, D is the midpoint of BC, which means BD = CD.
2. Draw an Auxiliary Altitude — To compare the heights, draw an altitude AM perpendicular to BC (or BC produced, if the triangle is obtuse-angled).
3. Express the Area of Both Sub-Triangles — Write the area equations: Area(ABD) = 1/2 BD AM. Area(ADC) = 1/2 CD AM.
4. Equate and Prove — Since D is the midpoint, substitute BD = CD into the area equations. You will see that Area(ABD) = Area(ADC). This proves that the median divides the triangle into two equal areas.

Crucial Exam Tips & Common Mistakes to Avoid

• Confusing Median with Altitude: Remember, a median bisects the opposite side but does NOT necessarily meet it at a 90-degree angle. Never assume the median is perpendicular unless the triangle is isosceles or equilateral.
• Failing to State Parallels: When writing proofs using the 'same base' theorem, always state clearly which lines are parallel (e.g., AB || CD) and why. Leaving this out can result in a loss of marks in CBSE board exams.
• Incorrect Area Notation: Always use the proper standard notation like 'ar(ABC)' to represent the area of triangle ABC in your proof statements to maintain clear, formal mathematical writing.

Practice Questions with Solutions

• Q: In a triangle ABC, E is the midpoint of the median AD. Show that ar(BED) = 1/4 ar(ABC). A: Step 1: AD is the median of triangle ABC. Therefore, by the median property, ar(ABD) = 1/2 ar(ABC). Step 2: Now, consider triangle ABD. Here, BE is the median because E is the midpoint of AD. Step 3: Since BE is the median of triangle ABD, it divides it into two equal areas: ar(BED) = 1/2 ar(ABD). Step 4: Substitute the value of ar(ABD) from Step 1 into Step 3: ar(BED) = 1/2 [1/2 ar(ABC)] = 1/4 ar(ABC). Final answer: Hence proved, ar(BED) = 1/4 * ar(ABC).
• Q: Show that the diagonals of a parallelogram divide it into four triangles of equal area. A: Step 1: Let ABCD be a parallelogram whose diagonals AC and BD intersect at O. In a parallelogram, diagonals bisect each other, so O is the midpoint of both AC and BD. Step 2: In triangle ABC, BO is the median because O is the midpoint of AC. Therefore, ar(AOB) = ar(BOC). Step 3: Similarly, in triangle BCD, CO is the median because O is the midpoint of BD. Therefore, ar(BOC) = ar(COD). Step 4: In triangle CDA, DO is the median, so ar(COD) = ar(AOD). Step 5: From steps 2, 3, and 4, we establish that ar(AOB) = ar(BOC) = ar(COD) = ar(AOD). Final answer: Hence, the diagonals of a parallelogram divide it into four triangles of equal area.
• Q: In a triangle ABC, if AD is a median, and P is any point on AD, show that ar(ABP) = ar(ACP). A: Step 1: Since AD is the median of triangle ABC, we know that ar(ABD) = ar(ACD). Step 2: Now consider the smaller triangle PBC. PD is the median of triangle PBC because D is the midpoint of BC. Therefore, ar(PBD) = ar(PCD). Step 3: Subtracting the area of the smaller triangle from the larger triangle: ar(ABD) - ar(PBD) = ar(ACD) - ar(PCD). Step 4: Simplifying the subtraction yields: ar(ABP) = ar(ACP). Final answer: Hence proved, ar(ABP) = ar(ACP).
• Q: D and E are points on sides AB and AC respectively of triangle ABC such that ar(DBC) = ar(EBC). Prove that DE || BC. A: Step 1: We are given that triangle DBC and triangle EBC have equal areas: ar(DBC) = ar(EBC). Step 2: Observe that both triangles DBC and EBC lie on the same base, which is BC. Step 3: According to the converse of the area theorem, if two triangles have the same base and equal areas, they must lie between the same parallel lines. Step 4: Therefore, the line joining the vertices opposite to the base must be parallel to the base line. This implies DE is parallel to BC. Final answer: Hence proved, DE || BC.

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