NCERT Solutions Class 9 Maths Areas of Parallelograms and Triangles

Core Concepts: Same Base and Same Parallels

In geometry, the area of a closed planar figure is the measure of the region enclosed by it. In Class 9, we focus on a special relationship: when do two figures share the exact same area without being congruent?

Two geometric figures are said to be on the same base and between the same parallels if they share a common side (the base), and their vertices opposite to this common base lie on a line parallel to the base.

For example, if a parallelogram ABCD and another parallelogram EFCD share the base CD, and their top vertices A, B, E, and F lie on a single line parallel to CD, then they are on the same base and between the same parallels. Under this condition, their areas are mathematically identical! Similarly, if a triangle and a parallelogram share a base and lie between the same parallels, the area of the triangle is exactly half the area of the parallelogram.

Proof of Theorem 9.1: Parallelograms on Same Base

1. Setup the Figures — Draw two parallelograms ABCD and EFCD sharing a common base CD, lying between the same parallel lines AF and CD. Here, A-B-E-F forms a straight line.
2. Identify Key Triangles — Look at the two triangles on the outer edges: triangle ADF and triangle BCE. We will prove these two triangles are congruent to equate their areas.
3. Establish Congruence (AAS Criteria) — 1. Angle DAF = Angle CBE (corresponding angles, as AD || BC and AF is a transversal line). 2. Angle AFD = Angle BCE (corresponding angles, as FD || EC and AF is a transversal line). 3. AD = BC (opposite sides of parallelogram ABCD). Therefore, by AAS congruence criterion, Triangle ADF is congruent to Triangle BCE.
4. Equate Areas and Add Common Region — Since congruent figures have equal areas: Area(ADF) = Area(BCE). Now, add the area of quadrilateral BCED to both sides: Area(ADF) + Area(BCED) = Area(BCE) + Area(BCED). This gives: Area(ABCD) = Area(EFCD). Hence, proved!

CBSE Class 9 Solved Examples

• Example 1: Prove that a median of a triangle divides it into two triangles of equal area.
  Solution:
  Let ABC be a triangle and AD be its median. So, D is the mid-point of BC (BD = DC).
  Draw AE perpendicular to BC.
  Area(ABD) = 0.5 base height = 0.5 BD AE.
  Area(ADC) = 0.5 base height = 0.5 DC AE.
  Since BD = DC, both areas are equal. Hence, Area(ABD) = Area(ADC).

• Example 2: In a parallelogram ABCD, CD = 12 cm. The altitude corresponding to base CD is AM = 8 cm. Find the area of ABCD. If the altitude corresponding to base AD is CN = 6 cm, find the length of AD.
  Solution:
  Area of parallelogram ABCD = Base Height = CD AM = 12 * 8 = 96 cm².
  We can also calculate the same area using base AD and altitude CN:
  Area = AD * CN
  96 = AD * 6
  AD = 96 / 6 = 16 cm.

Common Mistakes to Avoid in Exams

1. Strictly Common Base: Make sure the figures share the exact same base, not just equal base lengths. Theorem 9.1 is only applicable directly when the base is literally shared (identical side).
2. The Perpendicular Mistake: When using $Area = Base \times Height$, ensure you use the perpendicular altitude, not the slant side. Draw the altitude carefully in your board papers!
3. Triangle-Parallelogram Ratio: Always remember that if a triangle and a parallelogram lie on the same base and between same parallels, $ar(Triangle) = \frac{1}{2} ar(Parallelogram)$. Students often write $ar(Triangle) = ar(Parallelogram)$ by mistake under exam pressure.

Practice Questions with Solutions

• 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 point O.
  Step 2: We know that the diagonals of a parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
  Step 3: In Triangle ABC, BO is a median (since O is the mid-point of AC). Therefore, Area(AOB) = Area(BOC) ... (Equation 1).
  Step 4: In Triangle BCD, CO is a median. Therefore, Area(BOC) = Area(COD) ... (Equation 2).
  Step 5: Similarly, in Triangle ACD, DO is a median. Therefore, Area(COD) = Area(AOD) ... (Equation 3).
  Step 6: Combining Equations 1, 2, and 3, we get Area(AOB) = Area(BOC) = Area(COD) = Area(AOD).
  Final answer: Thus, the diagonals of a parallelogram divide it into four triangles of equal area.

• Q: In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 * ar(ABC).
  A: Step 1: AD is the median of Triangle ABC. Therefore, it divides the triangle into two equal areas: ar(ABD) = 1/2 * ar(ABC) ... (Equation 1).
  Step 2: In Triangle ABD, BE is a median because E is given as the mid-point of AD.
  Step 3: Since BE is a median, it bisects the area of Triangle ABD: ar(BED) = 1/2 * ar(ABD) ... (Equation 2).
  Step 4: Substitute the value of ar(ABD) from Equation 1 into Equation 2: ar(BED) = 1/2 (1/2 ar(ABC)).
  Final answer: Hence, ar(BED) = 1/4 * ar(ABC). Proved.

• Q: Triangles ABC and DBC are on the same base BC and between the same parallels AD and BC. Prove that ar(ABC) = ar(DBC).
  A: Step 1: Draw a line segment AL perpendicular to BC and DM perpendicular to BC.
  Step 2: Since both triangles lie between the same parallel lines AD and BC, the perpendicular distance between these parallel lines is constant everywhere. Therefore, AL = DM.
  Step 3: Calculate the area of Triangle ABC: ar(ABC) = 1/2 Base Height = 1/2 BC AL.
  Step 4: Calculate the area of Triangle DBC: ar(DBC) = 1/2 Base Height = 1/2 BC DM.
  Step 5: Since AL = DM, we substitute AL in place of DM: ar(DBC) = 1/2 BC AL.
  Final answer: Therefore, ar(ABC) = ar(DBC). Proved.

• Q: In a parallelogram ABCD, P is any point on side CD. If the area of Triangle APB is 24 cm², what is the area of parallelogram ABCD?
  A: Step 1: Observe that Triangle APB and parallelogram ABCD share the same base AB.
  Step 2: Since P lies on CD and ABCD is a parallelogram, AB || CD. This means Triangle APB and parallelogram ABCD lie between the same parallel lines AB and CD.
  Step 3: Use the theorem: ar(APB) = 1/2 * ar(ABCD).
  Step 4: Plug in the given value: 24 = 1/2 * ar(ABCD).
  Step 5: Solve for the area of ABCD: ar(ABCD) = 24 * 2 = 48 cm².
  Final answer: The area of parallelogram ABCD is 48 cm².

Frequently Asked Questions