Class 9 Notes

Class 9 Mathematics Notes Chapter 9 (Chapter 9) – Examplar Problem (Englisha) Book

Philoid

Philoid

— 9 min read

Examplar Problem (Englisha)
Alright class, let's focus on Chapter 9, "Areas of Parallelograms and Triangles," from your NCERT Exemplar. This chapter is fundamental, not just for your class exams but also forms the basis for many geometry problems you might encounter in government exams. The core idea is understanding how areas of these specific shapes relate to each other when they share a common base or lie between the same parallel lines.

Chapter 9: Areas of Parallelograms and Triangles - Detailed Notes

1. Introduction: Figures on the Same Base and Between the Same Parallels

  • Definition: Two geometric figures are said to be on the same base if they have a common side (base), and the vertices (or the vertex) opposite to the common base lie on a line parallel to the base.
  • Identification: Look for a shared line segment forming a side for both figures. Then check if the opposite vertices lie on a single line parallel to this shared base.
    • Example: A parallelogram ABCD and a triangle EBC are on the same base BC if A, D, and E lie on a line parallel to BC.
  • Significance: This condition is crucial for comparing the areas of figures without necessarily calculating them individually.

2. Key Theorems and Results

(a) Parallelograms on the same base and between the same parallels:

  • Theorem 9.1: Parallelograms on the same base and between the same parallels are equal in area.
    • Explanation: If parallelogram ABCD and parallelogram EBCF have the same base BC and lie between the parallel lines BC and AF, then ar(ABCD) = ar(EBCF).
    • Diagram: Imagine two parallelograms leaning against the same base BC, with their top sides AD and EF lying on the same line parallel to BC.
    • Application: Useful for proving equality of areas or finding the area of one parallelogram if the other's is known.

(b) Area of a Parallelogram:

  • Formula: Area of a parallelogram = Base × Corresponding Height (Altitude)
    • Explanation: The 'base' can be any side, and the 'height' is the perpendicular distance from the opposite side to that base. It's crucial to use the height corresponding to the chosen base.

(c) Triangle and Parallelogram on the same base and between the same parallels:

  • Theorem 9.2: 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.
    • Explanation: If triangle EBC and parallelogram ABCD are on the same base BC and between the parallels BC and AE, then ar(EBC) = ½ ar(ABCD).
    • Diagram: Visualize a triangle sitting inside or next to a parallelogram, sharing the base BC, with the triangle's apex E on the line AD (which is parallel to BC).
    • Application: Directly relates the area of a triangle to a related parallelogram.

(d) Triangles on the same base and between the same parallels:

  • Theorem 9.3: Triangles on the same base (or equal bases) and between the same parallels are equal in area.
    • Explanation: If triangles ABC and DBC have the same base BC and lie between the parallel lines BC and AD, then ar(ABC) = ar(DBC).
    • Extension: This also holds if the bases are equal in length (not necessarily the same segment) and the triangles lie between the same parallels.
    • Application: Proving equality of triangular areas.

(e) Area of a Triangle:

  • Formula: Area of a triangle = ½ × Base × Corresponding Height (Altitude)
    • Explanation: Similar to the parallelogram, the height must be the perpendicular distance from the opposite vertex to the chosen base.

(f) Converse Theorems (Important for reasoning):

  • Converse of Theorem 9.3: Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
    • Application: Used to prove that lines are parallel if certain area conditions are met.
  • Similar Converse for Parallelograms: Two parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.

3. Important Properties for Problem Solving (Frequently Tested)

  • Median of a Triangle: A median of a triangle divides it into two triangles of equal areas.
    • Explanation: If AD is a median of triangle ABC (D is the midpoint of BC), then ar(ABD) = ar(ACD). This is because they have equal bases (BD = DC) and the same height (perpendicular from A to BC).
    • Application: Very common in proof-based problems and problems involving area ratios.
  • Diagonals of a Parallelogram: The diagonals of a parallelogram divide it into four triangles of equal area.
    • Explanation: If diagonals AC and BD of parallelogram ABCD intersect at O, then ar(AOB) = ar(BOC) = ar(COD) = ar(DOA). This follows from the median property applied to triangles like ABC (BO is a median) etc., and the fact that diagonals bisect each other.
    • Application: Useful when dealing with the intersection point of diagonals.
  • Diagonals of a Trapezium: If the diagonals AC and BD of a trapezium ABCD (with AB || DC) intersect at O, then ar(AOD) = ar(BOC). This is because ar(ADC) = ar(BDC) (same base DC, between same parallels AB and DC), and subtracting ar(DOC) from both gives the result.

4. Focus for Government Exams:

  • Direct application of the theorems (Theorems 9.1, 9.2, 9.3).
  • Problems involving medians and their area-bisecting property.
  • Problems involving diagonals of parallelograms and their area properties.
  • Combining these theorems within composite figures (e.g., a triangle inside a parallelogram, properties within a trapezium).
  • Ratio of areas based on shared heights or bases.
  • Using the converse theorems to prove lines are parallel.

Multiple Choice Questions (MCQs)

Here are 10 MCQs to test your understanding. Choose the best option.

  1. Two parallelograms PQRS and PQMN stand on the same base PQ and between the same parallels PQ and RN. If ar(PQRS) = 48 cm², then ar(PQMN) is:
    (a) 24 cm²
    (b) 48 cm²
    (c) 96 cm²
    (d) 12 cm²

  2. In ΔABC, D is the mid-point of BC. Which of the following is true?
    (a) ar(ABD) = ½ ar(ABC)
    (b) ar(ADC) = ½ ar(ABC)
    (c) ar(ABD) = ar(ADC)
    (d) All of the above

  3. A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the area of the triangle to the area of the parallelogram is:
    (a) 1 : 1
    (b) 1 : 2
    (c) 2 : 1
    (d) 1 : 4

  4. ABCD is a parallelogram with area 80 cm². If P is any point on CD, then ar(APB) is:
    (a) 80 cm²
    (b) 60 cm²
    (c) 40 cm²
    (d) 20 cm²

  5. The area of a triangle is 30 cm². If its base is 10 cm, its corresponding height is:
    (a) 3 cm
    (b) 6 cm
    (c) 15 cm
    (d) 300 cm

  6. In parallelogram ABCD, diagonals AC and BD intersect at O. If ar(AOB) = 15 cm², then ar(ABCD) is:
    (a) 15 cm²
    (b) 30 cm²
    (c) 45 cm²
    (d) 60 cm²

  7. Triangles ABC and DEF have bases BC = EF = 5 cm and lie between the same parallels l and m. Then:
    (a) ar(ABC) > ar(DEF)
    (b) ar(ABC) < ar(DEF)
    (c) ar(ABC) = ar(DEF)
    (d) Cannot be determined

  8. ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect at O. Which pair of triangles necessarily have equal areas?
    (a) ΔAOB and ΔCOD
    (b) ΔABC and ΔABD
    (c) ΔAOD and ΔBOC
    (d) ΔADC and ΔABC

  9. If the area of parallelogram ABCD is 50 cm², and E is the mid-point of side BC, then ar(ADE) is:
    (a) 25 cm²
    (b) 12.5 cm²
    (c) 37.5 cm²
    (d) Cannot be determined uniquely

  10. The median of a triangle divides it into two:
    (a) Congruent triangles
    (b) Triangles of equal area
    (c) Right-angled triangles
    (d) Isosceles triangles

Answers to MCQs:

  1. (b) - Theorem 9.1

  2. (d) - Median property

  3. (b) - Theorem 9.2

  4. (c) - ΔAPB and ||gm ABCD are on same base AB and between same parallels AB and DC. ar(APB) = ½ ar(ABCD).

  5. (b) - Area = ½ × base × height => 30 = ½ × 10 × height => height = 6 cm.

  6. (d) - Diagonals divide a parallelogram into 4 triangles of equal area. ar(ABCD) = 4 × ar(AOB).

  7. (c) - Theorem 9.3 (Triangles on equal bases and between same parallels).

  8. (c) - Property of trapezium diagonals (derived from Theorem 9.3).

  9. (c) - Hint: Consider ar(ABE) + ar(DCE) = ½ ar(ABCD). Then ar(ADE) = ar(ABCD) - (ar(ABE) + ar(DCE)). Or, draw altitude from A to BC, altitude from D to BC extended. ar(ADE) = ar(ADC) + ar(ACE). Need more steps. Let's rethink. Draw DE. E is midpoint of BC. Consider ΔADC and ΔADE. Draw a line through E parallel to AD intersecting DC at F. This seems too complex for a direct MCQ. Let's use a simpler approach. ar(ABE) = 1/4 ar(ABCD) if E is midpoint? No. ar(ABE) = 1/2 ar(ABD)? No. Let's use ar(ADE) = ar(ABCD) - ar(ABE) - ar(DCE). ar(ABE) = 1/2 * BE * h (h is height of ||gm). ar(DCE) = 1/2 * CE * h. Since BE=CE=BC/2, ar(ABE) + ar(DCE) = 1/2 * (BE+CE) * h = 1/2 * BC * h = 1/2 * ar(ABCD). So, ar(ADE) = ar(ABCD) - 1/2 * ar(ABCD) = 1/2 * ar(ABCD) = 1/2 * 50 = 25 cm². Hmm, let me re-verify this logic.
    Let h be the height of the parallelogram w.r.t base AB (and DC). Let h' be the height w.r.t base BC (and AD).
    ar(ABCD) = DC * h = 50.
    ar(ADE) = ?
    Draw a line through E parallel to AB, meeting AD at F. ABEF is not necessarily a parallelogram.
    Let's use the property ar(ADЕ) = 1/2 * AD * (perpendicular from E to AD). This doesn't seem easy.
    Consider ΔADE. Let's try another approach.
    ar(ABD) = 1/2 ar(ABCD) = 25.
    ar(ACD) = 1/2 ar(ABCD) = 25.
    In ΔBCD, DE is a median (if we join BD). No, E is midpoint of BC.
    Consider ΔABC. AE is a median? No.
    Let's use the previous logic again: ar(ADE) = ar(ABCD) - ar(ABE) - ar(DCE).
    Let h be the perpendicular distance between AB and DC.
    Let h' be the perpendicular distance between AD and BC.
    ar(ABE) = 1/2 * BE * (perp from A to BC). This height is not h or h'.
    Let's draw altitude AM from A to DC and altitude DN from D to AB extended.
    Let's drop perpendiculars from A and D to BC. Let them be AX and DY. The height of the parallelogram w.r.t base BC is h'. Area = BC * h' = 50.
    Area(ABE) = 1/2 * BE * AX. Area(DCE) = 1/2 * CE * DY. Are AX and DY related to h'? Not directly.
    Let's use the base CD and height h. Area = CD * h = 50.
    Consider triangle ADE. Its area is not directly related in a simple way.
    What if we use coordinates? Let A=(0,h), B=(b,h), C=(b+a,0), D=(a,0). Base DC = b. Height = h. Area = b*h = 50.
    E is midpoint of BC. E = ( (b+b+a)/2, (h+0)/2 ) = ( (2b+a)/2, h/2 ).
    Area(ADE) = 1/2 | x_A(y_D-y_E) + x_D(y_E-y_A) + x_E(y_A-y_D) |
    = 1/2 | 0(0-h/2) + a(h/2 - h) + (2b+a)/2 * (h-0) |
    = 1/2 | a(-h/2) + (2b+a)/2 * h |
    = 1/2 | -ah/2 + bh + ah/2 |
    = 1/2 | bh | = 1/2 * Area(ABCD) = 1/2 * 50 = 25 cm².
    Okay, the coordinate geometry method gives 25 cm². Let me re-verify the logic: ar(ADE) = ar(ABCD) - ar(ABE) - ar(DCE).
    Let h_BC be the height of the parallelogram relative to base BC. Area = BC * h_BC = 50.
    ar(ABE) = 1/2 * BE * h_BC = 1/2 * (BC/2) * h_BC = 1/4 * (BC * h_BC) = 1/4 * ar(ABCD) = 12.5.
    ar(DCE) = 1/2 * CE * h_BC = 1/2 * (BC/2) * h_BC = 1/4 * ar(ABCD) = 12.5.
    ar(ADE) = ar(ABCD) - ar(ABE) - ar(DCE) = 50 - 12.5 - 12.5 = 50 - 25 = 25 cm².
    Where did I get 37.5 from earlier? Let's check the standard result.
    There is a result: If E is midpoint of BC in ||gm ABCD, then ar(ADE) = 3/8 ar(ABCD)? No, that's for something else.
    Let's check the logic ar(ABE) + ar(DCE) = 1/2 ar(ABCD).
    Draw altitudes from A and D to BC. Let them be AX and DY. In a parallelogram, these altitudes are equal. Let AX = DY = h'.
    ar(ABE) = 1/2 * BE * AX = 1/2 * (BC/2) * h' = (1/4) * BC * h'.
    ar(DCE) = 1/2 * CE * DY = 1/2 * (BC/2) * h' = (1/4) * BC * h'.
    Sum = (1/2) * BC * h'.
    Area(ABCD) = Base * Height = BC * h'.
    So, ar(ABE) + ar(DCE) = 1/2 * ar(ABCD).
    Therefore, ar(ADE) = ar(ABCD) - (ar(ABE) + ar(DCE)) = ar(ABCD) - 1/2 * ar(ABCD) = 1/2 * ar(ABCD).
    So, ar(ADE) = 1/2 * 50 = 25 cm².
    My initial answer (c) 37.5 cm² was incorrect. The calculation leads to (a) 25 cm². Let me double check common results.
    Ah, I found a standard result/problem type: If E is midpoint of BC, then ar(ADE) = ar(ABE) + ar(DCE)? No.
    Let's consider ΔADF where F is midpoint of AD. Then BEFC is a parallelogram? No.
    Consider ΔDEC and ΔAEB. Bases are equal (EC=EB). Heights are equal (perpendicular distance between AD and BC). So ar(DEC) = ar(AEB)? No, heights are from D to BC and A to BC, which are equal. Yes, ar(DEC) = ar(AEB).
    ar(ADE) = ?
    Join diagonal AC. O is midpoint. E is midpoint of BC. In ΔABC, AE is not median.
    Join diagonal BD. O is midpoint. In ΔBCD, DE is a median. So ar(DEC) = ar(DEB).
    ar(ABE) = ?
    ar(ABD) = 1/2 ar(ABCD) = 25.
    ar(BCD) = 1/2 ar(ABCD) = 25.
    Since DE is median of ΔBCD, ar(DEC) = 1/2 ar(BCD) = 1/2 * 25 = 12.5.
    Now, how to find ar(ABE)?
    Consider ΔABC. Area = 25. E is midpoint of BC. Is AE a median? No.
    Let's reconsider the sum: ar(ABE) + ar(DCE) = 1/2 ar(ABCD).
    ar(DCE) = 12.5.
    So, ar(ABE) must be 12.5. Let's prove it.
    In ΔABC, E is midpoint of BC. Drop altitude from A to BC, say AX=h'. ar(ABE) = 1/2 * BE * h' = 1/2 * (BC/2) * h' = 1/4 * BC * h'.
    Since ar(ABCD) = BC * h', ar(ABE) = 1/4 ar(ABCD) = 1/4 * 50 = 12.5.
    Yes, ar(ABE) = 12.5 and ar(DCE) = 12.5.
    ar(ADE) = ar(ABCD) - ar(ABE) - ar(DCE) = 50 - 12.5 - 12.5 = 25 cm².
    The correct answer should be 25 cm². Option (a).

  10. (b) - Definition/Property of a median.

Corrected Answers:

  1. (b)
  2. (d)
  3. (b)
  4. (c)
  5. (b)
  6. (d)
  7. (c)
  8. (c)
  9. (a)
  10. (b)

Make sure you understand the reasoning behind each answer, especially the application of the theorems and properties. This chapter relies heavily on visualization and applying the correct theorem to the given situation. Good luck with your preparation!

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