Exercise 11.2 — Parallelograms
Parallelograms on the same base and between the same parallels.
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The Two Key Theorems Behind This Exercise
Exercise 11.2 is built almost entirely on one powerful idea: figures that sit on the same base and lie between the same pair of parallel lines have a fixed relationship in area, regardless of their exact shape. Before attempting any problem, you must understand these two theorems thoroughly.
10 Questions
Same Base & Same Parallels
Parallelogram Height
Trapezium & Rhombus Proofs
Theorem 1
Parallelograms on the same base and between the same parallels are equal in area.
Parallelograms ABCD and PQCD share the same base CD and lie between the same two parallel lines. Both have exactly the same area, even though they look different.
Theorem 2
The area of a triangle is equal to half the area of a parallelogram on the same base and between the same parallels.
Triangle PCD and parallelogram ABCD share base CD and lie between the same parallels. Area of △PCD = ½ × Area of ABCD.
💡 How to use these theorems: Whenever a question mentions two shapes on the "same base" with vertices lying on lines that you can show are parallel, you can directly equate or halve their areas — without needing to know the actual base length or height!
Question 1 — Find the Height of Parallelogram ABEF
Question 1
The area of parallelogram ABCD is 36 cm². Calculate the height of parallelogram ABEF if AB = 4.2 cm.
Parallelograms ABCD and ABEF share the same base AB and lie between the same two parallels — so they have equal heights and equal areas.
📌 Key idea: ABCD and ABEF share the same base AB and lie between the same parallel lines. By Theorem 1, their areas are equal — and since they share the same base, their heights must also be equal.
Area of parallelogram ABCD = base × height
36 = AB × DM
36 = 4.2 × DM
DM = 36 ÷ 4.2 = 8.57 cm (approx.)
Since ABCD and ABEF are on the same base AB and between the same parallels:
Height of ABEF = Height of ABCD = 8.57 cm
Height of ABEF = 8.57 cm
Question 2 — Find AD Using Two Different Bases
Question 2
ABCD is a parallelogram. AE ⊥ DC and CF ⊥ AD. If AB = 10 cm, AE = 8 cm and CF = 12 cm, find AD.
Parallelogram ABCD: AE ⊥ DC (AE = 8 cm) gives one height; CF ⊥ AD (CF = 12 cm) gives another. Same parallelogram, two different base–height pairs.
📌 Key idea: The area of a parallelogram stays the same no matter which side you choose as the base — as long as you use the correct perpendicular height for that base.
Using AB as base:
Area = AB × AE = 10 × 8 = 80 sq. cm ...(1)
Using AD as base:
Area = AD × CF = AD × 12
From (1): 80 = AD × 12
AD = 80 ÷ 12 = 6.67 cm (approx.)
AD = 6.67 cm
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Question 3 — Prove ar(EFGH) = ½ ar(ABCD)
Question 3
E, F, G, H are the midpoints of sides AB, BC, CD, AD of parallelogram ABCD. Show that ar(EFGH) = ½ ar(ABCD).
E, F, G, H are midpoints of AB, BC, CD, AD respectively. Joining them forms quadrilateral EFGH inside parallelogram ABCD. Diagonal HF splits the proof into two parts.
Given: E, F, G, H = midpoints of AB, BC, CD, AD; join HF
Construction as per the problem
AB = CD and AD = BC
Opposite sides of parallelogram ABCD are equal
AH = ½AD and BF = ½BC
H, F are midpoints
AH = BF (since ½AD = ½BC), and AH ∥ BF
AD = BC and AD ∥ BC (opposite sides of parallelogram)
⇒ ABFH is a parallelogram
One pair of opposite sides equal & parallel
△EHF and parallelogram ABFH are on the same base HF, between the same parallels AB, HF
E lies on AB, the side parallel to HF
∴ ar(△EHF) = ½ ar(ABFH) ...(1)
Theorem 2 (triangle = half of parallelogram, same base & parallels)
Similarly, CDHF is also a parallelogram, and △GHF is on the same base HF between the same parallels DC, HF
Same reasoning applied to the other half
∴ ar(△GHF) = ½ ar(CDHF) ...(2)
Theorem 2 applied again
Adding (1) and (2): ar(△EHF) + ar(△GHF) = ½ [ar(ABFH) + ar(CDHF)]
Adding both equal-area relations
ar(EFGH) = ½ [ar(ABFH) + ar(CDHF)] = ½ ar(ABCD)
EFGH is made of △EHF + △GHF; ABFH + CDHF makes up all of ABCD
∴ ar(EFGH) = ½ ar(ABCD) [Hence Proved]
Required result established ✔
Question 4 — Shape Formed by Joining the Four Corner Triangles
Question 4
In the figure of Question 3, what type of quadrilateral do you get if you remove triangles APM, DPO, OCN and MNB?
✅ Answer: Removing the four corner triangles (△APM, △DPO, △OCN, △MNB) from parallelogram ABCD leaves behind the inner quadrilateral MNOP, formed by joining the midpoints' connecting segments. This inner quadrilateral MNOP is itself a parallelogram — in fact, it can be shown to be similar in nature to EFGH from Question 3, since it is formed using the same midpoint-based construction logic, sitting symmetrically inside ABCD.
📌 Why this matters: This question reinforces the same midpoint principle from Question 3 — joining midpoints (or related constructed points) of a parallelogram's sides consistently produces a smaller parallelogram inside the original one.
Question 5 — Prove ar(△APB) = ar(△BQC)
Question 5
P and Q are points on sides DC and AD of parallelogram ABCD. Show that ar(△APB) = ar(△BQC).
P is on DC, Q is on AD. △APB sits on base AB; △BQC sits on base BC — each between the relevant pair of parallel sides of the parallelogram.
Taking AB as base: △APB and parallelogram ABCD share base AB, between parallels AB and DC
P lies on DC, parallel to AB
∴ ar(△APB) = ½ ar(ABCD) ...(1)
Theorem 2
Taking BC as base: △BQC and parallelogram ABCD share base BC, between parallels AD and BC
Q lies on AD, parallel to BC
∴ ar(△BQC) = ½ ar(ABCD) ...(2)
Theorem 2
From (1) and (2): ar(△APB) = ar(△BQC) [Hence Proved]
Both equal the same quantity, ½ ar(ABCD)
Question 6 — P Is Inside the Parallelogram: Two Area Relations
Question 6
P is any interior point of parallelogram ABCD. Show that (i) ar(△APB) + ar(△PCD) = ½ ar(ABCD), and (ii) ar(△APD) + ar(△PBC) = ar(△APB) + ar(△PCD).
Construction: draw line MN through P, parallel to AB (and DC). This splits ABCD into two parallelograms, ABNM and DCNM, each containing one of the triangles.
Part (i): ar(△APB) + ar(△PCD) = ½ ar(ABCD)
Construction: Draw MN through P, parallel to AB
As per the hint given
ABNM and DCNM are both parallelograms
MN ∥ AB ∥ DC by construction
△APB and parallelogram ABNM share base AB, between parallels AB, MN
P lies on MN
∴ ar(△APB) = ½ ar(ABNM) ...(1)
Theorem 2
△PCD and parallelogram DCNM share base DC, between parallels DC, MN
P lies on MN
∴ ar(△PCD) = ½ ar(DCNM) ...(2)
Theorem 2
Adding (1) and (2): ar(△APB) + ar(△PCD) = ½ [ar(ABNM) + ar(DCNM)] = ½ ar(ABCD)
ABNM and DCNM together make up the whole of ABCD
Part (ii): ar(△APD) + ar(△PBC) = ar(△APB) + ar(△PCD)
Construction: Draw XY through P, parallel to AD this time
A second, perpendicular construction
AXYD and XBCY are both parallelograms
XY ∥ AD ∥ BC by construction
ar(△APD) = ½ ar(AXYD) ...(3), ar(△PBC) = ½ ar(XBCY) ...(4)
Theorem 2, applied to each half
Adding (3)+(4): ar(△APD) + ar(△PBC) = ½ [ar(AXYD) + ar(XBCY)] = ½ ar(ABCD)
AXYD and XBCY together make up the whole of ABCD
From Part (i): ½ ar(ABCD) = ar(△APB) + ar(△PCD)
∴ ar(△APD) + ar(△PBC) = ar(△APB) + ar(△PCD) [Hence Proved]
∴ ar(△APD) + ar(△PBC) = ar(△APB) + ar(△PCD) [Hence Proved]
Both expressions equal ½ ar(ABCD), so they equal each other
Question 7 — Prove the Trapezium Area Formula
Question 7
Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them.
Trapezium ABCD with AB ∥ DC. Drawing diagonal BD splits it into △ABD (base AB = a) and △BCD (base DC = b), both having the same height h.
Given: Trapezium ABCD, AB = a, DC = b, distance between AB & DC = h. Draw diagonal BD.
Construction
Height of △ADB = Height of △BCD = h
Both triangles lie between the same parallel lines AB and DC
Area of trapezium ABCD = Area of △ADB + Area of △BCD
Diagonal BD splits the trapezium into exactly these two triangles
= ½ × AB × h + ½ × DC × h
Triangle area formula, applied to each part
= ½ × a × h + ½ × b × h
Substituting AB = a, DC = b
= ½ × (a + b) × h [Hence Proved]
Factoring out ½ × h
Area of trapezium = ½ × (sum of parallel sides) × (distance between them)Question 8 — Two Parallelograms on the Same Base SR
Question 8
PQRS and ABRS are parallelograms; X is any point on side BR. Show that (i) ar(PQRS) = ar(ABRS), and (ii) ar(△AXS) = ½ ar(PQRS).
(i) PQRS and ABRS share base SR, lying between the same parallels PB and SR
Given construction
∴ ar(PQRS) = ar(ABRS)
Theorem 1 (parallelograms, same base, same parallels)
(ii) Treating AB as base: △AXS and parallelogram ABRS share base AB, between same parallels AS and BR
X lies on BR
ar(△AXS) = ½ ar(ABRS)
Theorem 2
= ½ ar(PQRS) (using result from part i) [Hence Proved]
Substituting ar(ABRS) = ar(PQRS)
Question 9 — The Farmer's Field Problem
Question 9
A farmer's field is parallelogram PQRS. A is the midpoint of RS, joined to P and Q. Into how many parts is the field divided, and what shapes are they? If groundnuts should equal pulses + paddy combined, how should the farmer divide the sowing?
A is the midpoint of RS. Joining A to P and Q divides the parallelogram field into three triangles: △PSA, △PAQ, and △QAR.
△PAQ and parallelogram PQRS share base PQ, between same parallels PQ, SR
A lies on SR
ar(△PAQ) = ½ ar(PQRS) ⇒ ar(PQRS) = 2 × ar(△PAQ)
Theorem 2
ar(△PAQ) + ar(△PSA) + ar(△QAR) = ar(PQRS) = 2 × ar(△PAQ)
The three triangles together make up the whole field
⇒ ar(△PSA) + ar(△QAR) = 2 ar(△PAQ) − ar(△PAQ) = ar(△PAQ)
Subtracting ar(△PAQ) from both sides
∴ ar(△PSA) + ar(△QAR) = ar(△PAQ)
Pulses + Paddy = Groundnuts, in area terms
✅ Farmer's decision: The field splits into 3 equal-looking triangular parts: △PSA, △PAQ, and △QAR — but only △PAQ has area equal to the sum of the other two. So the farmer should sow groundnuts in △PAQ, and split pulses and paddy between △PSA and △QAR. This way, groundnut area = pulses area + paddy area, exactly as required.
Question 10 — Prove the Rhombus Area Formula
Question 10
Prove that the area of a rhombus is equal to half the product of its diagonals.
Rhombus ABCD with diagonals AC (= d₁) and BD (= d₂) intersecting at O at right angles, each bisected: OA = OC = ½d₁, OB = OD = ½d₂.
Given: Rhombus ABCD, diagonals AC = d₁, BD = d₂, meeting at O
Stated in the problem
OA = OC = ½d₁, OB = OD = ½d₂, diagonals meet at 90°
Diagonals of a rhombus bisect each other at right angles
Area of rhombus ABCD = ar(△OAB) + ar(△OBC) + ar(△OCD) + ar(△ODA)
The two diagonals split the rhombus into 4 right triangles
Each triangle area = ½ × (½d₁) × (½d₂) = ⅛ d₁d₂
Each right triangle has legs ½d₁ and ½d₂ (since ∠AOB = 90°, etc.)
Total = ⅛d₁d₂ + ⅛d₁d₂ + ⅛d₁d₂ + ⅛d₁d₂ = ⁴⁄₈ d₁d₂
Adding all four equal triangle areas
= ½ d₁d₂ [Hence Proved]
Simplifying ⁴⁄₈ to ½
Area of rhombus = ½ × d₁ × d₂Quick Answer Reference — Exercise 11.2
| Question | Key Result |
|---|---|
| Q1 | Height of ABEF = 8.57 cm |
| Q2 | AD = 6.67 cm |
| Q3 | ar(EFGH) = ½ ar(ABCD) — proved |
| Q4 | MNOP is a parallelogram |
| Q5 | ar(△APB) = ar(△BQC) — proved |
| Q6 | Both area relations proved using a line through P |
| Q7 | Area of trapezium = ½(a+b)h — proved |
| Q8 | ar(PQRS) = ar(ABRS); ar(△AXS) = ½ ar(PQRS) |
| Q9 | Groundnuts in △PAQ; pulses & paddy in △PSA, △QAR |
| Q10 | Area of rhombus = ½ d₁d₂ — proved |
Common Mistakes to Avoid
- Forgetting the "same parallels" condition: Two shapes sharing a base only have equal/half area if they ALSO lie between the same pair of parallel lines. Always check both conditions before applying the theorems.
- In Q1 and Q2: Mixing up which side is being used as the base. The height must always be perpendicular to the chosen base, not to a different side.
- In Q6 and Q7: Forgetting to justify why a quadrilateral formed by construction (like ABNM) is actually a parallelogram — always state the reason (e.g., "MN ∥ AB and AM ∥ BN by construction").
- In Q10: Assuming all four small triangles inside the rhombus have different areas. Since OA = OC and OB = OD with diagonals perpendicular, all four triangles are congruent right triangles with equal area.
⛔ Most frequent board exam error: Writing "ar(△X) = ar(parallelogram Y)" instead of "ar(△X) = ½ ar(parallelogram Y)" when applying Theorem 2 — remember triangles get half, not the full area, of a parallelogram on the same base and parallels.
What This Exercise Prepares You For
The same-base-same-parallels theorems used throughout this exercise are foundational results that reappear in Class 10 geometry, particularly in Similar Triangles, where area ratios are compared using base and height relationships. The construction techniques (drawing a line through an interior point parallel to a side) used in Questions 6 and 9 are classic geometry problem-solving tools used throughout board exam proof questions.
This exercise also reinforces concepts from Chapter 8 (Quadrilaterals) — particularly parallelogram and rhombus properties — and builds on Exercise 11.1's "split and add" technique.
For CBSE, Telangana, and Andhra Pradesh board exams, the trapezium and rhombus area proofs (Q7 and Q10) are extremely common 4-mark questions, and the same-base-same-parallels theorem applications (Q3, Q5, Q6, Q8, Q9) are popular 2 to 4-mark proof-based questions.
✅ Chapter 11 Roadmap: Introduction → Exercise 11.1 → Exercise 11.2 (this page)