Exercise 11.1 — Area of Rectangle
Area of rectangle and area of planar regions.
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What Is This Exercise About?
Exercise 11.1 of Class 9 Mathematics (Chapter 11 — Areas) applies the formulas you learned in the introduction to real geometric figures. The big idea running through all four questions is "split and add" — break a complex figure into simple triangles or rectangles, find each part's area using ½ × base × height or length × breadth, and combine the results. You'll also use the Pythagoras theorem to find a missing side and a congruence proof to show two triangles have equal area.
4 Questions
Median Splits a Triangle Equally
Pythagoras Theorem
Split Into Known Shapes
SSS Congruence Proof
💡 Core formula used throughout: Area of a triangle = ½ × base × height. Every solution in this exercise eventually reduces to this one formula, applied once or twice.
Area of triangle = ½ × base × heightQuestion 1 — Area of △ADB When D Is the Midpoint of AC
Question 1
In △ABC, ∠ABC = 90°, AD = DC, AB = 12 cm and BC = 6.5 cm. Find the area of △ADB.
Given: right angle at B, D is the midpoint of AC (since AD = DC), AB = 12 cm, BC = 6.5 cm
Right triangle ABC with the right angle at B. D is the midpoint of hypotenuse-side AC, so BD is a median, splitting △ABC into two equal-area triangles: △ADB and △BDC.
Step 1 — Find the Area of the Whole Triangle ABC
In △ABC: base = BC = 6.5 cm, height = AB = 12 cm
Area of △ABC = ½ × base × height
= ½ × 6.5 × 12
= ½ × 78
= 39 sq. cm
Step 2 — Use the Median Property
📌 Key idea: Since AD = DC, point D is the midpoint of AC, which means BD is a median of △ABC. A median always divides a triangle into two triangles of equal area — because they share the same height and have equal bases (AD = DC).
Area of △ADB = Area of △BDC = ½ × Area of △ABC
= ½ × 39
∴ Area of △ADB = 19.5 sq. cm
Area of △ABC = 39 sq. cm
Area of △ADB = 19.5 sq. cm
Question 2 — Area of Quadrilateral PQRS Using Pythagoras Theorem
Question 2
Find the area of quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm, PS = 9 cm, QR = 8 cm and SR = 17 cm.
Hint given in the textbook: PQRS has two parts — split it along diagonal QS into △PQS and △QRS.
Quadrilateral PQRS with right angles at P and Q. Drawing diagonal QS splits it into right triangle △PQS (legs PQ, PS) and triangle △QRS (base QR, height QS).
Step 1 — Area of Right Triangle PQS
In △PQS, ∠QPS = 90°, so PQ and PS are the two legs
Area of △PQS = ½ × PQ × PS
= ½ × 12 × 9
= ½ × 108
= 54 sq. cm
Step 2 — Find the Diagonal QS Using Pythagoras Theorem
📌 Why Pythagoras? To find the area of △QRS, we need its base QR and height QS — but QS is not given directly. Since ∠QPS = 90°, △PQS is right-angled, so QS is its hypotenuse, found by Pythagoras theorem.
In right triangle PQS:
QS² = PQ² + PS²
= 12² + 9²
= 144 + 81
= 225
QS = √225 = 15 cm
Step 3 — Area of Triangle QRS
In △QRS, ∠SQR = 90°, so QR and QS act as base and height
Area of △QRS = ½ × QR × QS
= ½ × 8 × 15
= ½ × 120
= 60 sq. cm
Step 4 — Add Both Areas
Area of quadrilateral PQRS = Area of △PQS + Area of △QRS
= 54 + 60
∴ Area of PQRS = 114 sq. cm
QS = 15 cm
Area of △PQS = 54 sq. cm
Area of △QRS = 60 sq. cm
Total Area of PQRS = 114 sq. cm
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Question 3 — Area of Trapezium ABCD Using a Rectangle + Triangle
Question 3
Find the area of trapezium ABCD where ADCE is a rectangle. EC = 8 cm, AE = BE = 3 cm.
Hint given in the textbook: ABCD has two parts — a rectangle ADCE and a triangle BEC.
Trapezium ABCD with rectangle ADCE inside it (AE ∥ DC, both perpendicular to AD), plus right triangle BEC attached along the shared side EC.
Step 1 — Area of Rectangle ADCE
Area of rectangle ADCE = length × breadth = AD × AE
= 8 × 3 (AD = EC = 8 cm, AE = 3 cm)
= 24 sq. cm
Step 2 — Area of Triangle BEC
Area of △BEC = ½ × base × height = ½ × EC × BE
= ½ × 8 × 3
= ½ × 24
= 12 sq. cm
Step 3 — Add Both Areas
Area of trapezium ABCD = Area of rectangle ADCE + Area of △BEC
= 24 + 12
∴ Area of trapezium ABCD = 36 sq. cm
Rectangle ADCE = 24 sq. cm
Triangle BEC = 12 sq. cm
Trapezium ABCD = 36 sq. cm
💡 Why this method works: This is the "split and add" technique from the introduction lesson — any irregular or composite shape's area equals the sum of the areas of its non-overlapping parts.
Question 4 — Prove ar(△AOD) = ar(△BOC) in a Parallelogram
Question 4
ABCD is a parallelogram. The diagonals AC and BD intersect at O. Prove that ar(△AOD) = ar(△BOC).
Hint given in the textbook: Congruent figures have equal area.
Parallelogram ABCD with diagonals AC and BD meeting at O. The two shaded triangles, △AOD and △BOC, are congruent (and so have equal area).
Step-by-Step Proof
Given: ABCD is a parallelogram; diagonals AC, BD meet at O
Stated in the problem
To Prove: ar(△AOD) = ar(△BOC)
Required result
In △AOD and △COB:
Comparing the two triangles formed at O
AO = CO
Diagonals of a parallelogram bisect each other
BO = DO
Diagonals of a parallelogram bisect each other
AD = BC
Opposite sides of a parallelogram are equal
∴ △AOD ≅ △COB
SSS Congruence Rule (all three corresponding sides equal)
∴ ar(△AOD) = ar(△BOC)
Congruent figures always have equal area [Hence Proved]
✅ Key concept used: Two figures that are congruent (identical in shape and size) automatically have equal area — even before computing any actual numbers. This is a powerful shortcut: proving congruence is often easier than calculating areas directly.
Common Mistakes to Avoid
- In Q1: Forgetting that BD is a median only because AD = DC. If this condition weren't given, you couldn't assume the two triangles have equal area.
- In Q2: Using the wrong sides as base and height for △QRS. Since the right angle is at Q (∠SQR = 90°), the legs are QR and QS — not QR and SR.
- In Q2: Forgetting to take the square root after computing QS² = 225. Many students stop at 225 instead of finding QS = 15.
- In Q3: Mixing up which sides belong to the rectangle and which belong to the triangle. Always identify the shared side (EC here) since it appears in both the rectangle and the triangle's formulas.
- In Q4: Writing AAA or SAS instead of SSS. Here, all three pairs of sides (AO=CO, BO=DO, AD=BC) are used — making SSS the correct congruence rule, not SAS or ASA.
⛔ Most frequent error: In split-figure problems (Q2 and Q3), students sometimes add the wrong two areas or forget one part entirely. Always re-read which two shapes make up the original figure before adding.
Quick Answer Reference — Exercise 11.1
| Question | Final Answer | Key Concept Used |
|---|---|---|
| Q1 | Area of △ADB = 19.5 sq. cm | Median splits a triangle into two equal-area triangles |
| Q2 | Area of PQRS = 114 sq. cm | Split into 2 right triangles + Pythagoras theorem for QS |
| Q3 | Area of ABCD = 36 sq. cm | Split into rectangle + triangle, then add |
| Q4 | ar(△AOD) = ar(△BOC) — proved | SSS congruence ⇒ congruent figures have equal area |
What This Exercise Prepares You For
The "split and add" technique used in Questions 2 and 3 is the foundation for solving area problems involving any irregular or composite figure — a skill that continues into Exercise 11.2 (Area of Parallelogram & Rhombus), where similar diagonal-splitting techniques are used.
The congruence-based proof in Question 4 connects directly to Chapter 8 (Quadrilaterals) and to Chapter 7 (Triangles), where SSS, SAS, ASA, and RHS congruence rules are first introduced. The Pythagoras theorem application in Question 2 reappears throughout Class 10, especially in Similar Triangles and Coordinate Geometry (distance formula).
For CBSE, Telangana, and Andhra Pradesh board exams, expect 2 to 4-mark questions in this exact style — composite area problems and "prove equal area" questions are exam favourites because they test multiple concepts (Pythagoras, congruence, area formulas) in a single question.
✅ Chapter 11 Roadmap: Introduction → Exercise 11.1 (this page) → Exercise 11.2 — Area of Parallelogram & Rhombus