Area Theorem Class 9 ICSE — Maths Chapter 16
The area theorem class 9 ICSE chapter shows how to compare areas of triangles, parallelograms and rectangles without calculating each one. Figures on the same base and between the same parallels are the core idea, together with triangles that share one vertex. You will prove each result step by step before applying it to the exercises. The book’s treatment stays aligned with the syllabus published on the CISCE website.
Download Area Theorems Proof and Use Class 9 ICSE Mathematics Chapter 16 PDF
Download the Area Theorems Proof and Use Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 12 pages, 3 sections, 35 questions.
What is inside Chapter 16
12 pages · 3 sections · 5 worked examples · 35 exercise questions
| Section | Title |
|---|---|
| 16.1 | INTRODUCTION |
| 16.2 | FIGURES BETWEEN THE SAME PARALLELS |
| 16.3 | TRIANGLES WITH THE SAME VERTEX AND BASES ALONG THE SAME LINE |
- EXERCISE 16(A): 18 questions
- EXERCISE 16(B): 8 questions
- EXERCISE 16(C): 9 questions
16.1 INTRODUCTION
The area of a plane figure is the region bounded by it. This ICSE chapter starts from that definition and recalls three formulas you already know. A triangle has area \frac{1}{2} \times \text{base} \times \text{height}. A rectangle has area \text{length} \times \text{breadth}. A parallelogram has area \text{base} \times \text{height}.
The chapter then compares areas of different figures, not just measures them. Two expressions matter here. Equal figures means figures equal in area. Congruent figures are always equal in area, because congruent shapes match in size and shape. The converse is not always true.
Two triangles with the same base and height show why. A scalene triangle and an isosceles triangle, both with base 6 cm and height 4 cm, each cover 12 cm². They are equal in area, yet not congruent. Keep that difference clear, because every theorem below compares areas, not shapes.
16.2 FIGURES BETWEEN THE SAME PARALLELS
Figures lie between the same parallels when their bases sit on one straight line and their other vertices sit on a parallel line. A parallelogram, a rectangle and a triangle drawn this way all have equal altitudes. When their bases are also equal, the figures are on equal bases and between the same parallels.
Theorem 19 states that parallelograms on the same base and between the same parallels are equal in area. The proof shows \Delta ADF \cong \Delta BCE by ASA. Congruent triangles have equal areas, so adding the shared region to both sides leaves the two parallelograms equal in area.
Theorem 20 states that a triangle on the same base and between the same parallels has half the area of the parallelogram. Complete the parallelogram, and the diagonal divides it into two equal triangles. One of them is the triangle you started with.
Theorem 21 states that triangles on the same base and between the same parallels are equal in area. Complete each triangle into a parallelogram. The parallelograms are equal by Theorem 19, and each triangle is half of its parallelogram.
The corollaries extend the three theorems to equal bases. Parallelograms on equal bases between the same parallels are equal in area. A triangle on an equal base still has half the parallelogram’s area. Triangles on equal bases between the same parallels are equal in area.
Two more results close the section. If two triangles have equal areas and stand on the same base, their altitudes are equal. A median divides a triangle into two equal areas. This ICSE chapter proves that result from the area formula. More generally, if AD divides BC in the ratio m : n, then \frac{\text{Area}(\Delta ABD)}{\text{Area}(\Delta ADC)} = \frac{m}{n}.
Worked pattern: let parallelogram ABCD have area 120 cm². A triangle on the same base AB and between the same parallels has area \frac{1}{2} \times 120 = 60 cm², by Theorem 20. The same result holds if the triangle stands on an equal base instead of the same base.
| Theorem | Statement | Corollary |
|---|---|---|
| 19 | Parallelograms on the same base and between the same parallels are equal in area. | A parallelogram equals a rectangle on the same base and between the same parallels. |
| 20 | A triangle has half the area of a parallelogram on the same base and between the same parallels. | A triangle has half the area of a parallelogram on equal bases and between the same parallels. |
| 21 | Triangles on the same base and between the same parallels are equal in area. | Triangles on equal bases and between the same parallels are equal in area. |
- Equal in area does not mean congruent. Triangles on the same base and between the same parallels can be equal in area but shaped differently.
- Congruent figures are always equal in area, so use congruence only when you need both size and shape to match.
If you need to revise how to construct parallelograms before proving these results, the construction of polygons chapter explains the drawing steps. That groundwork makes the proofs in Theorem 19 easier to follow.
16.3 TRIANGLES WITH THE SAME VERTEX AND BASES ALONG THE SAME LINE
Suppose triangles ABD, BCD and ACD share vertex D, and their bases lie on the same straight line AC. Every triangle then has the same height, the distance from D to line AC. When triangles share a vertex and have bases along the same line, their areas are in the ratio of their bases.
Written out, the ratios are \frac{\text{Area}(\Delta ABD)}{\text{Area}(\Delta BCD)} = \frac{AB}{BC}, \frac{\text{Area}(\Delta ABD)}{\text{Area}(\Delta ACD)} = \frac{AB}{AC}, and \frac{\text{Area}(\Delta BCD)}{\text{Area}(\Delta ACD)} = \frac{BC}{AC}. The shared height cancels in each ratio. Only the bases remain, which is why the areas follow the bases exactly.
Worked pattern: in triangle ABC, let D lie on BC with BD : DC = 2 : 3. Triangles ABD and ADC share vertex A, and their bases lie along BC. Their areas are in the ratio 2 : 3, so \text{Area}(\Delta ABD) : \text{Area}(\Delta ADC) = 2 : 3. Since BC divides into 5 equal parts, \text{Area}(\Delta ABD) = \frac{2}{5} \times \text{Area}(\Delta ABC).
When the base is split equally, the ratio becomes 1 : 1. That is the median result from section 16.2: a median divides the triangle into two equal areas. The same vertex-and-base idea also works on quadrilateral diagonals, where triangles on the same line through the intersection point keep the ratio of their segments.
A quick way to remember the rule: same vertex, same line — areas follow the base sign. This principle drives most questions in Exercise 16(B), and the harder problems in Exercise 16(C) combine it with the theorems of section 16.2.
What to practise from Area Theorem Class 9 ICSE
The chapter gives you three exercises, and each one targets a different skill. Work through them in order, because the later exercises assume the earlier results. These exercises cover the full area theorem material for Class 9 ICSE.
- Exercise 16(A) — 18 questions: proves Theorems 19, 20 and 21 with the corollaries. Expect parallelograms, rectangles and triangles on the same base and between the same parallels.
- Exercise 16(B) — 8 questions: works with area ratios, medians and the ratio of areas for triangles of the same height or the same base.
- Exercise 16(C) — 9 questions: combines the section 16.2 theorems with base divisions, including the median results and the perimeter-altitude problem.
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| Exercise | Questions | Concepts it tests |
|---|---|---|
| 16(A) | 18 | Theorems 19–21 and corollaries: equal areas on the same base, triangle as half a parallelogram |
| 16(B) | 8 | Area ratios, medians, same-height and same-base comparisons |
| 16(C) | 9 | Combined applications of the theorems with advanced area-ratio problems |
Browse the full Class 9 Maths book to see how this chapter connects with the rest of the syllabus. The chapters around it use the same habit of proving equality by construction.
Area Theorem Class 9 ICSE frequently asked questions
What is the area theorem class 9 ICSE all about?
It explains how to compare areas of parallelograms, rectangles and triangles under fixed conditions. Figures on the same base and between the same parallels are equal in area, and triangles sharing a vertex have areas in the ratio of their bases. The proofs use congruence and the basic area formulas from section 16.1.
How do I prove that parallelograms on the same base and between the same parallels are equal in area?
Take parallelograms ABCD and ABEF on base AB. Triangles ADF and BCE are congruent by ASA, since AD = BC and their corresponding angles match. Congruent triangles have equal areas, so adding the shared region to both sides leaves the two parallelograms equal in area.
What does it mean when triangles have the same vertex and bases along the same line?
It means every triangle uses the same altitude from the shared vertex, so the heights cancel in the area formula. Their areas are proportional to their bases. This is the main tool in section 16.3 of this ICSE chapter.
How do I use the area ratio to solve exercise 16(B) questions?
Write the ratio of the bases first, then transfer it to the areas. A median gives equal bases, so the areas are equal. If BD : DC = m : n, then \text{Area}(\Delta ABD) : \text{Area}(\Delta ADC) = m : n.
Where can I download the chapter 16 area theorems PDF for class 9 ICSE?
The download block on this page provides the chapter 16 PDF for class 9 ICSE. You can read each section here in the same order as the book, then open the exercises alongside the PDF.
Reference: ICSE Class 9 Mathematics textbook material, chapter 16: Area Theorems Proof and Use — CISCE-aligned.