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Area

NCERT Class 8 · Mathematics Based on NCERT Class 8 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter explores the concept of area as the measure of the space enclosed by a shape, measured in unit squares. It builds on the idea that area is not related to perimeter and introduces formulas for rectangles, squares, triangles, parallelograms, rhombuses, and trapeziums. The big idea is that any polygon can be divided into triangles, and by knowing how to find the area of a triangle, we can find the area of any polygon. Students must understand that area is measured in square units and that different shapes can have the same area but different perimeters.
Key Concepts (12)
Area
The measure of a region found by counting the number of unit squares (which can also be a fraction) whose area equals that of the given region.
e.g. To find how much rangoli powder is needed to colour a rectangle, we count the number of 1 cm × 1 cm unit squares that fit inside it.
Unit square
A square of side length 1 cm (or 1 unit) used to measure area.
e.g. A rectangle of 7 cm by 4 cm contains 7 × 4 = 28 unit squares of side 1 cm.
Perimeter
The length of the boundary of a region.
e.g. A rectangle with sides 7 cm and 4 cm has perimeter 2×(7+4) = 22 cm.
Congruent triangles
Triangles that are identical in shape and size; the diagonal of a rectangle divides it into two congruent triangles.
e.g. In a rectangle of 7 cm by 4 cm, each triangle formed by a diagonal has area = ½ × 7 × 4 = 14 cm².
Base of a triangle
Any side of a triangle chosen as the reference side for calculating area.
e.g. In triangle ABC, if BC is taken as base, then the height is the perpendicular distance from A to BC.
Height (altitude) of a triangle
The perpendicular distance from the vertex opposite the base to the line containing the base.
e.g. In triangle ABC with base BC, the height is the length of the perpendicular from A to BC.
Dissection
The process of cutting a figure into pieces and rearranging them to get a different figure of equal area.
e.g. Cutting a parallelogram along a height and rearranging the pieces to form a rectangle of equal area.
Parallelogram
A quadrilateral with opposite sides parallel and equal.
e.g. A window grill shaped like a slanted rectangle is a parallelogram.
Rhombus
A parallelogram with all sides equal, and diagonals that are perpendicular bisectors of each other.
e.g. A kite shape or a diamond in a playing card is a rhombus.
Trapezium
A quadrilateral with at least one pair of parallel sides.
e.g. The shape of a table top with one side longer than the opposite side is a trapezium.
Diagonal of a rectangle
A line segment joining two opposite vertices of a rectangle; it divides the rectangle into two congruent triangles.
e.g. In rectangle ABCD, diagonal AC divides it into triangles ABC and ADC, each of equal area.
Midpoint
A point that divides a line segment into two equal parts.
e.g. In triangle XYZ, if M is the midpoint of XY, then line ZM divides the triangle into two triangles of equal area.
Formulas & Equations (9)
Area of a rectangle = length × width
Reads as: Area of a rectangle equals length times width
Variables: length (l) in cm or m, width (w) in cm or m, area in cm² or m²
Use when: Finding the area of any rectangular surface like a tabletop, a book cover, or a plot of land.
⚠ Students often confuse length and width with perimeter and write 2(l + w) instead of l × w.
Area of a triangle = ½ × base × height
Reads as: Area of a triangle equals half times base times height
Variables: base (b) in cm or m, height (h) in cm or m, area in cm² or m²
Use when: Finding the area of any triangle when the base and the perpendicular height are known.
⚠ Students forget to multiply by ½, or they use the slant height instead of the perpendicular height.
Area of a parallelogram = base × height
Reads as: Area of a parallelogram equals base times height
Variables: base (b) in cm or m, height (h) in cm or m, area in cm² or m²
Use when: Finding the area of any parallelogram, including rectangles and rhombuses (when using base and height).
⚠ Students use the slant side length instead of the perpendicular height.
Area of a rhombus = ½ × product of diagonals
Reads as: Area of a rhombus equals half times the product of its diagonals
Variables: diagonal 1 (d₁) in cm or m, diagonal 2 (d₂) in cm or m, area in cm² or m²
Use when: Finding the area of a rhombus when the lengths of both diagonals are known.
⚠ Students forget to multiply by ½, or they use side length instead of diagonals.
Area of a trapezium = ½ × height × (sum of parallel sides)
Reads as: Area of a trapezium equals half times height times the sum of the parallel sides
Variables: height (h) in cm or m, parallel sides a and b in cm or m, area in cm² or m²
Use when: Finding the area of a trapezium when the height and the lengths of the two parallel sides are known.
⚠ Students add all four sides instead of only the two parallel sides, or they forget to multiply by ½.
1 in = 2.54 cm
Reads as: One inch equals two point five four centimeters
Variables: in = inches, cm = centimeters
Use when: Converting lengths from inches to centimeters.
⚠ Students divide by 2.54 instead of multiplying when converting inches to cm.
1 ft = 12 in
Reads as: One foot equals twelve inches
Variables: ft = feet, in = inches
Use when: Converting lengths from feet to inches or vice versa.
⚠ Students use 10 instead of 12 for the conversion factor.
1 in² = 2.54² cm² = 6.4516 cm²
Reads as: One square inch equals six point four five one six square centimeters
Variables: in² = square inches, cm² = square centimeters
Use when: Converting area from square inches to square centimeters.
⚠ Students multiply by 2.54 instead of 2.54² when converting area.
1 acre = 43,560 ft²
Reads as: One acre equals forty-three thousand five hundred sixty square feet
Variables: acre = unit of land area, ft² = square feet
Use when: Converting land area from acres to square feet.
⚠ Students confuse acre with hectare or use the wrong conversion factor.
How It Works
Finding area by counting unit squares
  1. Step 1: Draw the shape on a grid of unit squares (e.g., 1 cm × 1 cm squares).
  2. Step 2: Count the number of full unit squares inside the shape.
  3. Step 3: Count partial squares and combine them to estimate the total area.
  4. Step 4: The total number of unit squares (including fractions) gives the area in square units.
💡 The area is the number of unit squares, not the length of the boundary. Two shapes with the same perimeter can have different areas.
Deriving area of a triangle from a rectangle
  1. Step 1: Draw a rectangle around the triangle such that the base of the triangle is one side of the rectangle.
  2. Step 2: The diagonal of the rectangle divides it into two congruent triangles.
  3. Step 3: Each triangle has area exactly half the area of the rectangle.
  4. Step 4: So, area of triangle = ½ × base × height.
💡 The height must be perpendicular to the base, not the slant side.
Converting a parallelogram into a rectangle by dissection
  1. Step 1: Draw a perpendicular (height) from one vertex to the opposite side.
  2. Step 2: Cut along this perpendicular to get a right triangle and a trapezium.
  3. Step 3: Slide the triangle to the other side of the trapezium to form a rectangle.
  4. Step 4: The area of the rectangle (base × height) equals the area of the original parallelogram.
💡 The base and height of the parallelogram are the same as the length and width of the resulting rectangle.
Finding area of a rhombus using diagonals
  1. Step 1: Draw both diagonals of the rhombus; they intersect at right angles and bisect each other.
  2. Step 2: The rhombus is divided into four right triangles.
  3. Step 3: Each triangle can be rearranged to form a rectangle with sides equal to one diagonal and half the other diagonal.
  4. Step 4: Area of rhombus = ½ × product of diagonals.
💡 This formula works only for rhombus (all sides equal, diagonals perpendicular). It does not work for a general parallelogram.
Finding area of a trapezium by splitting into rectangle and triangles
  1. Step 1: Draw perpendiculars from the ends of the shorter parallel side to the longer parallel side.
  2. Step 2: This divides the trapezium into a rectangle and two right triangles.
  3. Step 3: Find the area of the rectangle (base × height) and each triangle (½ × base × height).
  4. Step 4: Add all areas to get the total area of the trapezium.
  5. Step 5: The formula simplifies to ½ × height × (sum of parallel sides).
💡 The height is the perpendicular distance between the two parallel sides, not the slant height.
Finding area of a trapezium using two copies
  1. Step 1: Take two identical trapeziums.
  2. Step 2: Rotate one trapezium by 180° and join it to the other along the non-parallel side.
  3. Step 3: The resulting shape is a parallelogram with base = sum of parallel sides of trapezium and height = same height.
  4. Step 4: Area of parallelogram = height × (a + b).
  5. Step 5: Area of one trapezium = ½ × height × (a + b).
💡 The two trapeziums must be identical for this method to work.
Exam Traps — Don't Make These Mistakes
⚠ Using slant height instead of perpendicular height in triangle area
✗ Area = ½ × 5 × 4 = 10 cm² for a triangle with base 5 cm and slant side 4 cm.
→ The height must be the perpendicular distance from the vertex to the base, not the length of the slant side.
✓ First identify or calculate the perpendicular height. If not given, use Pythagoras theorem to find it.
⚠ Confusing perimeter with area
✗ Two rectangles with the same perimeter have the same area.
→ Perimeter is the length of the boundary, area is the space inside. Two shapes can have the same perimeter but very different areas (e.g., a long thin rectangle vs a square).
✓ Perimeter and area are independent measures. Always calculate area using the appropriate formula, not by comparing perimeters.
⚠ Forgetting the ½ in triangle, rhombus, and trapezium formulas
✗ Area of triangle = base × height; Area of rhombus = d₁ × d₂; Area of trapezium = height × (a + b).
→ All three formulas require multiplication by ½ because the triangle is half a rectangle, the rhombus area is half the product of diagonals, and the trapezium area is half the product of height and sum of parallel sides.
✓ Always write the formula with ½: Area of triangle = ½ × base × height; Area of rhombus = ½ × d₁ × d₂; Area of trapezium = ½ × height × (a + b).
⚠ Using wrong base-height pair in parallelogram
✗ Area = 5 cm × 6 cm = 30 cm² for a parallelogram with side 5 cm and slant side 6 cm.
→ The height must be perpendicular to the chosen base. The slant side is not the height unless it is perpendicular.
✓ Identify the perpendicular distance between the two parallel sides (height) and multiply by the length of the base (one of the parallel sides).
⚠ Adding all sides in trapezium area formula
✗ Area = ½ × height × (a + b + c + d) where a, b, c, d are all four sides.
→ The formula uses only the sum of the two parallel sides, not all four sides.
✓ Area = ½ × height × (sum of parallel sides). Identify which sides are parallel and add only those two.
Compare & Contrast
What they measure
Area of a rectangle
Perimeter of a rectangle
Area measures the space inside the rectangle (length × width, in square units), while perimeter measures the length of the boundary (2 × (length + width), in linear units). Two rectangles can have the same perimeter but different areas.
Formula and shape
Area of a parallelogram
Area of a rectangle
Both use base × height, but in a rectangle the height is the same as the side (since sides are perpendicular), while in a parallelogram the height is the perpendicular distance between the parallel sides, which is shorter than the slant side.
Method of calculation
Area of a rhombus using diagonals
Area of a rhombus using base and height
Using diagonals: Area = ½ × d₁ × d₂ (requires diagonal lengths). Using base and height: Area = base × height (requires one side and perpendicular height). Both give the same result, but the diagonal method is often easier when diagonals are given.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11 marks
Most asked: Area of triangle (½ × base × height) · Area of parallelogram (base × height) · Area of trapezium (½ × height × sum of parallel sides) · Finding missing side or height given area · Word problems involving paths around rectangles
Don't skip: Finding area of a composite figure (e.g., a path around a park) by breaking it into rectangles and triangles.; Application of area formulas to find missing dimensions (e.g., given area and base, find height).; Proof-based questions: Show that the diagonal of a rectangle divides it into two triangles of equal area.
One-Line Revision (read aloud before the exam)
  1. Area is measured in square units, such as cm² or m².
  2. Area of a rectangle = length × width.
  3. Area of a triangle = ½ × base × height.
  4. The diagonal of a rectangle divides it into two congruent triangles of equal area.
  5. Area of a parallelogram = base × height.
  6. Area of a rhombus = ½ × product of its diagonals.
  7. Area of a trapezium = ½ × height × (sum of parallel sides).
  8. Perimeter is not a measure of area; two shapes with the same perimeter can have different areas.
  9. The height in area formulas must always be perpendicular to the base.
  10. Any polygon can be divided into triangles to find its area.
  11. 1 inch = 2.54 cm, 1 foot = 12 inches, 1 acre = 43,560 ft².
  12. In a triangle, the line joining a vertex to the midpoint of the opposite side divides the triangle into two triangles of equal area.
  13. The diagonals of a rhombus are perpendicular bisectors of each other.
  14. A trapezium has at least one pair of parallel sides.

MCQs — 10 Questions with Answers

Q1. A rectangle has length 8 cm and width 5 cm. What is its area?

  • A. 13 cm²
  • B. 40 cm² ✓
  • C. 26 cm²
  • D. 64 cm²

Answer: B — Area of rectangle = length × width = 8 × 5 = 40 cm².

Q2. A triangle has a base of 10 m and a height of 6 m. What is its area?

  • A. 30 m² ✓
  • B. 60 m²
  • C. 16 m²
  • D. 15 m²

Answer: A — Area of triangle = 1/2 × base × height = 1/2 × 10 × 6 = 30 m².

Q3. Two rectangles have the same perimeter. Which statement is always true?

  • A. They must have the same area
  • B. They must have the same length and width
  • C. They could have different areas ✓
  • D. The one with larger length must have larger area

Answer: C — Perimeter does not determine area uniquely; two rectangles can have the same perimeter but different dimensions and therefore different areas.

Q4.

  • A. 1
  • B. 2 ✓
  • C. 3
  • D. 4

Answer: B — One diagonal creates two congruent triangles, each with area equal to half the rectangle.

Q5. Harsha is designing a rangoli on a rectangular floor measuring 12 m by 8 m. How much area does she need to cover with powder?

  • A. 40 m²
  • B. 96 m² ✓
  • C. 20 m²
  • D. 48 m²

Answer: B — Area = 12 × 8 = 96 m², which represents the total surface area to be colored.

Q6. A triangular field has a base of 20 m and the perpendicular distance from the opposite vertex to the base is 15 m. What is the area of the field?

  • A. 150 m² ✓
  • B. 300 m²
  • C. 35 m²
  • D. 75 m²

Answer: A — Area of triangle = 1/2 × 20 × 15 = 150 m².

Q7. In triangle ABC, M is the midpoint of side BC. If the area of triangle ABC is 40 cm², what is the area of triangle ABM?

  • A. 10 cm²
  • B. 20 cm² ✓
  • C. 30 cm²
  • D. 40 cm²

Answer: B — A median divides a triangle into two equal-area triangles, so area of ABM = 40 ÷ 2 = 20 cm².

Q8. Two rectangles have areas 24 cm² and 24 cm² respectively. One has dimensions 6 cm × 4 cm, the other has dimensions 8 cm × 3 cm. Which statement is correct?

  • A. Both have the same perimeter
  • B. The first has a larger perimeter
  • C. The second has a larger perimeter ✓
  • D. Perimeter cannot be compared for shapes with equal area

Answer: C — First rectangle: perimeter = 2(6+4) = 20 cm; second: perimeter = 2(8+3) = 22 cm, so the second has a larger perimeter despite equal areas.

Q9. A farmer has a triangular plot of land. If the base is 50 m and the perpendicular height is 24 m, and she wants to plant crops covering 450 m², how much area will be left uncovered?

  • A. 150 m² ✓
  • B. 200 m²
  • C. 250 m²
  • D. 300 m²

Answer: A — Total area of triangle = 1/2 × 50 × 24 = 600 m²; uncovered area = 600 − 450 = 150 m².

Q10. Consider a rectangle ABCD. When both diagonals are drawn, they intersect at point O. Comparing triangles AOB and BOC, which is true if the rectangle has length 10 m and width 6 m?

  • A. Triangle AOB has greater area
  • B. Triangle BOC has greater area
  • C. Both triangles have equal area ✓
  • D. Cannot be determined without more information

Answer: C — The diagonals of a rectangle bisect each other, and all four triangles formed have equal area of 15 m² each (half of 30 m² per triangle pair).

Flashcards

What is the formula for the area of a rectangle?

Area of rectangle = length × width (in square units).

How is the area of a triangle related to the area of a rectangle with the same base and height?

The area of a triangle is exactly half the area of the rectangle because the diagonal divides the rectangle into two congruent triangles.

Can two shapes with the same perimeter have different areas?

Yes, perimeter measures only the boundary length; it does not determine area, so shapes can have equal perimeters but unequal areas.

What is the formula for the area of a triangle?

Area of triangle = 1/2 × base × height, where height is perpendicular to the base.

What does it mean when we say a line is parallel to the base of a triangle in the context of area?

Any triangle with its vertex on a line parallel to the base will have the same area because the perpendicular distance (height) from the vertex to the base remains constant.

If the diagonals of a rectangle intersect at point O, how many equal-area triangles are formed?

Four equal-area triangles are formed because the diagonals bisect each other, making opposite triangles have equal bases and the same height.

What measurement is essential to find the area of a triangle?

Both the base and the perpendicular height from the opposite vertex are essential.

How does the median of a triangle relate to area?

A median (line from vertex to midpoint of opposite side) divides the triangle into two triangles of equal area.

Why does the triangle area formula work for obtuse triangles?

The formula still applies because the height can be drawn perpendicular to the base (or its extension) even outside the triangle.

If a rectangle has area 36 cm² and is divided into 4 equal parts, what is the area of each part?

Each part has area 9 cm² because 36 ÷ 4 = 9.

Important Board Questions

State the formula for the area of a rectangle and explain what each symbol represents. [1 mark]

Area = length × width; length is the longer dimension, width is the shorter, and the result is in square units.

A rectangular park measures 25 m by 15 m. A path of width 2 m is laid around the inside of the park. Find the area of the path. [2 marks]

Find the area of the outer rectangle (25 × 15), then the area of the inner rectangle after reducing each dimension by 2 m on each side (21 × 11), then subtract: outer area − inner area.

In a triangle XYZ, if you draw a line from vertex X to the midpoint M of side YZ, explain how this line (median) divides the triangle into two parts. What can you say about the areas of the two triangles XYM and XMZ? Give reasons. [3 marks]

A median divides a triangle into two triangles with equal area because both share the same height from X and have equal bases (M is the midpoint). Use the area formula 1/2 × base × height to show equality.

A field is in the shape of a triangle with a base of 60 m and a perpendicular height of 40 m. An engineer needs to lay a road that reduces the height of the field to 30 m but keeps the base the same. (a) Find the original area of the field. (b) Find the new area after the road is laid. (c) What is the loss in area? (d) If the original field cost ₹5,000 per square metre, what is the loss in value? [5 marks]

Use area formula for triangle: 1/2 × base × height. (a) Original area = 1/2 × 60 × 40 = 1,200 m². (b) New area = 1/2 × 60 × 30 = 900 m². (c) Loss = 1,200 − 900 = 300 m². (d) Loss in value = 300 × 5,000 = ₹15,00,000.

Assertion-Reason Questions for Area

Q1. Assertion (A):

The area of a triangle is always half the area of the rectangle that circumscribes it, regardless of the triangle's shape or orientation.

Reason (R):

A triangle can always be enclosed in a rectangle by drawing perpendiculars from one vertex to the opposite side, and this enclosing rectangle's diagonal divides it into two equal triangles.

Answer: A — Both A and R are true, and R is the correct explanation of A — The assertion is true—when a rectangle circumscribes a triangle with one side as the rectangle's base, the triangle occupies exactly half the rectangle's area. The reason correctly explains why: the enclosing rectangle's diagonal creates two congruent triangles, and the inscribed triangle has the same height and base as one of these, making it half the rectangle. This is the core concept demonstrated in the chapter.

Q2. Assertion (A):

Two rectangles with equal areas must have equal perimeters.

Reason (R):

The area of a rectangle is length × width, while the perimeter is 2(length + width), and these are independent properties that can vary separately.

Answer: D — A is false but R is true — The assertion is false. Two rectangles with the same area can have very different perimeters. For example, a 10 cm × 1 cm rectangle (area 10 cm², perimeter 22 cm) and a 5 cm × 2 cm rectangle (area 10 cm², perimeter 14 cm) both have area 10 but different perimeters. The reason is true but doesn't support the assertion—in fact, it explains why the assertion is wrong.

Q3. Assertion (A):

When a path of uniform width is laid around a rectangular park, the area of the path can be calculated as the difference between the outer rectangle's area and the inner park's area.

Reason (R):

A path around a rectangle can be decomposed into multiple smaller rectangles, and the sum of these smaller rectangles' areas must equal the total path area.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—subtracting the inner area from the outer area gives the path area. The reason is also true; the path can indeed be broken into smaller rectangles (four corner squares and four edge rectangles). However, the reason is not the best explanation for why the assertion works. The assertion works simply because the path is the region between two rectangles, so: Path area = Outer area - Inner area. The reason describes an alternative method, not the fundamental explanation.

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Case-Based Questions (CBQ) for Area

Case Study:

A shopping mall has a rectangular parking lot that is 100 m long and 60 m wide. A decorative floor tiling project covers a path around the inside of the parking lot boundary. The path has a uniform width of 3 m running along all four sides. Inside the path is the actual parking area. The mall management wants to calculate the costs of maintenance: tiling costs ₹150 per square meter for the path, and parking space allocation costs ₹50 per square meter. They also need to determine the remaining parking capacity after the path is constructed.

Q1. What is the total area of the decorative path?

  • A. 924 m²
  • B. 876 m²
  • C. 936 m² ✓
  • D. 960 m²

Answer: C

Q2. What is the area available for actual parking after the path is constructed?

  • A. 5184 m² ✓
  • B. 5400 m²
  • C. 5076 m²
  • D. 6000 m²

Answer: A

Q3. If the path width is increased from 3 m to 4 m on all sides, what would be the new parking area?

  • A. 4928 m² ✓
  • B. 5184 m²
  • C. 4800 m²
  • D. 5000 m²

Answer: A

Q4. What is the total cost of tiling the path at ₹150 per square meter?

  • A. ₹139,400
  • B. ₹140,400 ✓
  • C. ₹141,600
  • D. ₹138,000

Answer: B

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