Class 8 › Mathematics › Chapter
Area
Area of rectangles and triangles, why perimeter cannot measure area, area of polygons, and the formulas for parallelogram, rhombus and trapezium derived by dissection, plus units of area.
1. Area of a Rectangle
Area of a region = the number of unit squares (can be fractional) whose total area equals that of the region.
Rectangle
$$\text{Area} = \text{length} \times \text{width}$$
- Example: a $7 \times 4$ cm rectangle holds $28$ unit squares ($28\ \text{cm}^2$); an $8 \times 3$ cm one holds $24$. So the first is larger.
- Units: written as sq. cm or $\text{cm}^2$.
- Diagonal: a diagonal splits a rectangle into two congruent triangles, so each has area $\tfrac12 \times l \times w$ (e.g. $\tfrac12 \times 7 \times 4 = 14\ \text{cm}^2$).
- Same area, different shape: a square can be cut into 4 equal-area parts in infinitely many ways (compress one edge, expand another by the same amount).
- Paths and borders: area of a path around a park = area of the outer rectangle $-$ area of the inner rectangle (or split the path into rectangles). Moving the inner park inside the outer rectangle does not change the path's area.
2. Why Perimeter Cannot Measure Area
Perimeter
Length of the boundary
Measured in length units (cm, m). Tells how long the edge is.
Area
Space inside the boundary
Measured in square units ($\text{cm}^2$, $\text{m}^2$). Counts unit squares.
- Regions with the same perimeter can have different areas, and vice versa.
- It is even possible that Perimeter 1 $\gt$ Perimeter 2 but Area 1 $\lt$ Area 2.
- Example: a $1 \times 10$ rectangle has perimeter $22$, area $10$; a $4 \times 4$ square has perimeter $16$, area $16$.
- Parallelograms on the same base and between the same parallels have equal area, but their perimeters differ.
- A rectangle and a parallelogram with the same side lengths: the rectangle has greater area (its height is the full side; the parallelogram's height is shorter).
3. Area of a Triangle
Triangle
$$\text{Area} = \tfrac12 \times \text{base} \times \text{height}$$
Draw the outer rectangle on base BC through A: the triangle is exactly half of it.
Why it works
Any triangle in a rectangle: drop the altitude from the apex; it splits the rectangle into two parts, and the triangle takes exactly half of each. So area $= \tfrac12 \times \text{rectangle}$.
Obtuse triangle (no rectangle on BC): with foot $D$ of the altitude outside, $\text{Area}(ABC) = \tfrac12 h\,DC - \tfrac12 h\,DB = \tfrac12 h(DC - DB) = \tfrac12 h \times BC$.
- The formula holds for all triangles: acute, right, obtuse.
- Same base, same height $\Rightarrow$ same area: triangles with their apex anywhere on a line parallel to the base have equal area.
- Height is the perpendicular distance from the apex to the base (not a side, unless the angle is right).
4. Applications of the Triangle Formula
- Finding another altitude: area is the same whichever base you use. If $BC = 5$, $AX = 3$, $AC = 4$: $\tfrac12 \times 3 \times 5 = \tfrac{15}{2} = \tfrac12 \times BY \times 4 \Rightarrow BY = \tfrac{15}{4} = 3.75$.
- Median rule: a line from a vertex to the midpoint of the opposite side divides the triangle into two triangles of equal area (equal bases, same height).
- Diagonals of a rectangle make four triangles of equal area (not congruent) because the diagonals bisect each other.
- Midpoint triangle: if $M, N$ are midpoints of $XY, XZ$, then $\text{Area}(XMN) = \tfrac14 \text{Area}(XYZ)$.
- Minimum perimeter: among triangles on base $BC$ with apex on a parallel line $l$, the one with minimum perimeter is the isosceles one (apex on the perpendicular bisector of $BC$). Reason: reflect $B$ in $l$ to get $B'$; $AB = AB'$, so $AB + AC = AB' + AC$ is least when $A$ lies on the straight line $B'C$. The same idea gives the shortest path house $\to$ river $\to$ tank.
- Area-preserving constructions (Śulba-Sūtras / Euclid): a triangle can be turned into a rectangle of equal area (base $\times$ half height), and a rectangle into a triangle of equal area.
5. Area of Any Polygon
Key idea: any polygon can be divided into triangles, so its area is the sum of the areas of those triangles.
- Quadrilateral: join a diagonal to get two triangles. If $B$ and $D$ are at perpendicular distances $h_1, h_2$ from diagonal $AC$: $\text{Area} = \tfrac12 \times AC \times (h_1 + h_2)$.
- Example: $AC = 22$ cm, $BM = DN = 3$ cm: $\tfrac12 \times 22 \times 3 + \tfrac12 \times 22 \times 3 = 66\ \text{cm}^2$.
- Irregular shapes: split into rectangles, triangles and trapeziums; add (or subtract) the parts.
- The area of a regular polygon (e.g. hexagon) needs the side and the perpendicular distance from the centre, or can be found by cutting into triangles.
6. Parallelogram
Parallelogram
$$\text{Area} = \text{base} \times \text{height}$$
Dissection into a rectangle
1. Draw $AX \perp CD$ (the height). Cut off $\triangle AXD$.
2. Extend $XC$ to $Y$ with $BY \perp XY$. Then $ABYX$ is a rectangle.
3. $\triangle BYC \cong \triangle AXD$ by RHS ($BY = AX$, $BC = AD$, right angles), so the triangle fits exactly into $\triangle BYC$.
4. Area of parallelogram $=$ area of rectangle $ABYX = AX \times XY = \text{height} \times \text{base}$ (since $XY = DC$).
- Dissection: cutting a figure and rearranging the pieces into another figure of equal area.
- Any side can be the base, as long as you use its corresponding height.
- Parallelograms on the same base with the same height have equal area, whatever their slant.
- A diagonal splits a parallelogram into two triangles of equal area.
- Triangle $\to$ parallelogram/rectangle: area of a rectangle with twice a triangle's area $=$ base $\times$ height of that triangle.
7. Rhombus
Rhombus
$$\text{Area} = \tfrac12 \times d_1 \times d_2$$
Derivation
Diagonals of a rhombus are perpendicular bisectors of each other, so $\triangle ABD$ and $\triangle CBD$ are isosceles.
Dissect each into a rectangle and join them: rectangle $WXYZ$ with $XW = AC$ and $WZ = \tfrac{BD}{2}$.
Area $= AC \times \tfrac{BD}{2} = \tfrac12 \times AC \times BD$.
Check: $\tfrac12 AO \cdot BD + \tfrac12 CO \cdot BD = \tfrac12 BD(AO + CO) = \tfrac12 BD \cdot AC$.
- A rhombus is also a parallelogram, so base $\times$ height works too.
- Example: diagonals $20$ cm and $15$ cm: $\tfrac12 \times 20 \times 15 = 150\ \text{cm}^2$.
8. Trapezium
Trapezium
$$\text{Area} = \tfrac12 \times h \times (a + b)$$
Trapezium with parallel sides $a$ (top) and $b$ (bottom), height $h$.
Method 1: rectangle + two triangles
Drop perpendiculars to get a rectangle (width $a$) and two right triangles with bases $x, y$: Area $= \tfrac12 hx + ha + \tfrac12 hy = \tfrac12 h(x + y + 2a)$.
Since $b = x + y + a$, we get $x + y = b - a$. So Area $= \tfrac12 h(b - a + 2a) = \tfrac12 h(a + b)$.
Method 2: two copies
Rotate a second copy and join along the slant side. Since $x + y = 180^\circ$ and $u + v = 180^\circ$, the result is a parallelogram with base $(a + b)$ and height $h$.
Trapezium $= \tfrac12 \times \text{parallelogram} = \tfrac12 h(a + b)$.
- Other proofs: rectangle minus a triangle plus a triangle; or a parallelogram plus a triangle (draw $BG \parallel AD$).
- $h$ is the perpendicular distance between the parallel sides, not the slant side.
- Examples: $a = 24, b = 36, h = 14$ m $\Rightarrow \tfrac12 \times 14 \times 60 = 420\ \text{m}^2$. $a = 12, b = 18, h = 8$ ft $\Rightarrow 120\ \text{ft}^2$.
- An isosceles trapezium converts to a rectangle by cutting along the line joining midpoints of the slant sides and moving the cut triangles (rectangle width $= \tfrac{a+b}{2}$, height $h$).
9. Units & Real-Life Area
| Conversion | Value |
|---|---|
| Length | $1\ \text{in} = 2.54\ \text{cm}$, $1\ \text{ft} = 12\ \text{in}$ |
| Square inch | $1\ \text{in}^2 = 2.54^2 = 6.4516\ \text{cm}^2$ |
| Square foot | $1\ \text{ft}^2 = 144\ \text{in}^2$ |
| Acre | $1\ \text{acre} = 43{,}560\ \text{ft}^2$ |
| Square kilometre | $1\ \text{km}^2 = 10{,}00{,}000\ \text{m}^2$ ($1000^2$) |
- Squared units: convert the length first, then square the factor for area (not just multiply by $2.54$).
- Examples: $10\ \text{in}^2 = 64.516\ \text{cm}^2$; $161.29\ \text{cm}^2 = \tfrac{161.29}{6.4516} = 25\ \text{in}^2$.
- A4 sheet: $21 \times 29.7 = 623.7\ \text{cm}^2$.
- Choice of unit: $\text{cm}^2$ for small objects, $\text{ft}^2$ or $\text{m}^2$ for rooms and houses, acres for land, $\text{km}^2$ for towns and cities.
- Local units in India: bigha, gaj, katha, dhur, cent, ankanam.
10. All Formulas and Quick Revision
| Shape | Area | Needed |
|---|---|---|
| Rectangle | $l \times w$ | length, width |
| Triangle | $\tfrac12 \times b \times h$ | base, perpendicular height |
| Parallelogram | $b \times h$ | a side and its height |
| Rhombus | $\tfrac12 d_1 d_2$ | both diagonals |
| Trapezium | $\tfrac12 h(a + b)$ | parallel sides, height |
| Any polygon | sum of triangle areas | split into triangles |
Strategy
Choosing a method
Unknown shape? Split it into rectangles and triangles. Two altitudes given for one triangle? Equate $\tfrac12 b_1 h_1 = \tfrac12 b_2 h_2$.
Check
Sanity checks
Area is in square units. A triangle is half its enclosing rectangle; a trapezium lies between the triangle and rectangle on its longer side.
Common mistakes:
• Using a slant side instead of the perpendicular height.
• Forgetting the $\tfrac12$ in triangle, rhombus and trapezium formulas.
• Using one parallel side instead of the sum $a + b$ in a trapezium.
• Using a base with the height of a different base in a parallelogram or triangle.
• Judging area by perimeter.
• Converting areas by multiplying by the length factor instead of its square.
• Writing area in plain length units (cm instead of $\text{cm}^2$).
• Using a slant side instead of the perpendicular height.
• Forgetting the $\tfrac12$ in triangle, rhombus and trapezium formulas.
• Using one parallel side instead of the sum $a + b$ in a trapezium.
• Using a base with the height of a different base in a parallelogram or triangle.
• Judging area by perimeter.
• Converting areas by multiplying by the length factor instead of its square.
• Writing area in plain length units (cm instead of $\text{cm}^2$).