Area

1. What "area" really means

Think of the rangoli example from the chapter: to colour a region evenly you need to know how much surface it covers. We measure that by counting how many unit squares — squares of side $1$ unit — fit inside the region without gaps or overlaps. A unit square of side $1$ cm has area $1$ square centimetre, written $1\text{ cm}^2$ (or "sq. cm").

For a rectangle we don't count one-by-one: a rectangle $7$ cm $\times$ $4$ cm packs $7\times4=28$ unit squares, while $8$ cm $\times$ $3$ cm packs only $8\times3=24$. So the first rectangle needs more rangoli powder.

The diagonal of a rectangle splits it into two congruent triangles, so each triangle is exactly half the rectangle — a $7\times4$ rectangle gives two triangles of area $\tfrac12\times7\times4=14\text{ cm}^2$ each. This single idea grows into the triangle formula.

2. Why perimeter can't measure area

A natural question: since perimeter is easy to measure, can't we use it for area? No. Perimeter measures the boundary length; area measures the surface inside. They are unrelated quantities:

• Two regions can have the same perimeter but different areas (e.g. a $1\times5$ rectangle and a $3\times3$ square both have... no — but a $1\times5$ has perimeter $12$, area $5$; a $4\times2$ also has perimeter $12$, area $8$).
• One region can have a larger perimeter yet a smaller area than another — a long thin region has a big boundary but little inside.

So always use the unit-square (formula) method, never the perimeter, to find area.

3. Area of a triangle

To find a triangle's area, enclose it in a rectangle. Drop the height (altitude) from the top vertex to the base; the rectangle built on that base and height has the triangle as exactly half of it. Hence:

The height is the perpendicular distance from a vertex to the opposite side (the base). Base and height must be a matching perpendicular pair.

Does it work for every triangle? Yes. For an obtuse triangle where the foot of the height falls outside the base, write the triangle as a difference of two right-cornered triangles: $\text{Area}(\triangle ABC)=\tfrac12 h\cdot DC-\tfrac12 h\cdot DB=\tfrac12 h\,(DC-DB)=\tfrac12 h\cdot BC$. So $\tfrac12\,b\,h$ holds for all triangles.

4. Using the formula backwards — find a missing height

A triangle has only one area, but you can compute it using any base–height pair. Equating two such expressions lets you find an unknown altitude.

Median fact (from the diagonals problem): the line from a vertex to the midpoint of the opposite side (a median) splits a triangle into two triangles of equal area — they share the same height and have equal bases.

5. Triangles between two parallel lines

Take a fixed base $BC$ and let the third vertex slide along a line $l$ parallel to $BC$. Every such triangle has the same height (the distance between the parallels), so:

Among these, the one with minimum perimeter is the isosceles triangle (third vertex above the midpoint of $BC$) — the chapter proves this using the "mirror reflection / shortest path" argument. (Perimeter changes even though area does not — another reminder that the two are independent.)

6. Area of any polygon

Any polygon — quadrilateral, pentagon, hexagon, anything — can be cut into triangles by drawing diagonals. Find each triangle's area and add. This is the master strategy whenever no special formula fits.

7. Area of a parallelogram (by dissection)

Cut a right triangle off one end of the parallelogram and slide it to the other end — the parallelogram becomes a rectangle of the same area, with width equal to the base and length equal to the height (perpendicular distance between the two parallel sides).

Watch out: the height is the perpendicular distance, NOT the slanting side. Any side may be chosen as base, provided you use the matching perpendicular height.

Rectangle vs parallelogram, same sides: a rectangle $5\times4$ has area $20$; a parallelogram with the same sides $5$ and $4$ has height less than $4$ (it leans), so its area is less than $20$. The rectangle is the "fattest" — for given sides, slanting only loses area.

8. Area of a rhombus

A rhombus is a parallelogram, so $\text{base}\times\text{height}$ still works. But its diagonals are perpendicular bisectors of each other, which gives a neater formula. Dissecting the rhombus along its diagonals and rearranging gives a rectangle with sides $AC$ and $\tfrac12 BD$:

This also follows by adding the two triangles the diagonals create: $\tfrac12\,AO\cdot BD+\tfrac12\,CO\cdot BD=\tfrac12\,BD\,(AO+CO)=\tfrac12\,BD\cdot AC$.

9. Area of a trapezium

A trapezium has one pair of parallel sides (lengths $a$ and $b$) a perpendicular distance $h$ apart. Splitting it into a rectangle and two triangles (or using two rotated copies that form a parallelogram of base $a+b$) gives:

Two-copies idea: two identical trapeziums joined edge-to-edge make a parallelogram with base $(a+b)$ and height $h$, area $h(a+b)$; one trapezium is half of that — hence the $\tfrac12$.

10. Splitting a composite figure

Real shapes are mixes of rectangles and triangles. Break them up, find each piece, then add (or subtract a cut-out region).

11. Area units in real life

The chapter ends with everyday area sense:

• An A4 sheet is $21\text{ cm}\times29.7\text{ cm}=623.7\text{ cm}^2$.
• Unit links: $1\text{ in}=2.54$ cm, so $1\text{ in}^2=2.54^2=6.4516\text{ cm}^2$; $1\text{ ft}=12$ in.
• Land is measured in larger units: $1\text{ acre}=43{,}560\text{ ft}^2$; very large areas in $\text{km}^2$ ($1\text{ km}^2=1{,}000{,}000\text{ m}^2$).
• India also uses local units — bigha, gaj, katha, dhur, cent, ankanam.

12. The five formulas — quick table & common mistakes

• Using the slant side instead of the perpendicular height in parallelogram/triangle — height must make a right angle with the base.
• Forgetting the $\tfrac12$ in triangle, rhombus or trapezium.
• Trapezium: using $h\times(a+b)$ without halving, or putting a non-parallel side as $a$ or $b$.
• Mixing units — convert everything to one unit before multiplying; area unit is the length unit squared.
• Confusing perimeter with area — perimeter is boundary length, area is surface covered.

13. NCERT Figure-it-Out — selected solutions

• Rhombus, diagonals 20 & 15: $\tfrac12\times20\times15=150\text{ cm}^2$.
• Parallelogram, base 7 cm, height 4 cm: $7\times4=28\text{ cm}^2$.
• Parallelogram, base 5 cm, height 3 cm: $5\times3=15\text{ cm}^2$.
• Trapezium, parallel sides 10 in & (top) with height — (iii): use $\tfrac12 h(a+b)$ with the two parallel sides and perpendicular height $6$ in.
• Quadrilateral with $AC=22$, $BM=DN=3$: $66\text{ cm}^2$ (see §6).
• $\triangle$SUB (isosceles, $SE\perp UB$, area of $\triangle SEB=24$): by symmetry $\triangle SEU$ also $=24$, so $\triangle SUB=24+24=48$ sq. units.
• Midpoint triangle: if $M,N$ are midpoints of $XY,XZ$, then $\triangle XMN$ has $\tfrac14$ the area of $\triangle XYZ$ (sides halved $\Rightarrow$ area quartered).