Two Crossing Roads Do Not Cover Their Two Areas Added

Because the square where they cross would be counted twice.

A park $60$ m by $40$ m has two roads, each $3$ m wide — one running the full length, one the full breadth, crossing near the middle. The long road covers $60\times 3=180{\text{ m}}^2$ and the short one $40\times 3=120{\text{ m}}^2$. Adding gives $300{\text{ m}}^2$.

But the little $3$ m by $3$ m square where they meet belongs to both of those figures, so it has been paid for twice. Subtract it once:

$$180+120-9=291{\text{ m}}^2$$

That overlap is worth $9{\text{ m}}^2$ — small, and exactly the sort of thing a question is set to catch.

Every problem on this page works the same way: break an awkward region into pieces whose areas you know, then add or subtract carefully, watching for anything counted twice or left out. This page covers the second part of the ICSE Class 8 Mathematics chapter on the area of a trapezium and a polygon.

Draw diagonals or perpendiculars to cut it into triangles, trapeziums and rectangles, find each area, then add.

Worked example 1 — a house-shaped pentagon. A five-sided plot consists of a rectangle $10$ m wide and $6$ m high, with a triangle on top of base $10$ m and height $4$ m.

$$\text{rectangle}=10\times 6=60{\text{ m}}^2,\text{triangle}=\frac{1}{2}\times 10\times 4=20{\text{ m}}^2$$

$$\text{total}=80{\text{ m}}^2$$

Worked example 2 — splitting into two trapeziums. A hexagonal plot is cut by a line into two trapeziums. The first has parallel sides $14$ m and $20$ m with height $5$ m; the second has parallel sides $20$ m and $12$ m with height $8$ m.

$$\text{first}=\frac{1}{2}(14+20)\times 5=\frac{1}{2}\times 34\times 5=85{\text{ m}}^2$$

$$\text{second}=\frac{1}{2}(20+12)\times 8=\frac{1}{2}\times 32\times 8=128{\text{ m}}^2$$

$$\text{total}=85+128=213{\text{ m}}^2$$

The shared side of $20$ m appears in both trapeziums, and that is correct — it is a boundary between the pieces, not an area being double-counted. Only overlapping regions get subtracted, never shared edges.

Worked example 3 — the same polygon split differently. Take a quadrilateral $ABCD$ with diagonal $AC=24$ m, where $B$ is $8$ m from $AC$ and $D$ is $10$ m from it on the other side. Splitting along $AC$ gives two triangles:

$$\frac{1}{2}\times 24\times 8+\frac{1}{2}\times 24\times 10=96+120=216{\text{ m}}^2$$

Or, factoring first:

$$\frac{1}{2}\times 24\times (8+10)=\frac{1}{2}\times 24\times 18=216{\text{ m}}^2$$

The second version is one line shorter and is the form worth remembering: half the diagonal times the sum of the two offsets.

Choose the split that gives you measurements you actually have. A polygon can be cut up in many ways and every way gives the same total, so pick the one where each piece has a base and a perpendicular height you have been told. Drawing a diagonal that leaves you with a triangle of unknown height helps nobody, and redrawing the split is cheaper than guessing a length.

A field book records a long diagonal and a set of perpendicular offsets to each corner. Walk along the diagonal, note how far along each offset is taken, and the field splits into triangles at the ends and trapeziums in between.

Worked example. A field is surveyed along the diagonal $AD=30$ m. On the left of the line, $B$ has an offset of $8$ m taken at $6$ m from $A$, and $C$ has an offset of $10$ m taken at $20$ m from $A$. On the right, $E$ has an offset of $6$ m at $10$ m from $A$, and $F$ has an offset of $9$ m at $24$ m from $A$.

The left-hand side splits into a triangle, a trapezium and a triangle:

- Triangle from $A$ to the first offset: base $6$ m, height $8$ m, so $\frac{1}{2}\times 6\times 8=24{\text{ m}}^2$
- Trapezium between the offsets: parallel sides $8$ m and $10$ m, distance apart $20-6=14$ m, so $\frac{1}{2}(8+10)\times 14=126{\text{ m}}^2$
- Triangle from the last offset to $D$: base $30-20=10$ m, height $10$ m, so $\frac{1}{2}\times 10\times 10=50{\text{ m}}^2$

$$\text{left total}=24+126+50=200{\text{ m}}^2$$

The right-hand side, the same way:

- Triangle: base $10$ m, height $6$ m, so $30{\text{ m}}^2$
- Trapezium: parallel sides $6$ m and $9$ m, distance $24-10=14$ m, so $\frac{1}{2}(15)\times 14=105{\text{ m}}^2$
- Triangle: base $30-24=6$ m, height $9$ m, so $27{\text{ m}}^2$

$$\text{right total}=30+105+27=162{\text{ m}}^2$$

$$\text{area of the field}=200+162=362{\text{ m}}^2$$

At a levelling charge of ₹$25$ per square metre the cost would be $362\times 25=₹9050$.

The distance between two offsets is a subtraction, not a reading. The trapezium's height was $20-6=14$ m, because both figures were measured **from $A$**. Using $20$ m as the height is the standard error in field-book questions, and it inflates the answer badly.

Why the end pieces are triangles. At $A$ the offset is zero — the boundary meets the diagonal there — so the first piece has parallel sides $0$ and $8$, which is a triangle. The trapezium formula still works: $\frac{1}{2}(0+8)\times 6=24{\text{ m}}^2$, the same answer. A triangle is a trapezium with one parallel side of zero, so you can run the whole calculation with a single formula if you prefer.

Keep the two sides separate. Offsets on opposite sides of the diagonal belong to different regions, and a trapezium can only be formed between two offsets on the same side. Mixing an $8$ m left offset with a $6$ m right offset invents a shape that is not part of the field.

Subtract the inner rectangle from the outer one. The path is the difference between two areas, and the only care needed is over which dimensions grow or shrink.

Worked example 1 — a path outside. A rectangular garden measures $20$ m by $15$ m, with a path $2$ m wide running all around the outside.

The path adds $2$ m on each of two opposite sides, so the outer rectangle is

$$(20+2+2)\times (15+2+2)=24\times 19=456{\text{ m}}^2$$

$$\text{path}=456-20\times 15=456-300=156{\text{ m}}^2$$

Add the width twice, not once. Writing $22\times 17$ is the mistake this question exists for: the path runs along the left and the right, so the length grows by $4$ m in total.

Worked example 2 — a path inside. A rectangular plot $30$ m by $20$ m has a path $2.5$ m wide inside all round.

The inner rectangle shrinks by twice the width on each dimension:

$$(30-5)\times (20-5)=25\times 15=375{\text{ m}}^2$$

$$\text{path}=600-375=225{\text{ m}}^2$$

Worked example 3 — crossing roads. Returning to the park $60$ m by $40$ m with two roads each $3$ m wide:

$$\text{roads}=60\times 3+40\times 3-3\times 3=180+120-9=291{\text{ m}}^2$$

$$\text{remaining grass}=60\times 40-291=2400-291=2109{\text{ m}}^2$$

At a paving cost of ₹$150$ per square metre the roads would cost $291\times 150=₹43650$.

Worked example 4 — checking the overlap by a second method. The long road minus the crossing square is $180-9=171{\text{ m}}^2$, and adding the whole short road gives $171+120=291{\text{ m}}^2$. The same answer, reached by assigning the crossing to one road only — which is a useful way to see why it is subtracted exactly once.

Worked example 5 — a canal cross-section. A canal has a top width of $10$ m, a bottom width of $6$ m and a depth of $3$ m. Its cross-section is a trapezium:

$$\text{area}=\frac{1}{2}(10+6)\times 3=\frac{1}{2}\times 16\times 3=24{\text{ m}}^2$$

Worked example 6 — a border on a picture. A photograph $18$ cm by $12$ cm is set in a frame with a border $3$ cm wide all round.

$$\text{outer}=24\times 18=432{\text{ cm}}^2,\text{border}=432-216=216{\text{ cm}}^2$$

Here the border happens to have the same area as the photograph itself — a coincidence of these numbers, not a rule, and worth checking rather than assuming.

A road across a park is not the same as a path around it. A path around the outside makes both dimensions grow; a road across the middle makes neither change. Drawing the figure before calculating settles which situation you are in, and it takes about ten seconds.

Square the length conversion. This is where most errors in the chapter are made.

$$1\text{ m}=100\text{ cm}\Rightarrow 1{\text{ m}}^2=100\times 100=10000{\text{ cm}}^2$$

The rest follow the same way:

- $1\text{ hectare}=10000{\text{ m}}^2$
- $1\text{ are}=100{\text{ m}}^2$, so $1$ hectare $=100$ ares
- $1{\text{ km}}^2=1000000{\text{ m}}^2=100$ hectares

**$1{\text{ m}}^2$ is not $100{\text{ cm}}^2$.** Picture a square metre ruled into centimetre squares: it is $100$ squares across and $100$ down, so it holds $10000$ of them. Carrying the length factor across instead of the squared factor makes every answer wrong by a factor of one hundred, which is why this deserves its own section.

Worked example 1. Convert $3.5{\text{ m}}^2$ to square centimetres.

$$3.5\times 10000=35000{\text{ cm}}^2$$

Worked example 2. A rectangular field measures $200$ m by $125$ m. Find its area in hectares.

$$\text{area}=200\times 125=25000{\text{ m}}^2=\frac{25000}{10000}=2.5\text{ hectares}$$

Worked example 3. A trapezium-shaped field has parallel sides $80$ m and $120$ m with height $50$ m. Give its area in hectares.

$$\text{area}=\frac{1}{2}(80+120)\times 50=\frac{1}{2}\times 200\times 50=5000{\text{ m}}^2=0.5\text{ hectare}$$

Worked example 4 — converting before calculating. A room floor measures $450$ cm by $300$ cm. Find its area in square metres.

Convert the lengths first: $4.5$ m by $3$ m, giving $13.5{\text{ m}}^2$.

Or convert at the end: $450\times 300=135000{\text{ cm}}^2$, and $\frac{135000}{10000}=13.5{\text{ m}}^2$. The same answer, and converting the lengths first usually keeps the numbers smaller.

Worked example 5 — mixed units in one question. A path $50$ cm wide runs around the outside of a lawn $12$ m by $8$ m. Find the path's area in square metres.

First put everything in metres: the path is $0.5$ m wide, so the outer rectangle is $13\times 9=117{\text{ m}}^2$ and

$$\text{path}=117-96=21{\text{ m}}^2$$

Never subtract areas measured in different units. Working out the lawn in square metres and the path in square centimetres and then combining them produces nonsense — so the very first line of any mixed-unit question should convert everything to a single unit.