Class 8 Mathematics Chapter 14: Area

Four parts of equal area

How many ways can a square be cut into four parts of equal area? There are infinitely many. Take any one way of doing it, then squeeze a part along one edge while stretching it along another by the same amount. The four areas stay equal.

A rangoli question

Rangoli powder fills a region evenly, so the question of which design needs more powder is a question about area. Two rectangles are offered: one 7 centimetres by 4, and one 8 centimetres by 3. Which takes more?

Counting unit squares

Answer it by counting the unit squares that pack into each without overlapping. Here a unit square is one centimetre on a side. The first rectangle holds 7 times 4, which is 28 of them. The second holds 8 times 3, which is 24.

What area means

That is what area is: the number of unit squares, whole or fractional, whose total matches the region. For a rectangle this always comes to the length times the width. We write the two answers as 28 square centimetres and 24 square centimetres.

Could perimeter have done instead?

Here is a fair question. Why count squares at all? The perimeter is easier to measure, being just the length of the boundary. Could it serve as a measure of area?

The two rangoli rectangles again

Look at their perimeters. The first is 2 times 7 plus 4, which is 22 centimetres. The second is 2 times 8 plus 3, which is 22 centimetres as well. Equal perimeters, and yet the areas were 28 and 24.

Perimeter settles nothing

So perimeter is not indicative of area. Regions can have the same perimeter and different areas, and the same area with different perimeters. You can even find a region whose perimeter is larger than another’s while its area is smaller.

A diagonal cuts a rectangle in half

Now to triangles. A diagonal divides a rectangle into two congruent triangles, so each takes half the area. In the 7 by 4 rectangle each triangle has an area of half of 7 times 4, which is 14 square centimetres.

From the rectangle to a formula

That suggests a general formula. Given any triangle, draw a line through the far vertex parallel to the base. A rectangle can then be built on the base with its opposite side along that line, and the triangle sits inside it. Now drop the perpendicular from the far vertex. It splits both the triangle and the rectangle, and each piece of the triangle is half of the rectangle piece it sits in. So the whole triangle is half the rectangle.

The formula

The rectangle measures base by height, where the height is the perpendicular distance from the base to the far vertex. So the area of a triangle is half the base times the height.

Does it hold for every triangle?

For an obtuse triangle the foot of the perpendicular falls outside the base, and no rectangle can be drawn round it in that way. Take the triangle as the difference of two right-angled ones instead.

Working that case

If the foot of the perpendicular is D and the height h, the area is half h times DC minus half h times DB. Factor the half h out and that is half h times DC minus DB, which is half h times BC. The same formula again.

Using the formula backwards

The formula can also be run in reverse. Suppose a triangle is known to have an area of 15 over 2 square units, and one of its sides measures 4 units. What is the height drawn to that side? The area is half that side times the height, which is half of 4 times the height, that is 2 times the height. So 2 times the height equals 15 over 2, and the height is 15 over 4, which is 3.75 units.

Four triangles from two diagonals

Draw both diagonals of a rectangle. The four triangles that result are not congruent, so are their areas equal? Take two that sit side by side and choose their bases along one diagonal.

Why those four are equal

With those bases the two triangles share the same height. And the bases are equal, because the diagonals of a rectangle cut each other in half. Equal base and equal height means equal area, and the argument repeats round all four.

A general statement

That argument proves something wider. In any triangle, the line from a vertex to the midpoint of the opposite side divides it into two triangles of equal area, since they have equal bases and share a height.

Triangles between two parallels

Last, take a fixed base BC and a line parallel to it. Every triangle with BC as base and its third vertex somewhere on that line has the same height, so every one of them has the same area, however different they look.

Their perimeters are not equal

Their perimeters are another matter entirely. The shortest perimeter belongs to one particular position of the third vertex, found by treating the parallel line as a mirror and joining B to the reflection of C. Area and perimeter really are independent.

Scenario:

Two pupils plan rangoli designs for a school function. Asha outlines a rectangle 9 centimetres by 5, while Bina outlines one 11 centimetres by 3. Each will be filled evenly with powder, and the stall has to know which needs more. The art teacher then draws a diagonal across Asha’s rectangle and asks for the area of one of the two triangles. Finally a triangle of area 24 square centimetres is to be drawn with one side measuring 6 centimetres, and the height to that side is wanted.

Analysis:

One question is about area, one about perimeter, and the last runs the triangle formula backwards.

Title:

Triangles on one base

Safety:

Cut the string with scissors only under supervision, or tear it by hand.

What You Notice:

Every area should come to 20 square units. The perimeters will all differ, which is the whole point of the exercise.

A square can be cut into four parts of equal area in infinitely many ways. Squeeze one part along one edge and stretch it along another by the same amount, and the area is unchanged.

Of all the triangles standing on one base with the far vertex on a parallel line, every single one has the same area. Their perimeters, though, are all different.

Count the squares, not the fence — Area counts what fills the region. Perimeter measures only the boundary round it.

Finding the area of a quadrilateral

The triangle formula turns out to be worth more than it looks. Take a quadrilateral ABCD. Joining the diagonal BD cuts it into two triangles. Find each of their areas and add them, and the quadrilateral is done.

A worked case

Suppose the diagonal AC measures 22 centimetres, and from B and D perpendiculars of 3 centimetres each are dropped onto it. Each triangle has area half of 22 times 3, which is 33. So the quadrilateral has an area of 66 square centimetres.

And any polygon at all

A pentagon takes two diagonals from one vertex and falls into three triangles. Any polygon can be cut up this way. So knowing how to find the area of a triangle is enough to find the area of every straight-sided figure there is.

Special formulas are still worth having

Even so, the parallelogram, the rhombus and the trapezium turn up often enough to deserve formulas of their own. Each comes from the same idea, and the idea is worth naming first.

What dissection means

Taking a figure apart and setting its pieces down as a different figure of equal area goes by the name of dissection. Nothing is added and nothing thrown away, so the area cannot change, however unfamiliar the new shape looks.

A parallelogram, cut

Take a parallelogram ABCD and construct AX perpendicular to the side DC. That perpendicular is called a height of the parallelogram. Cutting along it leaves two pieces: the triangle AXD and the trapezium ABCX.

What completes the trapezium

The trapezium ABCX needs one triangle to make it a rectangle. Extend XC to the right, then drop a perpendicular to it through B. The triangle BYC formed there is exactly the piece needed, and the question is whether it matches the piece cut off.

Proving the pieces match

Compare the triangles BYC and AXD. Since ABYX is a rectangle, BY equals AX. Both the angles at Y and at X are right angles. And BC equals AD, because ABCD is a parallelogram.

The congruence

A right angle, a matching hypotenuse and a matching side is the RHS criterion, so the triangles BYC and AXD are congruent. The triangle cut off therefore fits the gap exactly, and the parallelogram has become the rectangle ABYX.

Reading off the formula

Now the area is easy. The rectangle ABYX measures AX by XY. The length AX is the height of the parallelogram. As for XY, notice that DX equals CY, so adding the common part XC to each gives DC equal to XY.

Base times height

And DC is the base of the parallelogram. So the area of a parallelogram is its base times its height. A base of 12 with height 7 gives 84 square units, and a base of 9 with height 6 gives 54.

Either side as the base

Nothing about the argument preferred one side. Cutting along the height to another side works just as well, so any side may be taken as the base, provided the height used is the one measured to it.

Which keeps the area the same

That gives a useful check. A parallelogram of area 60 standing on a base of 10 has a height of 6. If its other side is 8, then the height to that side must be 60 divided by 8, which is 7.5.

The rhombus

A rhombus is a parallelogram, so base times height holds for it already. But a rhombus has extra properties, and they allow a different dissection that appears in one of the Sulba-Sutras.

What those properties give

All four sides of a rhombus are equal, and its diagonals bisect each other at right angles. So the triangles ABD and CBD are both isosceles, and each can be turned into a rectangle of equal area.

Joining the two rectangles

Those two rectangles can then be set side by side to make one rectangle, call it WXYZ, whose area equals that of the rhombus. Its sides turn out to be the whole of the diagonal AC, with the other diagonal BD halved.

The rhombus formula

So the area is AC times BD over 2, which is half the product of the two diagonals. Diagonals of 20 and 15 give an area of 150 square centimetres, and diagonals of 16 and 12 give 96.

The same result the other way

There is a second route. The diagonal BD splits the rhombus into the triangles ADB and CDB. Their areas are half AO times BD and half CO times BD, and adding those gives half BD times AO plus CO, which is half BD times AC. The same formula again.

Scenario:

A school is laying patterned tiles in a courtyard and three shapes are being costed. The first is a quadrilateral flower bed whose long diagonal measures 18 metres, with perpendiculars of 4 metres and 6 metres onto it from the other two corners. The second is a parallelogram border that stands on a 14 metre base with a height of 5 metres to it, and whose other side measures 10 metres. The third is a rhombus-shaped panel with diagonals of 16 metres and 12 metres. The tiler wants an area for each.

Analysis:

One shape needs splitting into triangles, one needs the base paired with its own height, and one needs the diagonals halved.

Title:

Two formulas on one rhombus

Safety:

No hazards. Keep the diagonals crossing at their midpoints or the figure will not be a rhombus.

What You Notice:

The two answers must agree, because a rhombus is a parallelogram. A small gap means the height was measured by eye rather than perpendicular.

The dissection that turns a rhombus into a rectangle appears in one of the Sulba-Sutras, so the method is a very old one indeed.

Since a rhombus is also a parallelogram, both formulas apply to it. Base times height and half the product of the diagonals must give the same answer, which is a useful check on your working.

Cut, slide, nothing lost — Dissection moves area about without creating or destroying any of it.

Breaking a trapezium up

A trapezium has one pair of parallel sides. Take WXYZ with WX parallel to ZY, the shorter parallel side on top. Drop perpendiculars WM and XN from the top corners onto ZY, and the figure falls into three pieces.

The middle piece is a rectangle

That middle piece WXNM really is a rectangle. The angles at M and N are right angles by construction. And since WX is parallel to ZY, the two interior angles sharing a side of the transversal WM come to 180 degrees, which makes the angle at W a right angle too.

Adding the three areas

So the trapezium is a triangle, plus a rectangle, plus a triangle. In symbols that is half MZ times WM, plus WX times WM, plus half NY times XN. Naming the lengths will tidy it up.

Naming the lengths

Let MZ be x and NY be y, the two bits of the lower side that stick out past the rectangle. Let the common height WM and XN be h, and let the top side WX be a. The three areas become half hx, plus ha, plus half hy.

Collecting the terms

Take out the h. That gives h times the quantity half x plus a plus half y, which tidies into half h times the quantity x plus y plus 2a. It is nearly there, but x and y are not the measurements anyone would be given.

Replacing x and y

Call the lower parallel side b. Looking along it, b is made of x, then the rectangle’s width a, then y. So b equals x plus y plus a, and taking a from both sides gives x plus y equal to b minus a.

The formula

Substitute that in. Half h times the quantity b minus a plus 2a is half h times the quantity a plus b. So the area of a trapezium is half the height times the sum of the parallel sides.

Using it

Parallel sides of 24 metres and 36 metres with a height of 14 metres give half of 14 times 60, which is 420 square metres. Parallel sides of 10 and 14 inches with a height of 6 inches give half of 6 times 24, which is 72 square inches.

A second route: two copies

There is a neater derivation. Take two copies of the trapezium, rotate the second one, and join them along the slanting side BC. What shape results? Either a six-sided figure or a quadrilateral, and it is worth settling which.

Why it must be a quadrilateral

Look at the two angles meeting at each join. Since AB is parallel to CD, those two interior angles sharing a side of transversal BC come to 180 degrees. So the two edges line up straight, and the figure has four sides rather than six.

And a parallelogram at that

The other pair of angles adds to 180 degrees in the same way, which makes the remaining two sides parallel as well. So the two copies have made a parallelogram, with base a plus b and height h.

Halving it

That parallelogram has area h times the quantity a plus b. The trapezium is exactly half of it, so its area is half h times the quantity a plus b. The same formula, reached without splitting anything up.

Areas in real life

Formulas are only half the job. An A4 sheet measures 21 centimetres by 29.7, so its area is 623.7 square centimetres. Picturing how many such sheets would cover a tabletop is a good way to estimate a real area.

Inches, feet, and the squaring trap

Furniture is often measured in inches and feet. One inch is 2.54 centimetres and one foot is 12 inches. So 5 inches is 12.7 centimetres, and running it backwards, 5.08 centimetres is 2 inches. Now the important part. How many square centimetres is a square inch? Not 2.54. A square inch is a square 2.54 centimetres on each side, so its area is 2.54 squared, which is 6.4516 square centimetres.

Converting areas

With that in hand, 10 square inches is 10 times 6.4516, which is 64.516 square centimetres. Going the other way, every 6.4516 square centimetres makes one square inch, so 161.29 square centimetres is 161.29 divided by 6.4516, which is exactly 25 square inches.

Square feet

The same squaring applies to feet. Since a foot is 12 inches, a square foot is a square 12 inches on a side, so it is 144 square inches. Rooms, houses and classrooms are usually given in square feet or square metres.

Bigger and bigger units

Land is often measured in acres, and one acre is 43,560 square feet. Towns and cities go in square kilometres, and one square kilometre is a million square metres, since a kilometre is a thousand metres and a thousand squared is a million.

Local units

Different parts of India use their own units for land, among them the bigha, the gaj, the katha, the dhur, the cent and the ankanam. It is worth finding out which one is used where you live, and what it comes to in square metres.

Scenario:

A school is planning a hall and three measurements are in hand. The floor is a trapezium whose parallel walls measure 18 metres and 26 metres, with 10 metres between them. A notice board is to be ordered from a catalogue that quotes 3 square feet, and the carpenter works in square inches. A wall panel is listed as 129.032 square centimetres and the supplier wants the figure in square inches. The builder also asks which unit the hall floor should be quoted in.

Analysis:

One part needs the trapezium formula and two need a conversion factor squared.

Title:

Estimate in sheets, then in metres

Safety:

Keep the tape off the floor so nobody trips on it.

What You Notice:

The two answers should agree within a sheet or so. A desk sits awkwardly between square centimetres and square metres, which is itself worth noticing.

Two copies of any trapezium always join into a parallelogram, never into a six-sided figure. The angles on one side of the slanting edge add to 180 degrees, which forces the joins to come out straight.

An A4 sheet measures 21 centimetres by 29.7, so its area is 623.7 square centimetres. That makes it a handy unit for guessing the area of a tabletop.

Square the factor, not just the number — An inch is 2.54 centimetres, so a square inch is 2.54 SQUARED, which is 6.4516 square centimetres.