Part II, Chapter 7Area

Take a square of side \(s\), so its area is \(s^2\).

Each of the 4 parts must have area \(\dfrac{s^2}{4}\).

Four small squares: join the midpoints of opposite sides.

Each part is \(\dfrac s2\times\dfrac s2=\dfrac{s^2}{4}\).

Four strips: divide one side into 4 equal parts.

Each part is \(s\times\dfrac s4=\dfrac{s^2}{4}\).

Four triangles: draw both diagonals.

The diagonals bisect each other, so the 4 triangles have equal bases and the same height.

Each has area \(\dfrac{s^2}{4}\).

Tilted cross: draw two perpendicular lines through the centre.

A turn of 90° about the centre takes each part onto the next, so the 4 parts have equal area.

Region 1: a rectangle of length 9 and width 1.

Perimeter \(=2(9+1)=20\)

Area \(=9\times1=9\)

Region 2: a rectangle of length 4 and width 4.

Perimeter \(=2(4+4)=16\)

Area \(=4\times4=16\)

Region 1: a comb-shaped region made of 11 unit squares.

Area \(=11\) square units.

Perimeter \(=24\) units (counting the edges of the unit squares that lie on the boundary).

Region 2: a square of side 4.

Area \(=16\) square units.

Perimeter \(=16\) units.

We need the width of each path.

Let the horizontal path have width \(w_1\) and the vertical path have width \(w_2\).

Area of the horizontal path \(=14\times w_1\)

Area of the vertical path \(=12\times w_2\)

The small square where the two paths cross is counted twice, so subtract its area \(w_1w_2\).

Area of the crosspath \(=14w_1+12w_2-w_1w_2\)

Take \(w_1=2\) m and \(w_2=1\) m.

Area \(=14\times2+12\times1-2\times1\)

\(=28+12-2=38\) m²

Check: remaining plot is \((14-1)\times(12-2)=13\times10=130\) m².

\(14\times12-130=168-130=38\) m² ✓

Draw the two perpendicular lines through the centre of the square.

Cut along them to get 4 congruent pieces.

Each piece has a right angle at the centre of the square (call it O) and a right angle at one corner of the square.

Place the 4 pieces so that the right angles from O go to the 4 corners of a larger square.

The cut edges then form the outer sides of the larger square.

Each side is made of two cut edges, so it equals the length of one cut line.

The corners of the original square meet in the middle and enclose a square hole.

Check with areas

If the cut line crosses the square at a horizontal shift \(d\) over a height \(s\), its length is \(c=\sqrt{s^2+d^2}\).

The hole has side \(d\).

Area of the larger square \(-\) Area of the hole \(=c^2-d^2=s^2\), the area of the original square ✓

The rectangle ABCD has sidelengths 5 and 4, and X lies on AB.

The triangle XDC has base \(DC=5\) and height equal to the height of the rectangle, 4.

Area of ∆XDC \(=\dfrac12\times\text{base}\times\text{height}\)

\(=\dfrac12\times5\times4=10\) square units

Check: the rectangle has area \(5\times4=20\), and ∆XDC is exactly half of it ✓

Take BC as the base of the triangle ABC.

Through A, draw the line \(l\parallel BC\).

Through B and C, draw perpendiculars to \(l\).

They meet \(l\) at E and D.

The rectangle BCDE is the outer rectangle, and the triangle ABC sits inside it.

BE and CD are perpendicular to \(l\), by construction.

So \(\angle BED=90^\circ\) and \(\angle CDE=90^\circ\).

Since \(l\parallel BC\) and \(BE\perp l\), we get \(BE\perp BC\).

So \(\angle EBC=90^\circ\).

In the same way, \(CD\perp BC\), so \(\angle BCD=90^\circ\).

All four angles of BCDE are \(90^\circ\).

\(\angle EBX=90^\circ\), because it is an angle of the rectangle BCDE.

\(\angle BEA=90^\circ\), because it is an angle of the rectangle BCDE and A lies on ED.

\(\angle AXB=90^\circ\), because AX is the altitude of the triangle.

Since \(AX\perp BC\) and \(l\parallel BC\), we get \(AX\perp l\), so \(\angle EAX=90^\circ\).

All four angles of BXAE are \(90^\circ\), so BXAE is a rectangle.

So \(BE=AX\), the height of the triangle, and this is also the height of the rectangle BCDE.

In the same way, XCDA is a rectangle.

∆ABX is half of the rectangle BXAE, and ∆AXC is half of the rectangle XCDA.

So Area (∆ABC) \(=\dfrac12\times\) Area (BCDE) \(=\dfrac12\times\text{base}\times\text{height}\).

(i)

All these triangles have the same base BC.

Their third vertex lies on \(l\), and \(l\parallel BC\), so every triangle has the same height, the distance between \(l\) and BC.

Area \(=\dfrac12\times BC\times\text{height}\) is the same for all of them.

(ii)

BC is common, so we compare AB + AC.

Minimum: the triangle with A on the perpendicular bisector of BC has the least perimeter.

This was shown above using the mirror image.

Maximum: as A moves farther along \(l\), AB + AC keeps increasing and has no upper limit.

Let X and Y be the feet of the perpendiculars from B and C to \(l\).

Since \(l\parallel BC\), the distances BX and CY are equal.

Call each one \(d\).

C′ is the mirror image of C, so \(YC^{\prime}=CY=d\).

Compare ∆ABX and ∆AC′Y

\(\angle AXB=\angle AYC^{\prime}=90^\circ\)

\(BX=YC^{\prime}=d\)

\(\angle BAX=\angle C^{\prime}AY\) (vertically opposite angles, since B, A, C′ lie on one line and X, Y lie on \(l\) on either side of A)

So ∆ABX ≅ ∆AC′Y (AAS).

Hence \(AX=AY\) and \(AB=AC^{\prime}\).

Also \(AC=AC^{\prime}\), because C′ is the mirror image of C.

So \(AB=AC\).

A point that is equally far from B and C lies on the perpendicular bisector of BC.

Let the side of each square be \(s\).

Red region

The red region is ∆DCH.

Its base is \(DC=s\).

C, B and H lie on one line, so \(CH=CB+BH=s+s=2s\).

\(\angle DCH=90^\circ\), so CH is the height.

Area of the red region \(=\dfrac12\times s\times2s=s^2\)

Blue region

The blue region is ∆ADP, where P is the point where DH meets AB.

B is the midpoint of CH, and \(BP\parallel CD\).

So P is the midpoint of DH, and \(BP=\dfrac12\,CD=\dfrac s2\).

\(AP=AB-BP=s-\dfrac s2=\dfrac s2\)

∆ADP has a right angle at A, with \(AD=s\) and \(AP=\dfrac s2\).

Area of the blue region \(=\dfrac12\times s\times\dfrac s2=\dfrac{s^2}{4}\)

So the blue region is \(\dfrac14\) of the red region.

(i)

\(s^2=49\) (the red region)

Blue region \(=\dfrac{49}{4}=12.25\) sq. units

(ii)

Blue + red \(=s^2+\dfrac{s^2}{4}=\dfrac{5s^2}{4}=180\)

\(s^2=180\times\dfrac45=144\)

Area of each square \(=s^2=144\) sq. units (so \(s=12\) units).

Join NY.

N is the midpoint of XZ.

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

So Area (∆XYN) \(=\dfrac12\times\) Area (∆XYZ).

Now consider ∆XYN.

M is the midpoint of XY.

So NM divides ∆XYN into two triangles of equal area.

Area (∆XMN) \(=\dfrac12\times\) Area (∆XYN)

\(=\dfrac12\times\dfrac12\times\) Area (∆XYZ)

\(=\dfrac14\times\) Area (∆XYZ)

The house and the water tank are on the same side of the river.

Use the mirror idea from the previous section.

Treat the river as a mirror.

Mark T′, the mirror image of the water tank T.

T′ is on the other side of the river, at the same distance from it as T.

For any point P on the river, \(PT=PT^{\prime}\), so the path House → P → Tank has the same length as House → P → T′.

The shortest path from the house to T′ is the straight line.

Join the house to T′.

This line meets the river at P.

Gopal should walk from the house to P, and then from P to the water tank.

Divide the pentagon into triangles by drawing diagonals from one vertex.

From A, draw the diagonals AC and AD.

This gives 3 triangles: ABC, ACD and ADE.

Measure the base and the height of each triangle and find its area.

Area of the pentagon \(=\) Area (∆ABC) \(+\) Area (∆ACD) \(+\) Area (∆ADE)

Join the centre of the regular hexagon to its six vertices.

This gives 6 triangles of the same size and shape.

Each triangle has a base equal to a side of the hexagon, say \(s\).

Its height is the distance from the centre to that side, say \(a\) (the apothem).

Area of one triangle \(=\dfrac12\times s\times a\)

Area of the hexagon \(=6\times\dfrac12\times s\times a=3sa\)

So we need two measurements: the side \(s\) and the distance \(a\) from the centre to a side.

Example

Take \(s=4\) cm.

Measure \(a\approx3.46\) cm.

Area \(=3\times4\times3.46\approx41.5\) cm²

Let the rectangle have length \(l\) and breadth \(b\).

Let P be the point where the two blue triangles meet.

The top blue triangle has base \(l\) (the top side) and height \(h_1\), the distance of P from the top side.

The bottom blue triangle has base \(l\) (the bottom side) and height \(h_2\), the distance of P from the bottom side.

The heights add up to the breadth of the rectangle: \(h_1+h_2=b\).

Blue area \(=\dfrac12\,l\,h_1+\dfrac12\,l\,h_2\)

\(=\dfrac12\,l\,(h_1+h_2)\)

\(=\dfrac12\,l\,b\)

Area of the rectangle \(=l\,b\).

Let ABCD be the given quadrilateral, and draw the diagonal BD.

Mark M, the midpoint of BD.

Join AM and CM.

Consider the quadrilateral ABCM.

M is the midpoint of BD, so the line from vertex A to M divides ∆ABD into two triangles of equal area.

Area (∆ABM) \(=\dfrac12\times\) Area (∆ABD)

In the same way, Area (∆CBM) \(=\dfrac12\times\) Area (∆CBD).

Area (ABCM) \(=\) Area (∆ABM) \(+\) Area (∆CBM)

\(=\dfrac12\big[\text{Area}(\triangle ABD)+\text{Area}(\triangle CBD)\big]\)

\(=\dfrac12\times\) Area (ABCD)

The diagonals of a rhombus are perpendicular and bisect each other at O.

So AO is the height of ∆ADB on the base BD, and CO is the height of ∆CDB on the base BD.

Area (∆ADB) \(=\dfrac12\times AO\times BD\)

Area (∆CDB) \(=\dfrac12\times CO\times BD\)

Area of the rhombus \(=\) Area (∆ADB) \(+\) Area (∆CDB)

\(=\dfrac12\times AO\times BD+\dfrac12\times CO\times BD\)

\(=\dfrac12\times BD\times(AO+CO)\)

Since O lies on AC, \(AO+CO=AC\).

\(=\dfrac12\times BD\times AC\)

Take each side of a grid square as 1 unit.

Break each trapezium into a rectangle and triangles, using the dashed lines.

ABCD

Rectangle ABMD: \(4\times4=16\)

Triangle BMC: \(\dfrac12\times2\times4=4\)

Area \(=16+4=20\) square units

PQRS

Rectangle PQUT: \(6\times3=18\)

Triangle SPT: \(\dfrac12\times2\times3=3\)

Triangle QUR: \(\dfrac12\times2\times3=3\)

Area \(=18+3+3=24\) square units

WXYZ

Rectangle WXNM: \(5\times4=20\)

Triangle ZMW: \(\dfrac12\times2\times4=4\)

Triangle XNY: \(\dfrac12\times2\times4=4\)

Area \(=20+4+4=28\) square units

Let \(AB=a\), \(DC=b\) and let the height be \(h\).

Approach 1: rectangle and triangles

ABEF is a rectangle with sides \(a\) and \(h\).

Area \(=ah\).

Let \(FD=p\) and \(EC=q\).

Then \(DC=DE+EC=(FE-FD)+EC=(a-p)+q\), so \(b=a-p+q\) and \(q-p=b-a\).

Area ∆AFD \(=\dfrac12\,hp\) and Area ∆BEC \(=\dfrac12\,hq\).

Area ABED \(=\) Area ABEF \(-\) Area ∆AFD \(=ah-\dfrac12\,hp\)

Area ABCD \(=\) Area ABED \(+\) Area ∆BEC

\(=ah-\dfrac12\,hp+\dfrac12\,hq\)

\(=ah+\dfrac12\,h(q-p)\)

\(=ah+\dfrac12\,h(b-a)\)

\(=\dfrac12\,h(2a+b-a)\)

\(=\dfrac12\,h(a+b)\)

Approach 2: parallelogram and triangle

Draw \(BG\parallel AD\), with G on DC.

\(AB\parallel DG\) and \(AD\parallel BG\), so ABGD is a parallelogram.

\(DG=AB=a\), so \(GC=DC-DG=b-a\).

Area of the parallelogram ABGD \(=a\times h=ah\)

Area ∆BGC \(=\dfrac12\times(b-a)\times h\)

Area ABCD \(=ah+\dfrac12\,h(b-a)\)

\(=\dfrac12\,h(2a+b-a)\)

\(=\dfrac12\,h(a+b)\)

Let AB and DC be the parallel sides, with DC the longer one.

Draw \(BG\parallel AD\), with G on DC.

Since \(AB\parallel DG\) and \(AD\parallel BG\), ABGD is a parallelogram.

So \(DG=AB=a\lt b\), which means that G always falls between D and C.

The triangle BGC has base \(GC=b-a\) and the same height \(h\).

This does not depend on how the legs lean, so the argument is the same for a right trapezium, an isosceles trapezium or a leaning one.

Finding the vertices

Mark I, the midpoint of AD, and J, the midpoint of BC.

Through I, draw a line perpendicular to AB and DC.

It meets AB at H and DC at G.

Through J, draw a line perpendicular to AB and DC.

It meets AB at E and DC at F.

Then H, E, F and G are the vertices of the rectangle.

Why the areas are equal

In ∆AHI and ∆DGI: \(AI=DI\) (I is the midpoint of AD).

\(\angle AHI=\angle DGI=90^\circ\)

\(\angle AIH=\angle DIG\) (vertically opposite angles)

So ∆AHI ≅ ∆DGI (AAS).

In the same way, ∆BEJ ≅ ∆CFJ.

Area HEFG \(=\) Area ABCD \(+\) ∆AHI \(+\) ∆BEJ \(-\) ∆DGI \(-\) ∆CFJ

Since the triangles cancel in pairs, Area HEFG \(=\) Area ABCD.

Step 1: start with a rectangle of area 144 cm²

Take a rectangle EFGH with \(HE=16\) cm and \(EF=9\) cm.

Area \(=16\times9=144\) cm²

Step 2: reverse the conversion

Mark the midpoints I of HG and J of EF.

Through I, draw any slanting line.

It meets the line HE at A and the line GF at D, and I is automatically the midpoint of AD.

In the same way, through J draw a slanting line meeting HE at B and GF at C.

J is the midpoint of BC.

ABCD is a trapezium with \(AB\parallel DC\).

By the previous question, Area ABCD \(=\) Area EFGH \(=144\) cm².

Example

Take the parallel sides to be 12 cm and 20 cm, with a distance of 9 cm between them.

Area \(=\dfrac12\times9\times(12+20)=\dfrac12\times9\times32=144\) cm² ✓

Area \(=\text{length}\times\text{width}\)

\(=29.7\times21\)

\(=623.7\) cm²

The answer depends on your table.

Here is an example.

Take a tabletop 120 cm long and 60 cm wide.

Area \(=120\times60=7200\) cm²

Area of one A4 sheet \(=623.7\) cm²

Number of sheets by area \(=\dfrac{7200}{623.7}\approx11.5\)

So at most 11 sheets can fit.

Lay the sheets with the 21 cm side along the length: \(120\div21\approx5.7\), so 5 sheets fit along the length.

Across the width, the 29.7 cm side gives \(60\div29.7\approx2.02\), so 2 sheets fit.

Total \(=5\times2=10\) sheets.

\(1\) in \(=2.54\) cm

(i)

\(5\) in \(=5\times2.54=12.7\) cm

(ii)

\(7.4\) in \(=7.4\times2.54=18.796\) cm

\(1\) in \(=2.54\) cm, so \(1\) cm \(=\dfrac1{2.54}\) in.

(i)

\(5.08\) cm \(=\dfrac{5.08}{2.54}=2\) in

(ii)

\(11.43\) cm \(=\dfrac{11.43}{2.54}=4.5\) in

Every \(6.4516\) cm² makes 1 in².

\(161.29\) cm² \(=\dfrac{161.29}{6.4516}\) in²

\(6.4516\times25=161.29\)

So \(\dfrac{161.29}{6.4516}=25\).

\(1\) ft \(=12\) in

\(1\) ft² \(=12\text{ in}\times12\text{ in}\)

\(=144\) in²

The answer depends on your classroom.

Measure its length and breadth, and multiply.

Example

Take a classroom 8 m long and 6 m wide.

Area \(=8\times6=48\) m²

\(1\) m \(=\dfrac{100}{2.54}\approx39.37\) in \(\approx3.28\) ft

\(1\) m² \(\approx3.28\times3.28\approx10.76\) ft²

Area \(\approx48\times10.76\approx516\) ft²

This question is open-ended, because the answer depends on your school.

Method

Break the school grounds into rectangles and triangles, as you did for polygons.

Pace out or measure the lengths.

Find the area of each part and add.

Compare your estimate with the area given in the school records.

This question is open-ended, because the answer depends on your region.

Ask an elder or check the land records of your area for the local unit.

Then find how many square feet or square metres make one such unit.

Acre in other units

\(1\) acre \(=43{,}560\) ft²

\(43{,}560\) ft² \(\approx43{,}560\times0.0929\approx4047\) m²

This question is open-ended, because the answer depends on your place.

Estimate the length and breadth of the area in km, and multiply.

Compare your estimate with the area given in official data or an atlas.

\(1\) km \(=1000\) m

\(1\) km² \(=1000\text{ m}\times1000\text{ m}\)

\(=1{,}000{,}000\) m²

Divide the area of your village/town/city by the area of your school.

Convert both areas to the same unit first.

Use \(1\) km² \(=1{,}000{,}000\) m².

Example

City area \(=60\) km² \(=60{,}000{,}000\) m²

School area \(=20{,}000\) m²

\(\dfrac{60{,}000{,}000}{20{,}000}=3000\)

This is a research question.

The answer depends on how a city is defined (city limits or the whole urban region) and on the year of the data.

Look up the list of cities by area in an atlas or an almanac, or an official source such as the Census of India.

Note the source and the year with your answer.