Chapter 7: Area (Mensuration)

In infinitely many ways. Some simple ones:

• four strips
• four small squares
• four triangles made by the two diagonals
• four L-shapes or zig-zag pieces

Starting from any equal division, push one edge of a part inwards and let it bulge outwards by the same amount elsewhere. The areas stay equal, so there is no end to the possibilities.

Infinitely many: e.g. strips, quarters, the four diagonal triangles, or any shapes made by equal "push-in, push-out" changes.

Rectangles:

• 1 cm × 10 cm: perimeter 22 cm, area 10 cm².
• 4 cm × 4 cm: perimeter 16 cm, area 16 cm².

Other shapes: a long thin zig-zag strip has a huge perimeter but a tiny area. A plain circle drawn around it can have a much larger area with a smaller perimeter.

So perimeter cannot be a measure of area.

E.g. 1 × 10 (perimeter 22, area 10) against 4 × 4 (perimeter 16, area 16).

(i)

• 21 in² rectangle: width 7 in, so height $=21÷7=$ 3 in.
• 28 in² rectangle: its height is $4+3=7$ in, so its width $=28÷7=$ 4 in.
• 35 in² rectangle: its width is $3+4=7$ in, so its height $=35÷7=$ 5 in.
• 14 in² strip: its height is $5+2=7$ in, so the missing width $=14÷7=$ 2 in.

(ii)

• Dotted rectangle: $29÷4=$ 7.25 m wide.
• Hatched rectangle: $11÷4=$ 2.75 m wide.

So the top row is 10 m long altogether.

The bold rectangle (the dotted part together with the part below it) has area 50 m². Its height is then $50÷7.25\approx 6.9$ m, so the part below the dotted rectangle is about $6.9-4=$ 2.9 m high (exactly $\frac{84}{29}$ m).

(i) The missing width is 2 in (other sides: 3 in, 4 in, 5 in). (ii) 7.25 m and 2.75 m; lower height ≈ 2.9 m.

(i) Measure the length and breadth of ABCD (L, B) and of EFGH (l, b). Then:

$$\text{Area of path}=LB-lb$$

For example, if ABCD is 30 m × 20 m and EFGH is 24 m × 14 m, the path is $600-336=264$ m².

(ii) The widths alone are not enough: we also need the park's length $l$ and breadth $b$. With widths $p$ (left), $q$ (right), $r$ (bottom) and $s$ (top):

$$\text{Area}=(l+p+q)(b+r+s)-lb$$

You can also break the path into four rectangles. For example, with the park 24 m × 14 m and every width 3 m: $30\times 20-24\times 14=264$ m².

(iii) No. The path is always the area of ABCD minus the area of EFGH, and neither of these changes when the outer rectangle is moved.

Area of path = area of ABCD − area of EFGH; the widths alone need the park's size as well; moving the outer rectangle does not change the area.

We need the widths of the two paths: $a$ (the path along the 14 m side) and $b$ (the path along the 12 m side).

The two strips cover $14a+12b$, but their crossing square $a\times b$ is counted twice:

$$\text{Area}=14a+12b-ab$$

For example, with $a=b=1$ m: $14+12-1=25$ m². With $a=2$, $b=1.5$: $28+18-3=43$ m².

We need the two path widths; area = 14a + 12b − ab (e.g. 25 m² for 1 m wide paths).

Hint: an L-bend with outer arms of 5 and 5 has area $5+5-1=9$, because the corner square is counted in both arms. So it equals a straight tube of length 9.

The spiral's arms, measured along their outer edges, are 20, 20, 20, 15, 15, 10, 10, 5 and 5, a total of 120. There are 8 bends, and each bend's corner square is counted twice:

$$\text{Area}=120-8=112\text{ square units}$$

112 square units, the same as a straight tube 112 units long (and 1 unit wide).

The diagonal cuts the square in half, which gives region 3: $\frac{s^2}{2}$. The line from the corner to the centre halves the other half, so regions 1 and 2 are $\frac{s^2}{4}$ each.

Doubling the side multiplies every area by $2\times 2=4$. So each region increases by 3 times its old area:

• regions 1 and 2 grow by $\frac{3s^2}{4}$ each
• region 3 grows by $\frac{3s^2}{2}$

Every region becomes 4 times as large (increase = 3 × its original area), since both length and width double.

The lines meet at the centre, so the 4 pieces are congruent. Each piece has two right angles: one is a corner of the old square, and the other is at the centre.

Turn each piece so that its centre angle becomes an outer corner of a new square, with the pieces going round in a windmill pattern. The old corners now point inwards, and a square hole is left in the middle.

The pieces' total area is unchanged, so the larger square's area equals the old square plus the hole.

Rotate the four congruent pieces so their centre right angles form the corners of a bigger square; a square hole appears in the middle.

The triangles have equal areas. Each has a base equal to a side of the rectangle and a height equal to the other side, so each is half the rectangle.

Fig. 7.1: the rectangle is 5 × 4, so $\text{Area}(△XDC)=\frac{1}{2}\times 5\times 4=10$ square units.

Equal: each is half the rectangle. Area of XDC = 10 square units.

BCDE: line $l$ through A is parallel to BC. EB and DC are drawn perpendicular to BC, so they are also perpendicular to $l$. All four angles are $90°$.

BXAE: AX ⊥ BC, so $\angle X=90°$. Angles B and E are also $90°$ from the rectangle BCDE, so the fourth angle at A is $90°$ too.

In both cases, all four angles are right angles.

All four angles are 90° in each (perpendiculars to a pair of parallel lines).

(i) All have the same area, $\frac{1}{2}\times BC\times h$, since the height is always the distance between the lines. There is no largest or smallest.

(ii) There is no maximum perimeter: as A slides further along $l$, AB + AC grows without limit. The minimum comes when A lies on the straight line from B to $C^{\prime }$, the reflection of C in $l$.

That point is the midpoint of BC′ in the horizontal direction, so it lies directly above the midpoint of BC. So A is on the perpendicular bisector of BC, and then $AB=AC$: the triangle is isosceles.

(i) All have equal area. (ii) No maximum perimeter; the minimum is the isosceles triangle with A on the perpendicular bisector of BC.

1. $\frac{1}{2}\times 4\times 3=6$ cm²
2. $\frac{1}{2}\times 5\times 3.2=8$ cm²
3. $\frac{1}{2}\times 3\times 4=6$ cm²

(i) 6 cm² (ii) 8 cm² (iii) 6 cm²

$\text{Area}=\frac{1}{2}\times BC\times AX=\frac{1}{2}\times 6\times 4=12$.

Also $\text{Area}=\frac{1}{2}\times AC\times BY=4\times BY$.

So $BY=12÷4=3$.

BY = 3 units

In an isosceles triangle, the altitude from the apex bisects the base. So $UE=EB$, and triangles SUE and SEB have equal bases and the same height. Their areas are equal.

48 sq. units

Take a rectangle ABCD with base AB = $l$ and height $w$.

1. Extend AB to E so that $BE=AB$, making the base $AE=2l$.
2. Join D (or any point on the opposite side) to E.

The triangle on base $2l$ with height $w$ has area $\frac{1}{2}\times 2l\times w=lw$.

As a dissection: cut the rectangle along a diagonal and put the two right triangles side by side along their common leg. They form a triangle with base $2l$ and height $w$.

Double the base: a triangle on base 2l with height w has area lw. (Dissection: cut along a diagonal and rejoin the halves.)

1. Join the midpoints of the two slanting sides. This midline is parallel to the base, at half the height.
2. Cut along the midline. Then cut the small top triangle along its altitude into two right triangles.
3. Place these two pieces into the two lower corners of the trapezium below.

They fill a rectangle with the same base and half the height: area $b\times \frac{h}{2}=\frac{1}{2}bh$.

A rectangle on the same base with half the height; cut at the midline and fit the top pieces into the corners.

Place D at the origin with side $s$. The line DH goes from $(0,0)$ to $(s,2s)$.

• Red is the triangle DCH: $\frac{1}{2}\times s\times 2s=s^2$.
• Blue: DH crosses AB (height $s$) at $x=\frac{s}{2}$. So blue is a right triangle with legs $s$ and $\frac{s}{2}$: area $\frac{s^2}{4}$.

(i) $s^2=49$, so blue $=\frac{49}{4}=$ 12.25 sq. units.

(ii) $s^2+\frac{s^2}{4}=\frac{5s^2}{4}=180$, so $s^2=$ 144 sq. units.

(i) 12.25 sq. units (ii) each square is 144 sq. units (side 12).

Join NY.

• N is the midpoint of XZ, so the median YN halves the triangle: $\text{area}(XNY)=\frac{1}{2}\text{area}(XYZ)$.
• In triangle XNY, NM is a median (M is the midpoint of XY), so $\text{area}(XMN)=\frac{1}{2}\text{area}(XNY)$.

Together: $\text{area}(XMN)=\frac{1}{4}\text{area}(XYZ)$.

One-quarter.

Use the mirror idea:

1. Reflect the tank in the near bank of the river to get $T^{\prime }$.
2. Join the house to $T^{\prime }$ with a straight line. It meets the bank at P.

The shortest route is house → P → tank, because $PT=P{T}^{\prime }$ and the straight line is the shortest.

Reflect the tank in the river bank and join the house to the reflection; where this line meets the bank is where Gopal should fetch water.

Quadrilateral: join the diagonal BD, then measure BD and the perpendicular distances from A and from C to BD:

$$\text{Area}=\frac{1}{2}\times BD\times (h_A+h_C)$$

Pentagon: draw the two diagonals from one vertex. This makes 3 triangles; find their areas and add them.

Every polygon can be cut into triangles by diagonals. An $n$-sided polygon gives $n-2$ triangles.

Split into triangles with diagonals (a quadrilateral into 2, a pentagon into 3) and add their areas.

$$\text{Area}=\frac{1}{2}\times 22\times 3+\frac{1}{2}\times 22\times 3=33+33$$

66 cm²

Subtract the two unshaded triangles from the rectangle:

$$180-\frac{1}{2}\times 10\times 6-\frac{1}{2}\times 8\times 10=180-30-40$$

110 cm²

Just one: the side $s$. A regular hexagon splits into 6 equilateral triangles of side $s$, so

$$\text{Area}=6\times \frac{1}{2}\times s\times h$$

where $h=\sqrt{s^2-(s/2{)}^2}=\frac{\sqrt{3}}{2}s$ is the height of each triangle. This works out to about $2.6s^2$.

Alternatively, measure the side and the distance across between opposite sides ($2h$): the area is $3\times s\times h$.

Only the sidelength (it makes 6 equilateral triangles); or the side and the distance across.

Each blue triangle has a whole side of the rectangle as its base (the top side for one, the bottom side for the other), and both meet at the point P inside. If P is $h_1$ below the top and $h_2$ above the bottom, then $h_1+h_2=H$.

$$\text{Blue}=\frac{1}{2}W{h}_1+\frac{1}{2}W{h}_2=\frac{1}{2}WH$$

One-half.

Join the midpoints of the four sides. The inner quadrilateral (a parallelogram) has exactly half the area.

Why: the diagonal AC splits ABCD into triangles ABC and ADC. The corner triangles at B and D are one-quarter of these (as in Q7 above), so together they are $\frac{1}{4}$ of ABCD. In the same way, using the diagonal BD, the corner triangles at A and C are another $\frac{1}{4}$. The four corners make $\frac{1}{2}$, so the middle part is the other half.

Join the midpoints of its sides: the inner parallelogram has half the area.

Cut off the right triangle CZD (with its right angle at Z). Slide it across and place DC along AB: the slanted sides match, because $AB=DC$ and $AB\parallel DC$.

The right angle at Z then completes a rectangle with sides AD and CZ. So area = (side) × (its height), with any side as the base.

Yes: move triangle CZD to the opposite side to make a rectangle AD × CZ.

(i) Every one has a base of 5 units and a height of 3 units on the grid, so all have the same area, 15 square units.

(ii) The perimeters are different. The more a parallelogram leans, the longer its slanting sides. Measured from the grid:

• (a): the smallest perimeter, about 16 units, since it is almost a rectangle.
• (g): the largest, $2\left(5+\sqrt{3^2+3^2}\right)\approx 18.5$ units, since it leans the most.

(i) All equal (15 square units). (ii) Perimeters differ: (a) is smallest and (g) is largest.

1. $7\times 4=28$ cm²
2. $5\times 3=15$ cm²
3. $5\times 4.8=24$ cm²
4. $2\times 4.4=8.8$ cm²

(i) 28 cm² (ii) 15 cm² (iii) 24 cm² (iv) 8.8 cm²

$\text{Area}=SR\times QM=12\times 6=72$ cm². Also $\text{Area}=PS\times QN=7.6\times QN$.

So $QN=72÷7.6\approx 9.47$ cm.

QN = 72 ÷ 7.6 ≈ 9.47 cm

On the same 5 cm base, the rectangle's height is 4 cm. The parallelogram's slanting 4 cm side gives a height less than 4 cm. So its area is $5\times (\text{less than }4)<20$.

The rectangle (20 cm²) has the greater area.

• Enclosing rectangle: draw a line through the top vertex parallel to the base, and drop perpendiculars from the ends of the base. The rectangle (base × height) is twice the triangle.
• Two copies: put two copies of the triangle together to make a parallelogram. Then cut and move a corner triangle to make it a rectangle.
• Same base, same height in any other position: the area is still $b\times h=2\times \frac{1}{2}bh$.

E.g. the rectangle on the same base with the same height (draw through the apex a line parallel to the base).

Use the same base and half the height:

1. Join the midpoints of the two slanting sides (the midline).
2. Drop perpendiculars from its ends to the base.

The two small triangles above the midline exactly fill the two gaps at the bottom corners.

Rectangle on the same base with half the height (cut at the midline, fit the top pieces into the corners).

1. Cut along the altitude AD. In an isosceles triangle, D is the midpoint of BC, so this gives two congruent right triangles, ADB and ADC.
2. Turn one over and join the two along their hypotenuses (AB with AC).

Together they form a rectangle with sides BD $\left(=\frac{BC}{2}\right)$ and AD.

Cut along the altitude and join the two right triangles along their hypotenuses: a rectangle (½ base) × height.

This is the reverse of Q7:

1. Cut the rectangle along a diagonal into two congruent right triangles.
2. Place them back to back along one of the equal legs (the width).

The result is an isosceles triangle with base $2\times$ length and height = width.

Cut along a diagonal and put the two halves side by side along the width: an isosceles triangle with twice the length as base.

The height of an equilateral triangle of side $s$ is $\sqrt{s^2-{\left(\frac{s}{2}\right)}^2}=\frac{\sqrt{3}}{2}s\approx 0.87s$, which is less than $s$.

• One triangle: $\frac{1}{2}\times s\times 0.87s\approx 0.43s^2$, less than $s^2$.
• Two triangles together make a rhombus with base $s$ and height $0.87s$: area $\approx 0.87s^2$, still less than $s^2$.

The square is larger in both cases (one triangle ≈ 0.43 s², two ≈ 0.87 s²).

The diagonals are perpendicular, so AO and CO are the heights to BD:

$$\frac{1}{2}\times AO\times BD+\frac{1}{2}\times CO\times BD=\frac{1}{2}\times BD\times (AO+CO)=\frac{1}{2}\times BD\times AC$$

Area = ½ × BD × (AO + CO) = ½ × product of the diagonals.

• ABCD: a 4 × 4 square plus a right triangle with legs 2 and 4: $16+4=$ 20.
• PQRS: height 3, middle rectangle 6 × 3, and two end triangles with base 2: $18+3+3=$ 24.
• WXYZ: height 4, middle rectangle 5 × 4, and two end triangles with base 2: $20+4+4=$ 28.

Each one matches $\frac{1}{2}h(a+b)$, e.g. $\frac{1}{2}\times 4\times (4+6)=20$.

20, 24 and 28 square units.

Approach 1:

$$\text{Area }ABED=\text{Area }ABEF-\text{Area }△AFD=ah-\frac{1}{2}(FD)h$$

$$\text{Area }△BEC=\frac{1}{2}(EC)h$$

Since $FD+b=a+EC$, adding gives

$$ah+\frac{1}{2}h(EC-FD)=ah+\frac{1}{2}h(b-a)=\frac{1}{2}h(a+b)$$

Approach 2: draw BG ∥ AD. Then ABGD is a parallelogram with area $ah$, and triangle BGC has base $b-a$ and height $h$:

$$ah+\frac{1}{2}h(b-a)=\frac{1}{2}h(a+b)$$

This works for any trapezium in which $b>a$ (take $b$ as the longer parallel side). G then lies on the longer side.

Yes: both approaches give ½h(a + b); Approach 2 works for every trapezium (taking b as the longer parallel side).

$\frac{1}{2}\times 20\times 15=150$

150 cm²

For a rectangle $l\times w$:

1. Cut it in half by a line through the midpoints of the two long sides, giving two rectangles $l\times \frac{w}{2}$.
2. Cut each of these along a diagonal. This gives 4 right triangles with legs $l$ and $\frac{w}{2}$.
3. Put the four right angles together at one point.

The result is a rhombus with diagonals $2l$ and $w$. Its area is $\frac{1}{2}\times 2l\times w=lw$ ✓

Cut into 4 right triangles (legs l and w/2) and put their right angles together: a rhombus with diagonals 2l and w.

All four are trapeziums:

1. $\frac{1}{2}\times 16\times (10+7)=136$ ft²
2. $\frac{1}{2}\times 14\times (24+36)=420$ m²
3. $\frac{1}{2}\times 10\times (14+6)=100$ in²
4. $\frac{1}{2}\times 8\times (12+18)=120$ ft²

(i) 136 ft² (ii) 420 m² (iii) 100 in² (iv) 120 ft²

1. Drop perpendiculars from the two top corners to the base. This cuts off two congruent right triangles, one at each end (by RHS).
2. Cut one of them off, turn it upside down, and fit it against the other end triangle. The two together make a rectangle, height $h$ by width $\frac{b-a}{2}$.
3. Place that rectangle beside the middle rectangle.

The result is one rectangle $h\times \frac{a+b}{2}$. (Or use the midpoint method of Q5.)

Cut off the two congruent end triangles and rearrange them into a rectangle beside the middle part; the final rectangle is h by (a + b)/2.

1. Mark I and J, the midpoints of the slanting sides AD and BC.
2. Through I and through J, draw lines perpendicular to DC.
3. These lines meet the line AB at H and E, and meet DC at G and F.

Triangle AHI ≅ triangle DGI (ASA: AI = DI, vertically opposite angles at I, right angles at H and G). In the same way, BEJ ≅ CFJ. So each triangle cut off from the trapezium is replaced by an equal triangle, and the areas are equal.

The rectangle's width is the midline IJ $=\frac{a+b}{2}$, which explains $\frac{1}{2}h(a+b)$.

Draw perpendiculars to DC through the midpoints I and J of the slanting sides; they cut AB (extended) and DC at H, E, G, F.

Any trapezium with $\frac{1}{2}h(a+b)=144$ works, i.e. $h(a+b)=288$. For example:

• $h=12$ cm, with parallel sides $a=10$ cm and $b=14$ cm.
• $h=16$ cm, with $a=8$ cm and $b=10$ cm.

To construct the first one:

1. Draw $b=14$ cm.
2. Draw a parallel line 12 cm above it.
3. Mark 10 cm on that line.
4. Join the ends.

Equivalently, start from a 12 cm × 12 cm rectangle (the midline method in reverse).

E.g. parallel sides 10 cm and 14 cm with height 12 cm (½ × 12 × 24 = 144).

The hexagon is made of 6 identical small equilateral triangles meeting at its centre:

• Trapezium: half the hexagon, cut off by a long diagonal = 3 small triangles.
• Equilateral triangle: 1 small triangle.
• Rhombus: 2 small triangles.

Trapezium : triangle : rhombus = 3 : 1 : 2

In triangles ZYA and BXA:

• $YA=XA$ (A is the midpoint)
• $\angle ZAY=\angle BAX$ (vertically opposite)
• $\angle ZYA=\angle BXA$ (alternate angles, since ZY ∥ WB)

So $△ZYA\cong △BXA$ (ASA), and they have equal areas. Therefore

$$\text{Area}(ZYXW)=\text{Area}(ZAXW)+\text{Area}(ZYA)=\text{Area}(ZAXW)+\text{Area}(BXA)=\text{Area}(ZWB)$$

∆ZYA ≅ ∆BXA (ASA), so swapping one for the other turns the trapezium into ∆ZWB without changing the area.

$21\times 29.7=$ 623.7 cm².

A school desk of 120 cm × 60 cm = 7200 cm² holds about $7200÷623.7\approx 11.5$ A4 sheets.

623.7 cm² (a 120 cm × 60 cm desk ≈ 11–12 A4 sheets).

• $5\text{ in}=5\times 2.54=$ 12.7 cm
• $7.4\text{ in}=$ 18.796 cm
• $5.08\text{ cm}=5.08÷2.54=$ 2 in
• $11.43\text{ cm}=$ 4.5 in
• $161.29÷6.4516=$ 25 in² (a 5 in × 5 in square)
• $1{\text{ ft}}^2=12\times 12=$ 144 in²
• $1{\text{ km}}^2=1000\times 1000=$ 10,00,000 m²

12.7 cm, 18.796 cm, 2 in, 4.5 in, 25 in², 144 in² per ft², 10,00,000 m² per km².

These depend on where you live. Some guidance:

• Classroom: for example, 8 m × 6 m = 48 m² (about 517 ft², since 1 m² ≈ 10.76 ft²).
• School campus: often 1 to 5 acres (1 acre = 43,560 ft² ≈ 4047 m²).
• Local units: bigha, gaj, katha, dhur, cent, ankanam and others. Their sizes differ from state to state. For example, 1 cent is $\frac{1}{100}$ acre, about 40.5 m².
• Comparison: a town of 50 km² is about 50,00,000 m². That is about 2500 times a school of 2000 m².
• Smallest in the world: Vatican City is the smallest country and city, only about 0.44 km².
• Largest: the answer depends on how a city's boundary is counted. Look up your state's largest municipal area and compare.

Use real measurements: e.g. a classroom of about 48 m², a town of a few km²; Vatican City (≈ 0.44 km²) is the world's smallest.