Area

You have met triangles named in two ways.

• By their sides: an equilateral triangle has all three sides equal, an isosceles triangle two sides equal, and a scalene triangle no two sides equal.
• By their angles: acute-angled (every angle less than 90∘°), right-angled (one angle 90∘°) and obtuse-angled (one angle more than 90∘°).

And you know the area of any triangle:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

where the height is the perpendicular from the opposite corner to the base.

On paper, finding a height is easy: draw a perpendicular with a set square and measure it. On land it is not. A field is too big for a set square, the foot of the height may fall in the middle of a crop, and on a slope it is hard to keep a tape level and straight. The three sides, though, can always be measured: walk along each with a tape. So surveyors want the area from the three sides alone.

Finding the height from the sides

Let ABC be a triangle with BC = a, CA = b and AB = c. Draw the height $AD \perp BC$, and call AD = h and DC = x; then BD = a − x.

The perimeter is a + b + c, and half of it is the semi-perimeter:

$$s = \frac{a + b + c}{2}$$

The height makes two right-angled triangles, and Pythagoras’ theorem works in each.

Three lengths make a triangle only when each is shorter than the other two put together; then s − a, s − b and s − c are all positive. If one side equals the other two together, the triangle is flat and Heron’s formula gives 0.

Question 1

Answer these questions.

Question 2

Find the area of each triangle.

Question 3

Find the area of each quadrilateral.

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A room is a box: a cuboid. Painters, tilers and carpenters price their work by the area of its flat surfaces, so this is where areas earn money.

The formulas

The four walls are lh + bh + lh + bh = 2lh + 2bh, so

$$\text{Area of four walls} = 2h(l + b)$$

Notice that 2(l + b) is the perimeter of the floor. So the walls’ area is the floor’s perimeter times the height, whatever the room’s length and breadth.

$$\text{Floor} = \text{ceiling} = lb, \qquad \text{floor and ceiling} = 2lb$$
$$\text{Four walls, floor and ceiling} = 2h(l + b) + 2lb = 2(lb + bh + lh)$$

For a cube with edge l (since l = b = h):

$$\text{Floor} = \text{ceiling} = l^2, \quad \text{four walls} = 4l^2, \quad \text{all six} = 6l^2$$

and with windows of area a₁ and doors of area a₂, all six without them $= 6l^2 - a_1 - a_2$.

Question 1

Answer these questions.

Question 2

Name the surfaces that are the floor, the four walls and the ceiling in each model.

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Question 4

A room is 8.5 m long, 6 m wide and 2.4 m high. Find:

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Sonam’s school has built a hall 30 m long, 25 m wide and 4 m high (inside). It has 5 windows of 2.5 m × 1.5 m and 2 doors of 5 m × 3 m.

Question 9

A classroom is 8 m long, 6 m wide and 3 m high, with 2 doors of 2 m × 1.5 m and 2 windows of 1.5 m × 1 m.

Question 10

Sijan’s father has rented a dark room for photo shoots. It is a cube with sides 2.7 m, and has one door of 2 m × 1 m.

A painter, a tiler or a carpet shop quotes a rate: so many rupees for each square meter (or square foot). The cost is then

$$\text{Total cost } (T) = \text{area } (A) \times \text{rate } (R)$$

and if the cost and rate are known, the area is $A = \frac{T}{R}$.

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An assembly hall is 32 m long, 32 m wide and 4 m high. It has 6 windows of 2.5 m × 1.8 m and 2 doors of 4 m × 3 m.

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A park laid out by a municipality is 50 m × 40 m. It is to be covered with square pieces of dubo turf (dubo grass cut with its soil), each 20 cm by 20 cm.

When land is bought, sold, divided between brothers and sisters, or taxed, its area must be measured and recorded. In Nepal this is the work of the Survey Department, set up in Jestha 2014 BS (1957 AD). Its cadastral survey (kitta napi) measures every plot and gives it a number on a map sheet; it also runs the geodetic survey, which fixes exact points across the country, and the topographic survey, which maps the land’s shape for the country’s maps.

Here are three people and their land. Discuss how land is measured where you live.

1. Durga Kumari lives in a hill district. She has 3 ropani of farmland, 1 ropani of dry hillside field (pakho bari), a house on 8 aana 3 paisa of land, and a triangular plot of 4 aana beside the main road.
2. Ravilal lives in the Terai. He has 3 bigha of farmland, 1 bigha 3 kattha of bari, a house on 1 kattha 2 dhur, and a four-sided plot of 4 kattha beside the main road.
3. Bhim Bahadur is an amin, a land surveyor. He follows the Survey Department’s rules: a four-sided plot is cut into two triangles by a diagonal, each triangle’s area is found from its sides, and the two are added.

Units of land in Nepal

Square feet and square meters are used everywhere in the world. But Nepal also keeps its own units:

• in the hills (and the mountains and Kathmandu Valley): ropani, aana, paisa, daam;
• in the Terai: bigha, kattha, dhur.

Changing between units

To change an area into hill units, divide by one ropani (5,476 ft² or 508.72 m²); the whole number is the ropani. Multiply what is left by 16: the whole number is the aana. Multiply what is left by 4: the paisa. Multiply what is left by 4: the daam, kept with its decimal. The answer is written R-A-P-D, for example 0-4-2-0.93.

For the Terai, divide by one bigha (72,900 ft² or 6,772.63 m²), then multiply what is left by 20 for kattha, and by 20 again for dhur. The answer is written B-K-D.

To go from hill units to Terai units or back, go through square feet: those values are exact.

A land ownership certificate, the Lalpurja, gives each plot’s number, its map sheet, and its area in the local units and in square meters. In the Terai the area has four places, B-K-D and a fourth for the kanwa, a still smaller unit; in the examples here it is 0.

Area of triangular and quadrilateral land

Blueprint maps of land are drawn to a scale, such as 1 : 500 (1 cm on the map is 500 cm, or 5 m, on the ground). But for the area, an amin uses the sides measured on the ground.

Question 1

Answer from the conversion table.

Question 2

Change into square feet.

Question 3

Change into square meters.

Question 4

Change into aana (through square feet).

Question 5

Change into dhur.

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Question 7

Write each area a. in R-A-P-D, b. in B-K-D.

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Question 9

A Lalpurja gives a plot’s area only in square meters: 677.26 m². Write it:

Question 10

A piece of land has been cut into plots, each marked on its map with its area in R-A-P-D. Some of them are:

Plot no.	1	2	4	6	8	9	18	20	22
R-A-P-D	0-3-2-3	0-4-0-2	0-4-2-0	0-4-0-0	0-3-1-0	0-5-1-3	0-3-2-2	0-4-0-0	0-7-0-0

Question 11

Find the area of each plot a. in R-A-P-D, b. in B-K-D.

Question 12

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Mensuration is 13 marks in three questions; the grid’s model area question for Grade 9 is 5 marks on one room:

Part	Level	Marks	What it asks
a.	knowledge	1	the formula for the four walls
b.	understanding	1	the height (or a side) from the walls’ area
c.	application	2	the cost of painting without the door and windows
d.	higher ability	1	painting against wallpaper: which costs less, by how much

Here is one in that shape, with how it is marked.