You have studied in earlier classes about figures of different shapes such as squares, rectangles, triangles and quadrilaterals.
You have also calculated perimeters and the areas of some of these figures like rectangle, square etc. For instance, you can find the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length `10 m` and width `8 m`, its perimeter would be `2(10 m + 8 m) = 36 m` and its area would be `10 m × 8 m`, i.e., `80 m^2`.
Unit of measurement for length or breadth is taken as metre (m) or centimetre (cm) etc.
Unit of measurement for area of any plane figure is taken as square metre `(m^2)` or square centimetre `(cm^2)` etc.
Suppose that you are sitting in a triangular garden. How would you find its area? From Chapter 9 and from your earlier classes, you know that:
`text(Area of a triangle ) =1/2 xx text(base) xx text (height) `
We see that when the triangle is `text ( right angled )`, we can directly apply the formula by using two sides containing the right angle as base and height.
For example, suppose that the sides of a right triangle `ABC` are `5 cm, 12 cm `and `13 cm`; we take base as `12 cm` and height as `5 cm` (see Fig. 12.1). Then the
area of `Delta ABC` is given by
`1/2 xx text (base) xx text(height) = 1/2 xx 12 xx 5 cm^2` , i.e., `30 cm^2`
Note that we could also take `5 cm` as the base and` 12 cm` as height.
Now suppose we want to find the area of an equilateral triangle `PQR` with side `10cm` (see Fig. 12.2). To find its area we need its height. Can you find the height of this triangle?
Let us recall how we find its height when we know its sides. This is possible in an equilateral triangle. Take the mid-point of `QR` as `M` and join it to `P`.
We know that `PMQ` is a right triangle. Therefore, by using Pythagoras Theorem, we can find the length `PM` as shown below:
`PQ^2 = PM^2 + QM^2`
i.e., `(10)^2 = PM^2 + (5)^2`, since `QM = MR`.
Therefore, we have `PM^2 = 75`
i.e., `PM = sqrt (75) cm = 5 sqrt (3) cm` .
Then area of `Delta PQR = 1/2 xx text(base) xx text (height) = 1/2 xx 10 xx 5 sqrt 3 cm^2 = 25 sqrt 3 cm^2` .
Let us see now whether we can calculate the area of an`text ( isosceles triangle )` also with the help of this formula. For example, we take a triangle `XYZ` with two equal sides `XY` and `XZ` as `5 cm` each and unequal side `YZ` as `8 cm` (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from `X` we draw a perpendicular `XP` to side `YZ`. You can see that this perpendicular `XP` divides the base `YZ` of the triangle in two equal parts.
Therefore, `YP = PZ = 1/2 YZ =4 cm`
Then, by using Pythagoras theorem, we get
`XP^2 = XY^2 – YP^2`
`= 5^2 – 4^2 = 25 – 16 = 9`
So, `XP = 3 cm`
Now, area of `Delta XYZ = 1/2 xx text (base) YZ xx text (height) XP`
`= 1/2 xx 8 xx 3 cm^2 = 12 cm^2`
Now suppose that we know the lengths of the sides of a scalene triangle and not the height. Can you still find its area? For instance, you have a triangular park whose sides are `40 m, 32 m`, and `24 m`.
How will you calculate its area? Definitely if you want to apply the formula, you will have to calculate its height. But we do not have a clue to calculate the height. Try doing so. If you are not able to get it, then go to the next section.