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### Area of a Triangle

♦ Area of a Triangle

In your earlier classes, you have studied how to calculate the area of a triangle when

its base and corresponding height (altitude) are given. You have used the formula :

`text (Area of a triangle) = 1/2 xx text (base) xx text (altitude )`

Now, if the coordinates of the vertices of a triangle are given. Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula.

But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.

Let `ABC` be any triangle whose vertices are `A(x_1, y_1), B(x_2, y_2)` and `C(x_3, y_3)`. Draw `AP, BQ` and `CR` perpendiculars from `A, B` and `C`, respectively, to the `x`-axis. Clearly `ABQP`, `APRC` and `BQRC `are all trapezia (see Fig. 7.13).

Now, from Fig. 7.13, it is clear that area of

`Delta ABC = text ( area of trapezium ) ABQP + text (area of trapezium) APRC – text (area of trapezium) BQRC`.

You also know that the

`text (area of a trapezium ) = 1/2xx "sum of parallel sides" xx "distance between them"`

Therefore, Area of

`Delta ABC =1/2 (BQ +AP) QP +1/2 (AP +CR) PR -1/2 (BQ +CR) QR`

` =1/2 (y_2 +y_1) (x_1 - x_2) +1/2 (y_1 +y_3) (x_3 -x_1) -1/2 (y_2 +y_3) (x_3 -x_2 )`

` =1/2 [x_1 (y_2 -y_3) +x_2 (y_3 -y_1) + x_3 ( y_1 -y_2) ]`

Thus, the area of `Delta ABC` is the numerical value of the expression

`color{red}{= 1/2 [ x_1 (y_2 - y_3) +x_2 (y_3 -y_1) + x_3 (y_1 - y_2)]}`

its base and corresponding height (altitude) are given. You have used the formula :

`text (Area of a triangle) = 1/2 xx text (base) xx text (altitude )`

Now, if the coordinates of the vertices of a triangle are given. Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula.

But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.

Let `ABC` be any triangle whose vertices are `A(x_1, y_1), B(x_2, y_2)` and `C(x_3, y_3)`. Draw `AP, BQ` and `CR` perpendiculars from `A, B` and `C`, respectively, to the `x`-axis. Clearly `ABQP`, `APRC` and `BQRC `are all trapezia (see Fig. 7.13).

Now, from Fig. 7.13, it is clear that area of

`Delta ABC = text ( area of trapezium ) ABQP + text (area of trapezium) APRC – text (area of trapezium) BQRC`.

You also know that the

`text (area of a trapezium ) = 1/2xx "sum of parallel sides" xx "distance between them"`

Therefore, Area of

`Delta ABC =1/2 (BQ +AP) QP +1/2 (AP +CR) PR -1/2 (BQ +CR) QR`

` =1/2 (y_2 +y_1) (x_1 - x_2) +1/2 (y_1 +y_3) (x_3 -x_1) -1/2 (y_2 +y_3) (x_3 -x_2 )`

` =1/2 [x_1 (y_2 -y_3) +x_2 (y_3 -y_1) + x_3 ( y_1 -y_2) ]`

Thus, the area of `Delta ABC` is the numerical value of the expression

`color{red}{= 1/2 [ x_1 (y_2 - y_3) +x_2 (y_3 -y_1) + x_3 (y_1 - y_2)]}`

Q 3129780611

Find the area of a triangle whose vertices are` (1, –1), (– 4, 6)` and

`(–3, –5)`.

Class 10 Chapter 7 Example 11

`(–3, –5)`.

Class 10 Chapter 7 Example 11

The area of the triangle formed by the vertices` A(1, –1), B(– 4, 6)` and

`C (–3, –5)`, by using the formula above, is given by

`1/2 [ 1 (6+5) + (-4) (-5 +1 ) + (-3) (-1-6) ] `

`= 1/2 (11 +16 +21 ) = 24`

So, the area of the triangle is 24 square units.

Q 3139780612

Find the area of a triangle formed by the points `A(5, 2), B(4, 7)` and `C (7, – 4)`.

Class 10 Chapter 7 Example 12

Class 10 Chapter 7 Example 12

The area of the triangle formed by the vertices` A(5, 2), B(4, 7) `and

`C (7, – 4) `is given by

`1/2 [ 5 (7+4) +4 (-4-2) + 7 (2-7) ]`

` = 1/2 (55-24 -35) = (-4)/2 = -2`

Since area is a measure, which cannot be negative, we will take the numerical value

of `– 2`, i.e., `2`. Therefore, the area of the triangle `= 2` square units.

Q 3159780614

Find the area of the triangle formed by the points` P(–1.5, 3), Q(6, –2)`

and `R(–3, 4)`.

Class 10 Chapter 7 Example 13

and `R(–3, 4)`.

Class 10 Chapter 7 Example 13

The area of the triangle formed by the given points is equal to

`1/2 [-1.5 (-2-4) + 6 (4-3) +(-3) (3+2) ]`

`=1/2 (9 + 6 -15 ) = 0`

Can we have a triangle of area 0 square units? What does this mean?

If the area of a triangle is 0 square units, then its vertices will be collinear.

Q 3169780615

Find the value of `k` if the points `A(2, 3), B(4, k)` and `C(6, –3) `are

collinear.

Class 10 Chapter 7 Example 14

collinear.

Class 10 Chapter 7 Example 14

Since the given points are collinear, the area of the triangle formed by them

must be 0, i.e.,

`1/2 [2 ( k+3) +4 (-3-3 ) + 6 (3- k) ] = 0`

i.e., `1/2 (-4k) = 0`

Therefore , ` k = 0`

Let us verify our answer.

area of `Delta ABC = 1/2 [2 (0+3 ) +4 ( -3-3 ) + 6 (3 - 0 ) ] = 0`

Q 3179780616

If `A(–5, 7), B(– 4, –5), C(–1, –6)` and` D(4, 5)` are the vertices of a

quadrilateral, find the area of the quadrilateral `ABCD` .

Class 10 Chapter 7 Example 15

quadrilateral, find the area of the quadrilateral `ABCD` .

Class 10 Chapter 7 Example 15

By joining `B` to `D`, you will get two triangles `ABD` and` BCD` .

Now the area of `Delta ABD = 1/2 [ -5 (-5-5) + (-4) (5-7) + 4 (7+5) ]`

` = 1/2 (50 + 8 + 48) = 106/2 = 53` square units

Also, the area of `Delta BCD = 1/2 [ -4 (-6-5) -1 ( 5+5) +4 (-5 +6) ]`

` = 1/2 (44 -10 +4) =19` square units

So, the area of quadrilateral `ABCD = 53 + 19 = 72` square units.

Note : To find the area of a polygon, we divide it into triangular regions, which have

no common area, and add the areas of these regions.