Class 10 Area of a Triangle

### Topic Covered

♦ 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)]}
Q 3129780611

Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and
(–3, –5).
Class 10 Chapter 7 Example 11
Solution:

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
Solution:

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
Solution:

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
Solution:

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

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
Solution:

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.