Area of a Triangle:
ABC is a triangle with vertices A(x1, y1 ), B(x2, y2) and C(x3, y3).
To find the area of the triangle we need to draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Now we can see that ABQP, APRC and BQRC are all trapeziums.
Area of triangle ABC = Area of trapezium ABQP + Area of trapezium APRC – Area of trapezium BQRC.
Area of trapezium = 1/2 ( sum of parallel sides) x ( distance between them )
∴ Area of triangle ABC = 1/2 (BQ + AP ) +1/2(AP + CR) PR − 1/2(BQ + CR) QR
= 1/2 (y2+y1 ) ( x1 −x2) + 1/2(y1 +y3) ( x3 −x1) − 1/2(y2 +y3) (x3 −x2)
= 1/2 [x1 (y2−y3) + x2 (y3−y1 ) + x3 (y1−y2 )]
Remark: If the area of the triangle is zero then the given three points must be collinear.
Area of a Polygon:
Like the triangle, we can easily find the area of any polygon if we know the coordinates of all the vertices of the polygon
Centroid of a Triangle:
Centroid of a triangle is the point where all the three medians of the triangle meet with each other.
Here ABC is a triangle with vertices A (x1, y1), B (x2, y2) and C (x3 , y3 )). The centroid of the triangle is the point with the coordinates (x, y).
The coordinates of the centroid will be calculated as
x = (x1 + x2 + x3 )/3, y = (y1 + y2 + y3) /3 .
Example : Name the type of triangle formed by the points A (–5, 6), B (–4, –2) and C (7, 5).
Solution: The points are A (–5, 6), B (–4, –2) and C (7, 5).
Using distance formula,
d =√(x1 − x2)2 +(y1− y2)2
AB=√−4 + 5)2 + (−2−6)2
= √(1 + 64)
= √65
BC=√(7 + 4)2 + (5 + 2)2
= √(121 + 49)
= √170
AC=√(7 + 5)2 )+ (5 − 6)2
= √(144 + 1)
= √145
Since all sides are of different length, ABC is a scalene triangle.
EXERCISE 7.3
1. Find the area of the triangle whose vertices are:
(i) (2, 3), (−1, 0), (2, −4)
(ii) (−5, −1), (3, −5), (5, 2)
Sol: Area of a triangle formula= 1/2 [x1 (y2 − y3) + x2 (y3 − y1) + x3 (y1 −y2)]
(i) Putting all the values in the above formula, we get
Area of triangle = 1/2 [2 {0− (−4)} + (−1) {(−4) – (3)} + 2 (3 – 0)]
= 1/2{8 + 7 + 6} = 21/2
So area of triangle is 21/2 square units.
(ii) Here,area of the triangle = 1/2 [−5 { (−5) − (2)} + 3[2− (−1)] + 5{−1 – (−5)}]
= 1/2 {35 + 9 + 20} = 32
Therefore, area of the triangle is 32 square units.
2. In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, −2), (5, 1), (3, −k)
(ii) (8, 1), (k, −4), (2, −5)
Solution: (i) For collinear points, area of triangle formed by them is always zero.
Let points (7, −2) (5, 1), and (3, k) be the vertices of a triangle.
Area of triangle = 1/2 [x1 (y2 −y3 ) + x2 (y3 − y1 ) + x3 (y1 −y2 )]
= 1/2 [7 { 1− k} + 5(k− (−2)) + 3{(−2) – 1}] = 0
7 – 7k + 5k + 10 −9 = 0
−2k + 8 = 0
k = 4
(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0
1/2 [8 { −4(−5)} + k{(−5) − (1)} + 2{1 − (−4)}] = 0
8 – 6k + 10 = 0
6k = 18
k = 3
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, −1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution: Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle.
Coordinates of D, E, and F are given by
D = (1, 0)
E = (0, 1)
F = (1, 2)
Area of a triangle = 1/2 [x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 −y2 )]
Area of ΔDEF = 1/2{1(2−1) + 1(1−0) + 0(0−2)} = 1/2 (1 + 1) = 1
Area of ΔDEF is 1 square units
Area of ΔABC = 1/2 [0 (1−3) + 2 {3− (−1)} + 0 (−1−1)] = 1/2 {8} = 4
Area of ΔABC is 4 square units
Therefore, the required ratio is 1:4.
4. Find the area of the quadrilateral whose vertices, taken in order, are
(−4, −2), (−3, −5), (3, −2) and (2, 3).
Solution: Let the vertices of the quadrilateral be A (−4, – 2), B ( – 3, –5), C (3, –2), and D (2, 3). Join AC and divide quadrilateral into two triangles.
We have two triangles ΔABC and ΔACD.
Area of ΔABC = 1/2 [(−4) {(−5) − (−2)} + (−3) {(−2) – (−2)} + 3 {(−2) – (−5)}]
= 1/2 (12 + 0 + 9)
= 21/2 square units
Area of ΔACD = 1/2 [(−4) {(−2) – (3)} + 3{(3) – (−2)} + 2 {(−2) – (−2)}]
= 1/2 (20 + 15 + 0)
= 35/2 square units
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= ( 21/2 + 35/2) square units = 28 square units
Area of quadrilateral ABCD =28 square units
5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
Solution: Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.
Coordinates of point D = Midpoint of BC = (4, 0)
Area of a triangle = 1/2 [x1 (y2 − y3 )+ x2 (y3 − y1 )+ x3 (y1 −y2 )]
Now, Area of ΔABD = 1/2 [(4) {(−2) – (0)} + 3{(0) – (−6)} + (4) {(−6) – (−2)}]
= 1/2 (− 8 + 18 – 16)
= −3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔACD = 1/2 [(4) {0 – (2)} + 4{(2) – (−6)} + (5) {(−6) – (0)}]
= 1/2 (−8 + 32 – 30) = −3 square units
However, area cannot be negative. Therefore, area of ΔACD is 3 square units.
The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas.