170 MATHEMATICS
✁
1
2( 3) 4( 3 3) 6(3 )
2
kk✂ ✂ ✄ ✄ ✂ ✄ = 0i.e.,
1
(4) 0
2
✄ k ☎Therefore, k = 0
Let us verify our answer.
area of ✆ ABC = ✝ ✞1
2(0 3) 4( 3 3) 6(3 0) 0
2
✂ ✂ ✄ ✄ ✂ ✄ ☎
Example 15 : 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.
Solution : By joining B to D, you will get two triangles ABD and BCD.
Now the area of ✆ ABD = ✟ ✠
1
5( 5 5) ( 4)(5 7) 4(7 5)
2
✄ ✄ ✄ ✂ ✄ ✄ ✂ ✂
=
1106
(50 8 48) 53 square units
22✂ ✂ ☎ ☎
Also, the area of ✆ BCD = ✡ ☛
1
4(65)–1(55)4(56)
2
✄ ✄ ✄ ✂ ✂ ✄ ✂
=
1
(44 10 4) 19 square units
2✄ ✂ ☎
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.
EXERCISE 7.3
- Find the area of the triangle whose vertices are :
(i) (2, 3), (–1, 0), (2, – 4) (ii)(–5, –1), (3, –5), (5, 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) - 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. - Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5),
(3, – 2) and (2, 3). - You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides
it into two triangles of equal areas. Verify this result for ☞ ABC whose vertices are
A(4, – 6), B(3, –2) and C(5, 2).