Question

If vertices of a quadrilateral are at a(-5,7),b(-4,k),c(-1,-6) and d(4,5) and its area is 72 sq.Units find the value of k

Answer 1

a(-5,7),b(-4,k),c(-1,-6) and d(4,5)

let the diagonals is =AC and it divides it into two triangles =∆abc and ∆acd

now the area of the quadrilateral is the sum of the area of these two triangles....

now ...for the ∆abc ....

the vertices are ....

a(-5,7),b(-4,k),c(-1,-6)

therefore area of ∆abc is

$= \frac{1}{2} ( - 5(k + 6) + ( - 4)( - 6 - 7) + ( - 1)(7 - k)) \\ = \frac{1}{2} ( - 5k - 6 + 52 - 7 + k) \\ = \frac{1}{2} ( - 4k + 39)$

and for the ∆acd

the vertices are ....

a(-5,7),c(-1,-6) and d(4,5)

the area is

$= \frac{1 }{2} ( - 5( - 6 - 5) + ( - 1)(5 - 7) + 4(7 + 6)) \\ = \frac{1}{2}(55 + 2 + 52) \\ = \frac{1}{2} (109)$

therefore the area of the quadrilateral is

$\frac{109}{2} + \frac{1}{2} ( - 4k + 39) = 72 \\ = > \frac{109 - 4k + 39}{2} = 72 \\ = > 148 - 4k = 144 \\ = > 4k = 4 \\ = > k = 1$