Question

- 4, 2,-
3 - 3 - 53 - 2 and 2 are the vertices of a quadrilateral ABCD if the area of the quadrilateral is 28 square units if find the value of k​

Answer 1

Step-by-step explanation:

Given - 4, 2,-3 - 3 - 53 - 2 and 2 are the vertices of a quadrilateral ABCD if the area of the quadrilateral is 28 square units if find the value of k

• 1/2 [ - 4(- 5 + 2) – 3(-2 + 2) + 3(-2 + 5)]
•    = 21 / 2 sq units.
• Area of triangle ABC = 1/2 (- 4 (-2 – k) + 3 (k + 2) + 2(- 2 + 2))
•                                     =  7 k + 14 / 2 sq units
• Area of quadrilateral ABCD = 28 sq units.
•          21 / 2 + 7k + 14 / 2 = 28
•          7k + 35 / 2 = 28
•           7k = 21
• Therefore k = 3

Reference link will be

https://brainly.in/question/6055886

Answer 2

Answer:

Step-by-step explanation:

$A(-4,-2), B(-3,-5), C(3,-2), D(2, k)$

Area $\triangle A B C=$

$\frac{1}{2}[-4(-5+2)-3(-2+2)+3(-2+5)]$=$\frac{21}{2}[\tex]sq. units Area </p><p>=[tex]\frac{(7 k+14)}{2}$

sq. units Area of quadrilateral ABCD  sq. units

$\therefore \frac{21}{2}+\frac{7 k+14}{2}=28$

$\Rightarrow \frac{7 k+35}{2}=28$

$\Rightarrow 7 k=21$

\$therefore k=3$