Question

PLS ANSWER RN What is the area of AJKL?

Answer 1

Answer:

8.5

Step-by-step explanation:

slope of the line that contains KL

(y2 - y1)/(x2-x1)

(0-1)/(7-3)

m = -1/4

slope of the line that contains JK

(5-1)/(4-3)

m = 4

the linea are perpendicular so triangle is right

area = (leg1 x leg2)/2

KL = $\sqrt{(7-3)^2 + (0-1)^2} = \sqrt{16 + 1} = \sqrt{17}$

KJ = $\sqrt{(4-3)^2 + (5-1)^2}= \sqrt{1 + 16} = \sqrt{17}$

area = (√17)^2/2 = 17/2 = 8.5

Answer 2

8.5



….hope this helps!