Acity planner designs a park that is a quadrilateral with vertices at j(−1, 1), k(1, 3), l(5, −1), and m(−3, −1). there is an entrance to the park at the midpoint of each side of the park. a straight path connects each entrance to the entrance on the opposite side. assuming each unit of the coordinate plane represents 10 meters, what is the total length of the paths to the nearest meter? round your answer to the nearest whole number.

use the eqn below to find the mid-pts

x-mid-pt = (x₁+x₂)/2

y-mid-pt = (y₁+y₂)/2

J(−1, 1), K(1, 3), L(5, −1), and M(−3, −1)

So

midpoint JK = (-1+1/2, 1+3/2) = (0,2)

midpoint KL = (1+5/2, 3+⁻1/2) = (3,1)

midpoint LM = (5+⁻3/2, ⁻1+⁻1/2) = (1,-1)

midpoint MJ = (⁻3+⁻1/2, -1+1/2) = (-2,0)

use the eqn to find distance:

d = √(x₂ - x₁)² + (y₂ - y₁)²

Distance between JK and LM = √(0⁻1)² + (2 - ⁻1)² = √1 + 9 = √10

Distance between JM and KL = √(3-⁻2)² + (1-0)² = √25 + 1 = √26

Scale is 10

Actual distance = 10*(√10 + √26) = 82.6 = 83 units

answer: 298143

step-by-step explanation:

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐