.Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Part 1

The drawing of the quadrilateral is included here

Part 2

The area of the quadrilateral is 60.5 unit²

Step-by-step explanation:

Part 1

Please find attached the given drawing of the quadrilateral created with Microsoft Excel

Part 2

The given quadrilateral ABCD comprises of a trapezium AECD and triangle ABE

The height of the trapezium, h₁ = 4 units

For segment BC, we have;

The slope = (-5 - (-2))/(5 - (-4)) = -1/3

The equation of the line BC in point and slope form is y - (-5) = -1/3×(x - 5)

Which gives;

y = -1/3×(x - 5) - 5 = -x/3 + 5/3 - 5 = -x/3 - 10/3

The y-intercept is (0, -10/3)

The lengths of the opposite parallel sides of the trapezium AECD are;

a = 5 - (-2) = 7 and b = 7 - (-10/3) = 31/3

The area of the trapezium AECD = ((a + b)/2) × h₁ = ((7 + 31/3)/2) × 4 = 104/3 unit²

The height of triangle

The area of ABE, h₂ = 5

The base of triangle ABE = b = 31/3

The area of triangle ABE = 1/2 × b × h₂ = 1/2 × 31/3 × 5 = 155/6 unit²

The area of quadrilateral ABCD = The area of the trapezium AECD + The area of triangle ABE

∴ The area of quadrilateral ABCD = 104/3 unit² + 155/6 unit² = 60.5 unit².

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4