Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the rectangles formed by the sets of points to their corresponding areas.
A(-9, 8), B(-5,5), C(1, 13), D(-3, 16)
50 square units
E(30, 20), F(39, 29), G(49, 19), H(40, 10)
300 square units
I(-6, 2), J(2, 2), K(2, -8), L(-6, -8)
100 square units
M(5,5), N(11,5), O(11,-5), P(5,-5)
80 square units
Q(10, 0), R(15,5), S(25,-5), T(20,-10)
U(0,5), V(15, 20), W(25, 10), X(10,-5)

A(-9, 8), B(-5, 5), C(1, 13), D(-3, 16) → 50 square units

I(-6, 2), J(2, 2), K(2, -8), L(-6, -8) → 80 square units

Q(10, 0), R(15, 5), S(25, -5), T(20, -10) → 100 square units

U(0, 5), V(15, 20), W(25, 10), X(10, -5) → 300 square units

Step-by-step explanation:

The area of a rectangle, given the coordinates of the  vertices is found as follows;

1) The vertices of the rectangle ABCD are; A(-9, 8), B(-5, 5), C(1, 13), and D(-3, 16)

The length of side AB = √((-9 - (-5))² + (8 - 5)²) = 5

The length of side BC = √((1 - (-5))² + (13 - 5)²) = 10

The length of side CD = √((1 - (-3))² + (13 - 16)²) = 5

The length of side DA = √(((-9) - (-3))² + (8 - 16)²) = 10

The area of rectangle ABCD = 5 × 10  = 50 square units

2) The vertices of the rectangle EFGH are; E(30, 20), F(39, 29), G(49, 19), and H(40, 10)

The length of side EF= √((39 - 30)² + (29 - 20)²) = 9·√2

The length of side FG = √((39 - 49)² + (29 - 19)²) = 10·√2

The length of side GH = √((40 - 49)² + (10 - 19)²) = 9·√2

The length of side HE = √((40 - 30)² + (10 - 20)²) = 10·√2

The area of rectangle EFGH = 9·√2 × 10·√2  = 180 square units

3) The vertices of the rectangle IJKL are; I(-6, 2), J(2, 2), K(2, -8), and L(-6, -8)

The length of side IJ= √((-6 - 2)² + (2 - 2)²) = 8

The length of side JK= √((2 - 2)² + ((-8) - 2)²) = 10

The area of rectangle IJKL = IJ × JK

∴ The area of rectangle IJKL = 8 × 10 = 80 square units

4) The vertices of the rectangle MNOP are; M(5, 5), N(11, 5), O(11, -5), and P(5, -5)

The length of side MN = √((5 - 11)² + (5 - 5)²) = 6

The length of side NO = √((11 - 11)² + ((-5) - 5)²) = 10

The area of rectangle MNOP = 6 × 10 = 60 square units

5) The vertices of the rectangle QRST are; Q(10, 0), R(15, 5), S(25, -5), and T(20, -10)

The length of side QR = √((10 - 15)² + (0 - 5)²) = 5·√2

The length of side RS = √((25 - 15)² + ((-5) - 5)²) = 10·√2

The area of rectangle QRST = 5·√2 × 10·√2 = 100 square units

6) The vertices of the rectangle UVWX are; U(0, 5), V(15, 20), W(25, 10), and X(10, -5)

The length of side UV = √((0 - 15)² + (5 - 20)²) = 15·√2

The length of side VW = √((25 - 15)² + (10 - 20)²) = 10·√2

The area of rectangle UVWX = 15·√2 × 10·√2 = 300 square units

d. is the correct answer

step-by-step explanation:

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.