Question

find the perimeter of rhombus star

Answer 1

Answer:

$4\sqrt{10}$

Step-by-step explanation:

Perimeter of the rhombus, STAR, is the sum of the length of all it's 4 sides.

The coordinates of its vertices are given as,

S(-1, 2)

T(2, 3)

A(3, 0)

R(0, -1)

Length of each side can be calculated using the distance formula given as $d = \sqrt{x_2 - x_1)^2 + (y_2 - y_1)^2}$

Find the length of each side ST, TA, AR, RS, using the above formula by plugging in the coordinate values (x, y) of each vertices.

$ST = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

S(-1, 2) => (x1, y1)

T(2, 3) => (x2, y2)

$ST = \sqrt{(2 -(-1))^2 + (3 - 2)^2}$

$ST = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}$

$TA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

T(2, 3) => (x1, y1)

A(3, 0) => (x2, y2)

$TA = \sqrt{(3 - 2)^2 + (0 - 3)^2}$

$TA = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$

$AR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

A(3, 0) => (x1, y1)

R(0, -1) => (x2, y2)

$AR = \sqrt{(0 - 3)^2 + (-1 - 0)^2}$

$AR = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$

$RS = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

R(0, -1) => (x1, y1)

S(-1, 2) => (x2, y2)

$RS = \sqrt{(-1 - 0)^2 + (2 -(-1))^2}$

$RS = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}$

$Perimeter = ST + TA + AR + RS$

$Perimeter = \sqrt{10} + \sqrt{10} + \sqrt{10} + \sqrt{10} = 4\sqrt{10}$