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Semmy [17]
3 years ago
7

Find the perimeter of rhombus star

Mathematics
1 answer:
Degger [83]3 years ago
4 0

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}

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Answer:

Alrighty! So this is what we know:

Slope (m) = (-4)

Co-ordinate (0,3)

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t = ?

Let's find our equation...

y=mx+b <-- just plug what we know in...

3=(-4)0+b

3=0+b

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We just found b! woohoo

Now what is t? t is a value we need to know WHICH IS the y value of a co-ordinate that corresponds to when x is 6, as we know a co-ordinate is expressed as --> (x value, y value).

SO-

t=y

plug it in our equation...

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t=(-4)x+3

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

The answer is

<h2>9 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3>d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  }  \\</h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(6,-5) and (-3,-5)

The distance between them is

<h3>d =  \sqrt{ ({6 + 3})^{2} + ( { - 5 + 5})^{2}  }  \\   = \sqrt{ {9}^{2} +  0 }  \\  =  \sqrt{81}</h3>

We have the final answer as

<h3>9 units</h3>

Hope this helps you

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