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kicyunya [14]
3 years ago
12

What is the area of rectangle ABCD in square units ?

Mathematics
2 answers:
zloy xaker [14]3 years ago
7 0

<u><em>Answer:</em></u>

34 square units

<u><em>Explanation:</em></u>

In any rectangle, each two opposite sides are equal

<u>This means that, in the given rectangle:</u>

AB = CD and AD = BC

Area of the rectangle is the product of its dimensions (length and width)

<u>This means that:</u>

Area of ABCD = AB × BC

<u>1- getting the side length:</u>

To get the side length, we will use the distance formula:

D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

<u>For AB, we have:</u>

A = (-1,4) which means that x₁ = -1 and y₁ = 4

B = (3,3) which means that x₂ = 3 and y₂ = 3

<u>Substitute in the equation:</u>

AB = \sqrt{(3-(-1))^2+(3-4)^2} = \sqrt{17} units

<u>For BC, we have:</u>

B = (3,3) which means that x₁ = 3 and y₁ = 3

C = (1,-5) which means that x₂ = 1 and y₂ = -5

<u>Substitute in the equation:</u>

BC = \sqrt{(1-3)^2+(-5-3)^2} = 2\sqrt{17} units

<u>2- getting the area:</u>

Area of ABCD = AB × BC

Area of ABCD = \sqrt{17} *  2\sqrt{17} = 34 square units

Hope this helps :)

Licemer1 [7]3 years ago
5 0

Answer:

Step-by-step explanation:

Square units = √( 4^2 + 1^2) * √( 8^2 + 2^2)

= √17 * √68

= √1156

= 34

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Solve the system <br> 2x+2y+z=-2 <br> -x-2y+2z=-5<br> 2x+4y+z=0
drek231 [11]

The values of (x,y,z) are (3,-1,-2) , if the given are equations 2x+2y+z=- 2,-x-2y+2z=-5 and 2x+4y+z=0.

Step-by-step explanation:

The given is,

                          2x+2y+z=- 2.......................................(1)

                         -x-2y+2z=-5......................................(2)

                            2x+4y+z=0.........................................(3)

Step:1

           Equation (2) is multiplied by -1            ( Eqn(2) × -1 )

                                         x+2y-2z=5.............................(4)

          Subtract the equation (1) and (4)

                                        2x+2y+z=- 2

                                         x+2y-2z=5

                 ( - )

           (2x-x)+(2y-2y)+(z+2z)=(-2-5)

                                                  x+3z=-7......................(5)

Step:2

          Equation (2) is multiplied by -2                 ( Eqn(2) × -2)

                                        2x+4y-4z=10........................(6)

         Subtract equation (6) and (3),                  

                                        2x+4y-4z=10

                                         2x+4y+z=0

                   ( - )

       (2x-2x)+(4y-4y)+(-4z-z)= (10-0)

                                                     -5z=10

                                                         z = - \frac{10}{5}

                                                         z = -2

         From the equation (5),

                                                  x+3z=-7  

                                          x+(3)(-2)=-7

                                                           x = -7+6

                                                            x = -1

         From equation (1),

                                            2x+2y+z=- 2

          Substitute x and z values,

                               (2)(-1)+2y+(-2)=-2

                                                     2y - 4=2

                                                           2y=4-2

                                                           2y=2

                                                            y=\frac{2}{2}

                                                             y = 1

Step:3

                Check for solution,

                                  -x-2y+2z=-5

                Substitute x,y and z values,

               -(-1)-(2)(1)+(2)(-2)=-5

                                         1-2-4=-5

                                                    -5 = -5

Result:

              The values of (x,y,z) are (3,-1,-2) , if the given are equations 2x+2y+z=- 2,-x-2y+2z=-5 and 2x+4y+z=0.

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From there, you can just cross-multiply and divide to get:

\frac{1 * 378}{250}

A tenth of 250 is 25, so this is roughly \frac{375}{250} or \frac{17}{10}, so it'll be approximately 1.7 inches on the map.
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