40 POINTS
In the figure, the ratio of the area of rectangle ABEF to the area of rectangle ACDF is(a)_____ . If the coordinates of point A are (0,6), the area of rectangle ABEF is(b)____ square units, and the area of rectangle ACDF is(c)____ square units.  The perimeter of rectangle BCDE is(d)____ units.

(a)
2:1
2:3
3:4
3:4 

(b)
32.02
48.03
65.03
96.05

(c)
128.07
48.03
65.03
96.05

(d)
20.61
25.61
22.81
32.81

40 POINTS In the figure the ratio of the area of rectangle ABEF to the area of rectangle ACDF isa If the coordinates of point A are 06 the area of rectangle ABE class=

Respuesta :

we know that

The area of a rectangle is equal to

[tex]A=L*W[/tex]

where

L is the length side of the rectangle

W is the width side of the rectangle

The perimeter of the rectangle is equal to

[tex]P=2L+2W[/tex]

Step [tex]1[/tex]

Find the dimensions of the rectangles

we know that

the distance between two points is equal to the formula

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

Find the distance A-F

A-F is the width of the rectangles ABEF and ACDF

[tex]A(0,6)\\F(5,2)[/tex]

substitute the values in the formula

[tex]dAF=\sqrt{(2-6)^{2}+(5-0)^{2}}[/tex]

[tex]dAF=\sqrt{(-4)^{2}+(5)^{2}}[/tex]

[tex]dAF=\sqrt{41}\ units[/tex]

so

[tex]W=\sqrt{41}\ units[/tex]

Find the distance F-E

F-E is the length side of the rectangle ABEF

[tex]F(5,2)\\E(11,10)[/tex]

substitute the values in the formula

[tex]dFE=\sqrt{(10-2)^{2}+(11-5)^{2}}[/tex]

[tex]dFE=\sqrt{(8)^{2}+(6)^{2}}[/tex]

[tex]dFE=10\ units[/tex]

Find the distance F-D

F-D is the length side of the rectangle ACDF

[tex]F(5,2)\\D(14,14)[/tex]

substitute the values in the formula

[tex]dFD=\sqrt{(14-2)^{2}+(14-5)^{2}}[/tex]

[tex]dFD=\sqrt{(12)^{2}+(9)^{2}}[/tex]

[tex]dFD=15\ units[/tex]  

Step [tex]2[/tex]

the dimensions of the rectangle ABEF are

[tex]L=10\ units\\W=\sqrt{41}\ units[/tex]

Find the area of the rectangle ABEF

[tex]A=10*\sqrt{41}=10\sqrt{41}\ units^{2}[/tex]

[tex]A=64.03\ units^{2}[/tex]

Step [tex]3[/tex]

the dimensions of the rectangle ACDF are

[tex]L=15\ units\\W=\sqrt{41}\ units[/tex]

Find the area of the rectangle ACDF

[tex]A=15*\sqrt{41}=15\sqrt{41}\ units^{2}[/tex]

[tex]A=96.05\ units^{2}[/tex]

Step [tex]4[/tex]

The ratio of the area of rectangle ABEF to the area of rectangle ACDF is

[tex](10\sqrt{41}/15\sqrt{41})=(10/15)=2/3[/tex]

Step [tex]5[/tex]

Find the perimeter of the rectangle BCDE

we know that

[tex]W=\sqrt{41}\ units[/tex]

the length of the rectangle BCDE is equal to the distance E-D

[tex]ED=FD-FE[/tex]

[tex]ED=15-10=5\ units[/tex]

[tex]L=5\ units[/tex]

[tex]P=2*5+2*\sqrt{41}=22.81\ units[/tex]

therefore

the answer part a) is

[tex]2:3[/tex]

the answer part b) is

the area of the rectangle ABEF is [tex]64.03\ units^{2}[/tex]

the answer part c) is  

the area of the rectangle ACDF is [tex]96.05\ units^{2}[/tex]

the answer Part d) is  

the perimeter of the rectangle BCDE is [tex]22.81\ units[/tex]



If the coordinates of point A are (0,6), the area of rectangle ABEF is 2:3.

The area of rectangle ACDF is 65.03 square units.

The area of rectangle ACDF is 96.05 square units.

The perimeter of rectangle BCDE is 22.81 units.

Area of the rectangle;

The area of the triangle is given by multiplying by the length and width of the rectangle.

[tex]\rm Area \ of \ recatngle=Length \times Width[/tex]

The perimeter of the rectangle is given by;

[tex]\rm Perimeter\ of \ recatngle=2(Length + Width)[/tex]

1.  The area of the rectangle ACDF is;

[tex]\rm Distance\ of \ FE = \sqrt{(10-2)^2+(11-5)^2}\\\\ Distance\ of \ FE= \sqrt{(8)^2+(6)^2}\\\\ Distance\ of \ FE = \sqrt{64+36}\\\\ Distance\ of \ FE = \sqrt{20}\rm \\\\Distance\ of \ FE =10 \ units[/tex]

[tex]\rm Distance\ of \ FD = \sqrt{(14-2)^2+(14-5)^2}\\\\ Distance\ of \ FD=\sqrt{(12)^2+(9)^2}\\\\ Distance\ of \ FD = \sqrt{144+81}\\\\ Distance\ of \ FD= \sqrt{225}\rm \\\\Distance\ of \ FD =15 \ units[/tex]

2. The area of rectangle ABEF is;

[tex]\rm Area \ of \ recatngle=Length \times Width\\\\\rm Area \ of \ recatngle=10 \times \sqrt{41}\\\\Area \ of \ recatngle=65.03[/tex]

The area of the rectangle ABEF is; 65.03 square units.

3. The area of the rectangle ACDF is;

[tex]\rm Area \ of \ recatngle=Length \times Width\\\\\rm Area \ of \ recatngle=15 \times \sqrt{41}\\\\Area \ of \ recatngle=96.05[/tex]

The area of the rectangle ACDF is 96.05  square units.

4. The perimeter of rectangle BCDF is

[tex]\rm Perimeter\ of \ recatngle=2(Length + Width)\\\\\rm Perimeter\ of \ recatngle=2(5+\sqrt{41} )\\\\\rm Perimeter\ of \ recatngle=2(11.41)\\\\\rm Perimeter\ of \ recatngle=22.81[/tex]

The perimeter of the rectangle BCDF is 22.81 units.

To know more about the area of the rectangle click the link given below.

https://brainly.com/question/14383947

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