Mathematics, 14.07.2019 11:00, melmar76

40 points in the figure, the ratio of the area of rectangle abef to the area of rectangle acdf is( . if the coordinates of point a are (0,6), the area of rectangle abef is( square units, and the area of rectangle acdf is( square units. the perimeter of rectangle bcde is( 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

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answered: yasmin599

we know that

The area of a rectangle is equal to

A=L*W

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

P=2L+2W

Step

Find the dimensions of the rectangles

we know that

the distance between two points is equal to the formula

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance A-F

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

$A(0,6)\\F(5,2)$

substitute the values in the formula

$dAF=\sqrt{(2-6)^{2}+(5-0)^{2}}$

$dAF=\sqrt{(-4)^{2}+(5)^{2}}$

$dAF=\sqrt{41}\ units$

$W=\sqrt{41}\ units$

Find the distance F-E

F-E is the length side of the rectangle ABEF

$F(5,2)\\E(11,10)$

substitute the values in the formula

$dFE=\sqrt{(10-2)^{2}+(11-5)^{2}}$

$dFE=\sqrt{(8)^{2}+(6)^{2}}$

$dFE=10\ units$

Find the distance F-D

F-D is the length side of the rectangle ACDF

$F(5,2)\\D(14,14)$

substitute the values in the formula

$dFD=\sqrt{(14-2)^{2}+(14-5)^{2}}$

$dFD=\sqrt{(12)^{2}+(9)^{2}}$

$dFD=15\ units$

Step

the dimensions of the rectangle ABEF are

$L=10\ units\\W=\sqrt{41}\ units$

Find the area of the rectangle ABEF

$A=10*\sqrt{41}=10\sqrt{41}\ units^{2}$

$A=64.03\ units^{2}$

Step

the dimensions of the rectangle ACDF are

$L=15\ units\\W=\sqrt{41}\ units$

Find the area of the rectangle ACDF

$A=15*\sqrt{41}=15\sqrt{41}\ units^{2}$

$A=96.05\ units^{2}$

Step

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

$(10\sqrt{41}/15\sqrt{41})=(10/15)=2/3$

Step

Find the perimeter of the rectangle BCDE

we know that

$W=\sqrt{41}\ units$

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

ED=FD-FE

$ED=15-10=5\ units$

$L=5\ units$

$P=2*5+2*\sqrt{41}=22.81\ units$

therefore

the answer part a) is

2:3

the answer part b) is

the area of the rectangle ABEF is $64.03\ units^{2}$

the answer part c) is

the area of the rectangle ACDF is $96.05\ units^{2}$

the answer Part d) is

the perimeter of the rectangle BCDE is $22.81\ units$

answered: mirmir62

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.

$\rm Area \ of \ recatngle=Length \times Width$

The perimeter of the rectangle is given by;

$\rm Perimeter\ of \ recatngle=2(Length + Width)$

1. The area of the rectangle ACDF is;

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

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

2. The area of rectangle ABEF is;

$\rm Area \ of \ recatngle=Length \times Width\\\\\rm Area \ of \ recatngle=10 \times \sqrt{41}\\\\Area \ of \ recatngle=65.03$

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

3. The area of the rectangle ACDF is;

$\rm Area \ of \ recatngle=Length \times Width\\\\\rm Area \ of \ recatngle=15 \times \sqrt{41}\\\\Area \ of \ recatngle=96.05$

The area of the rectangle ACDF is 96.05 square units.

4. The perimeter of rectangle BCDF is

$\rm Perimeter\ of \ recatngle=2(Length + Width)\\\\\rm Perimeter\ of \ recatngle=2(5+\sqrt{41} )\\\\\rm Perimeter\ of \ recatngle=2(11.41)\\\\\rm Perimeter\ of \ recatngle=22.81$