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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Answer from: 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 $22.81\ units$

Answer from: 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

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

The perimeter of the rectangle BCDF is 22.81 units.

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

link

Answer from: Quest

the length of o'p' is $1.5\times op$ .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

$k=\frac{o'p'}{op}$

the scale factor is 1.5.

$1.5=\frac{o'p'}{op}$

multiply both sides by op.

$1.5\times op=o'p'$

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is $1.5\times op$ .

Answer from: Quest

answer: 1 + 1 = 11

step-by-step explanation:

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40 points in the figure, the ratio of the area of rectangle abef to the area of rectangle acdf is( ....