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In the figure, the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD is . The area of polygon ABCDE is s

Gekata [30.6K]

Step 1

Find the perimeter of rectangle ABDE

we know that

the perimeter of rectangle is equal to

$P=2b+2h$

In this problem

$b=ED=2\ units$

$h=AE=6\ units$

substitute

$P=2*2+2*6=16\ units$

Step 2

Find the perimeter of triangle BCD

we know that

the perimeter of triangle is equal to

$P=BD+DC+BC$

In this problem we have

$BD=AE=6\ units$

$DC=BC$

Applying the Pythagoras theorem

$DC^{2}=4^{2}+3^{2}$

$DC^{2}=25$

$DC=5\ units$

substitute

$P=6+5+5=16\ units$

Find the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD

we have

the perimeter of rectangle is equal to

$P=16\ units$

the perimeter of the triangle is

$P=16\ units$

so

the ratio is equal to

$\frac{16}{16} =1$

therefore

the answer Part 1) is the option B

$1$

Step 3

Find the area of polygon ABCDE

we know that

The area of polygon is equal to the sum of the area of rectangle plus the area of triangle

Area of rectangle is equal to

$A=AE*BD=6*2=12\ units^{2}$

Area of the triangle is equal to

$A=\frac{1}{2}AEh$

the height h of the triangle is equal to $4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

The area of polygon is

$12\ units^{2}+12\ units^{2}=24\ units^{2}$

therefore

the answer part 2) is the option C

$24\ units^{2}$

7 0

2 years ago

Read 2 more answers

herlene has 5.05 in quarters and dimes. The number of quarters is one more than twice the number of dimes. Find the number she h

VLD [36.1K]

Answer:

Herlene has 8 dimes and 17 quarters

Step-by-step explanation:

System of Equations

Let's call:

x = number of dimes Herlene has

y = number of quarters Herlene has

Since each dime has a value of $0.10 and each quarter has a value of $0.25, the total money Herlene has is 0.10x+0.25y.

We know this amount is $5.05, thus:

0.10x + 0.25y = 5.05 [1]

It's also given the number of quarters is one more than twice the number of dimes, i.e.:

y = 2x + 1 [2]

Substituting in [1]:

0.10x + 0.25(2x + 1) = 5.05

Operating:

0.10x + 0.5x + 0.25 = 5.05

0.6x = 5.05 - 0.25

0.6x = 4.8

x = 8

From [2]:

y = 2*8 + 1 = 17

y = 17

Herlene has 8 dimes and 17 quarters

4 0

1 year ago

Chelsea is graphing the function f(x) = 20(One-fourth)x. She begins by plotting the initial value. Which graph represents her fi

natta225 [31]

Answer:

C

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Which relationships have the same constant of proportionality between y and x as in the equation y=1/3x?

Marina CMI [18]

Y and X is urgent answer

4 0

2 years ago

Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than

Fittoniya [83]

Inequation 1:

$\frac{2}{3}x-3 \geq y$

to plot the pairs (x, y) for which the inequation holds, draw the line $y=\frac{2}{3}x-3$

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2:

$y+ \frac{2}{3}x\ \textless \ 4$

$y\ \textless \ - \frac{2}{3} x+ 4
$

draw the lines

i) $y=\frac{2}{3}x-3$ use points (0, -3), (3, -1)

ii) $y=- \frac{2}{3} x+ 4$ use points ( 0, 4), (3, 2)

let's use the point P(3, 3) to see what region of the lines need to be coloured:

$\frac{2}{3}x-3 \geq y$ ;
$\frac{2}{3}(3)-3 \geq 3$
$2-3 \geq 3$ , not true so we color the region not containing this point

$y+ \frac{2}{3}x\ \textless \ 4$
$(3)+ \frac{2}{3}(3)\ \textless \ 4$
$3+ 5\ \textless \ 4$ not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.

4 0

2 years ago