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A garden box has a perimeter of 18½ feet. If the length is 5 feet, what is the area of the garden box?

Travka [436]

Answer:

$area=\frac{85}{4}feet^{2}$

Step-by-step explanation:

1. Approach

Since it is given that the garden box is a rectangle, then the opposite sides are congruent. One can use this to their advantage, by setting up an equation that enables them to solve for the width of the rectangle. After doing so, one will multiply the width by the given length and solve for the area.

2. Solve for the width

It is given that the garden box is a rectangle. As per its definition, opposite sides in a rectangle are congruent. The problem gives the length and the perimeter of the rectangle, therefore, one can set up an equation and solve for the width.

$perimeter=18\frac{1}{2}\\\\length=5$

$2(length+width)=perimeter$

Substitute,

$2(5+width)=18\frac{1}{2}$

Conver the mixed number to an improper fraction. This can be done by multiplying the "number" part of the mixed number by the denominator of the fraction. Then add the result to the numerator.

$2(5+width)=\frac{37}{2}$

Inverse operations,

$2(5+width)=\frac{37}{2}\\/2\\\\5+width=\frac{37}{4}\\-5\\\\width=\frac{17}{4}$

3. Solve for the area

Now that one has solved for the width of the box, one must solve for the area. This can be done by multiplying the length by the width. Since the width is a fraction, one must remember, that when multiplying an integer by a fraction, one will multiply the integer by the numerator (the top of the fraction), and then simplify by reducing the fraction, if possible. Reducing the fraction is when one divides both the numerator and the denominator by the GCF (Greatest Common Factor).

$length=5\\\\width = \frac{17}{4}$

$length*width=area$

Substitute,

$5*\frac{17}{4}\\\\=\frac{85}{4}$

$area=\frac{85}{4}feet^{2}$

4 0

2 years ago

I need help with this please and fast

erik [133]

Step-by-step explanation:

what are you trying to find???

3 0

2 years ago

How did this change your nutrition level

choli [55]

Answer:

how did what?

Step-by-step explanation:

4 0

2 years ago

Give an example of an infinite set.

Montano1993 [528]

Infinite sets may be countable or uncountable. Some examples are: the set of all integers, {..., -1, 0, 1, 2, ...}, is a countably infinite set; and. the set of all real numbers is an uncountably infinite set.

5 0

3 years ago

Read 2 more answers

GIVING BRAINILEST-

Nutka1998 [239]

Answer:

To construct a dilation, we need the center of dilation, which is the point from which we make the image smaller or larger. We also need a scale factor, which is the number that determines how large or small we are making the new image. If the scale factor is greater than 1, we have an enlargement, because the new image will be larger than the original image. If the scale factor is less than 1, we have a reduction because the new image will be smaller than the original image.

Enlargement

Step-1

Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.

Step-2

First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.

Step-3

Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.

Step-4

Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.

Step-5

Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.

We now repeat the process for each of the other vertices of the triangle.

Step-6

These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.

Step-7

Finally, we use the straightedge to connect all of the new vertices.

Now,See how triangle P

′

Q

′

R

′

is twice the size of the original triangle.

Reduction

Step-1

Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of

2

1

Step-2

Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.

Step-3

The green lines are the new lines that we have drawn.

Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.

Step-4

Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.

Step-5

Now, we will take the straightedge and connect the points where the curves overlap.

4 0

2 years ago