ABCD is a square with 8-inch sides. Point E is the midpoint of side AB and point F is the midpoint of side BC. What is the area of triangle DEF? 8 4 A B F. C D​

The area of triangle DEF is equal to 8√3

Step-by-step explanation:

Firstly, side DF is the hypothenuse to the formed right triangle ADE. As side AD has a length of 8 and side AE has a length of 4, we can substitute these values into the Pythagorean theorem:

A^2 + B^2 = C^2

8^2 + 4^2 = C^2

64 + 16 = C^2

80 = C^2

√(80) = √(C^2)

4√5 = C

This means that side DE of triangle DEF is equal to a length of 4√5, and the side DF is also equal to this length because of how squares are congruent as shown in the diagram. This means all we have to do is find the length of side EF.

As side EB and BF is equal to 4, we can create the Pythagorean theorem once again:

A^2 + B^2 = C^2

4^2 + 4^2 = C^2

16 + 16 = C^2

32 = C^2

√(32) = √(C^2)

4√2 = C

Now, we have a triangle to work with! This triangle is triangle DEF and now we have to find the area of the newly defined triangle. To do this, we take the formula for the area of a triangle, A = (HB)/(2), and find values to substitute. Because we do not have height for this triangle, we have to use the Pythagorean theorem once again and hopefully for the last time!

We can bisect side EF with a perpendicular bisector to point D. By doing so, we are creating a line that can be used to solve the height of this triangle. We can now use the theorem:

A^2 + B^2 = C^2

A^2 + (2√2)^2 = (4√2)^2

A^2 + 8 = 32

A^2 = 24

√(A^2) = √24

A = 2√6

Finally! We can now substitute known values into the equation A = (HB)/(2).

A = (HB)/(2)

A = [(2√6)(4√2)]/(2)

A = (16√3)/(2)

A = 8√3

This means that the area of triangle DEF is equal to 8√3!

b c and d

Step-by-step explanation:

got it right on edge

won't that be plagrisim thought?

step-by-step explanation:

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: