Help please thank you very much

When writing this on a keyboard, you would say 2*sqrt(7)

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Explanation:

The formula we'll use is

A = 0.5*p*q*sin(R)

where R is the angle between sides p and q.

The diagram shows that p = x+4 is opposite angle P, and q = x is opposite angle Q. Convention usually has the uppercase letters be angles, and the lowercase letters as the sides.

Plug in the given values and simplify to get the following

A = 0.5*p*q*sin(R)

A = 0.5*(x+4)*x*sin(60)

A = 0.5*(x+4)*x*sqrt(3)/2 ..... use the unit circle

A = 0.25*(x^2+4x)*sqrt(3)

A = (0.25x^2+x)*sqrt(3)

Set this equal to the given area of A = sqrt(27) and solve for x

A = (0.25x^2+x)*sqrt(3)

(0.25x^2+x)*sqrt(3) = A

(0.25x^2+x)*sqrt(3) = sqrt(27)

(0.25x^2+x)*sqrt(3) = 3*sqrt(3) .... see note1 at the very bottom

0.25x^2+x = 3

(1/4)x^2+x = 3

x^2+4x = 12 .... see note2 at the very bottom

x^2+4x-12 = 0

(x+6)(x-2) = 0

x+6 = 0 or x-2 = 0

x = -6 or x = 2

A negative side length doesn't make sense, so we ignore x = -6. The only solution here is x = 2.

Since x = 2, this means we have the following side lengths so far

Now apply the law of cosines to find the missing side.

r^2 = p^2 + q^2 - 2*p*q*cos(R)

r^2 = 6^2 + 2^2 - 2*6*2*cos(60)

r^2 = 36 + 4 - 24*1/2

r^2 = 40 - 12

r^2 = 28

r = sqrt(28)

r = sqrt(4*7)

r = 2*sqrt(7)

Side PQ is exactly 2*sqrt(7) units long

This approximates to 2*sqrt(7) = 5.2915

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Footnotes:

i need the options

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.