They aren't similar because they're not proportional (the sides do not form equal ratios)<span />
Answer:
The answer is C
Step-by-step explanation:
The perimeter is the total of adding all of the side lengths together.
A square has 4 equal side lengths.
So the perimeter of a square is:
P = s + s + s + s or P = 4s
[P = perimeter s = side lengths of the square]
Since you know the side length of the square is (x + 2 1/4), you can replace s with (x + 2 1/4)
P = 4s
P = 4(x + 2 1/4) Multiply 4 into (x + 2 1/4)
P = 4x + 8 4/4
P = 4x + 9
Since you know the perimeter, you can plug it in.(you could have also plugged it in in the beginning)
P = 4x + 9
14 = 4x + 9 Subtract 9 on both sides
5 = 4x Divide 4 on both sides
5/4 = x
Now that you know x, find the side length of the square.
(x + 2 1/4)
(5/4 + 2 1/4)
2 6/4 = 3 2/4 = 3 1/2 units or 3.5 units
To find the area of a square, you multiply 2 of the sides together:
A = s · s
A = 3.5 · 3.5
A = 12.25 units²
Option B:
Solution:
In the given figure 
.
If two triangles are similar, then their corresponding sides and angles are equal.
By CPCTC, in 
,
– – – – (1)
– – – – (2)
– – – – (3)
– – – – (4)
– – – – (5)
– – – – (6)
Option A: 
By CPCTC proved in equation (2) .
Therefore 
. Option A is false.
Option B:
By CPCTC proved in equation (1) .
Therefore Option B is true.
Option C: 
By CPCTC proved in equation (4) .
Therefore 
. Option C is false.
Option D: 
By CPCTC proved in equation (5) .
Therefore 
. Option D is false.
Hence Option B is the correct answer.
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)
