since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)
Answer:
Given information: WY=ZX, A is the midpoint of WY and A is the midpoint of ZX.
To prove : WA =ZA
Proof:
A is the midpoint of WY, it means A divides WY in two equal parts.

So, WY is twice of WA.
.... (1)
A is the midpoint of ZX, it means A divides ZX in two equal parts.

So, ZX is twice of ZA.
... (2)
(Given)
Substitute the values from equation (1) and (2).

Divide both sides by 2.

Hence proved.
Answer:
X=90
Step-by-step explanation:
x=90 because it is a right angle.
hope that helps!
Answer:
see below
Step-by-step explanation:
I do not know what the 'given expression' is, bu the three UN-highlighted expressions in each question 1 and 2 are equivalent
Only the highlighted one is not equiv to the other three