Check the picture below.
now, let's notice the larger "yellow" semicircle, it has a gap, the gap on the right is of a semicircle with a diameter of 10, BUT it also has a descender on the left, a part that's hanging out, that part is also a semicircle.
so if we use the descending semicircle to fill up the gap on the right, we'll end up with a filled up larger semicircle, whose diameter is 20, and whose radius is 10 cm.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=10 \end{cases}\implies A=\pi 10^2\implies A=100\pi \\\\\\ \stackrel{\textit{half of that for a semicircle}}{A=\cfrac{100\pi }{2}}\implies A=50\pi \implies \stackrel{\pi =3.142}{A=157.1}](https://tex.z-dn.net/?f=%5Cbf%20%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%3D10%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2010%5E2%5Cimplies%20A%3D100%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semicircle%7D%7D%7BA%3D%5Ccfrac%7B100%5Cpi%20%7D%7B2%7D%7D%5Cimplies%20A%3D50%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.142%7D%7BA%3D157.1%7D)
Since a stick note i completely flat (paper) there isn't the 3rd dimension, height. So it cannot be measured with volume, but area. To find the area, multiple the leight by the width, to get the answer.
Answer:
0.33
Step-by-step explanation:
Answer:
Ur answer will be B.
Step-by-step explanation:
A dilation will always create an image smaller or bigger than the pre-image making it not congruent since it changes in size.
<h2>Step-by-step explanation :</h2>
Total shared part:
1/4 + 1/8
Taking LCM and adding:
2/8+1/8
Adding gives:
3/8
