if the diameter of a circle is 15, its radius is half that or 7.5.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=7.5 \end{cases} A=\pi (7.5)^2\implies A=56.25\pi \implies \stackrel{\pi =3.14}{A=176.625}](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%3D7.5%20%5Cend%7Bcases%7D%20A%3D%5Cpi%20%287.5%29%5E2%5Cimplies%20A%3D56.25%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7BA%3D176.625%7D%20)
Answer:
4
Step-by-step explanation:
the side Q-R corresponds to K-L
just as O-P corresponds to K-J
K-J= 36
O-P=6
36/6=6
so 24/6=4
20 is jenna a ssjsjsn a ans s s s s s s s s s s s. S s s s s s s s. S ss. S s
Answer:
Step-by-step explanation:
The ratio of corresponding sides DN and KI is 12 : 4 = 3 : 1. The same ratio applies to altitudes DQ and KO. Since the difference between these altitudes is 6 and the difference between their ratio units is 3-1 = 2, each ratio unit must stand for 6/2 = 3 units of linear measure. That is, ...
DQ = (3 units)·3 = 9 units
KO = (3 units)·1 = 3 units
Then the base lengths QN and OI can be found from the Pythagorean theorem:
KI² = KO² +OI²
4² = 3² +OI²
OI = √(16 -9)
OI = √7
QN = 3·OI = 3√7