heureka

From the figure shown below, DE is the diameter of circle A and BC is the radius of circle B. If DE = 60 cm and AC = 10 cm, find the area of the shaded region

$$h=\overline{AC}=10~\rm{cm}
\qquad r_{A} = \frac{ \overline{DE} }{2}= 30~\rm{cm}
\qquad r_{B} = \overline{CB} \\\\
r_A^2 = h\cdot(2\cdot r_B-h)\qquad \Rightarrow\qquad
r_B = \frac{h+\dfrac{r_A^2}{h} }{2}=\frac {10+90}{2}= 50 ~\rm{cm}\\\\
\small{\text{Angle DBE $=B
\qquad B\ensurement{^{\circ}} = 2\cdot\arcsin{ \left( \dfrac{r_A}{r_B} \right)} $}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{B\ensurement{^{\circ}} } { 360\ensurement{^{\circ}} } - \frac{1}{2}\cdot (2\cdot r_A)\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{ 2\cdot\arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 360\ensurement{^{\circ}} } - \frac{1}{2}\cdot (2\cdot r_A)\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{ \arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 180\ensurement{^{\circ}} } - r_A\cdot(r_B - h)
\right]
$}}$$

$$\small{ A_{Blue}= \pi\cdot r_B^2
\left(
1- \frac{ \arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 180\ensurement{^{\circ}} }
\right)
- \pi \frac{\cdot r_A^2}{2}+ r_A\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot 50^2
\left(
1- \frac{ \arcsin{ \left( \dfrac{30}{50} \right)} } { 180\ensurement{^{\circ}} }
\right)
- \pi \frac{\cdot 30^2}{2}+ 30\cdot(50 - 10)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot 50^2\cdot
0.7951672353008667 - \pi \frac{\cdot 30^2}{2}+ 1200\right]
$}} \\\\
\small{ A_{Blue}= 6031.5121678758663628914 ~\rm{cm^2}
$}}\\\\
A_{Blue} \approx 6031.5122 ~\rm{cm^2}$$

wie rechne ich den radius r einer kugel aus nur mit dem oberflächeninhalt?

$$\boxed{\mathbf{ O_{\rm{Kugel}} = 4\cdot \pi \cdot r^2 = \pi d^2 }}\\\\\small{\text{$\qquad O_{\rm{Kugel}} = \rm{~Oberfl\"ache~} \qquad d = \rm{~Durchmesser~} = 2\cdot r \qquad r = \rm{~Radius~}= ? $}}\\\\\\\begin{array}{rcl}d^2 &=& \dfrac{ O_{\rm{Kugel}} } { \pi }\\\\d &=& \sqrt{ \dfrac{ O_{\rm{Kugel}} } { \pi } }\\\\r &=& \dfrac{1}{2}\cdot d = \dfrac{1}{2}\cdot \sqrt{ \dfrac{ O_{\rm{Kugel}} } { \pi } }\\\\\end{array}\\\\ \boxed{\mathbf{ r = \dfrac{1}{2}\cdot \sqrt{ \dfrac{ O_{\rm{Kugel}} } { \pi } } }}\\\\\\\small{\text{$\pi = 3,1415926535897932 $}}\qquad\boxed{\mathbf{ r = \dfrac{1}{ 2\sqrt{\pi} }\cdot \sqrt{ O_{\rm{Kugel}} } \qquad r = 0,2820947917738782 \cdot \sqrt{ O_{\rm{Kugel}} } }}$$

If cot x = 2/3 , what is tan x ?

$$\boxed{\mathbf{ \tan{(x)}=\dfrac{1}{\cot{(x)}} }}\\\\\\
\tan{x}=\dfrac{1}{ \frac23 } = 1\cdot \frac32 = \frac32\\\\\\
\boxed{ \mathbf{ \cot{(x)}= \frac23 \qquad \tan{(x)}= \frac32 }}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{a}{b} \qquad \tan{(x)}= \frac{b}{a}
}}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{\cos{(x)}}{\sin{(x)}} \qquad \tan{(x)}= \frac{\sin{(x)}}{\cos{(x)}}
}}$$

invers sin of negativ number ?

$$\boxed{\mathbf{
\arcsin{(-x)} = {-\arcsin (x)}
}}$$

7 days is how many hours ?

$$\small{\text{$
\begin{array}{rcl}
7~ \rm{days}
\cdot \left( \dfrac{ 24 ~\rm{hours} } { 1~\rm{day} } \right)
&=& 7\cdot 24 ~\rm{hours} = 168 ~\rm{hours}
\end{array}
$}}$$

The volume of a water tank with radius 10m is 4000m3. Calculate the height of the tank to the nearest metre

$$\boxed{\mathbf{
V_{\rm{Tank}}=(\pi \cdot r^2)\cdot h
}
}
\small{\text{$
\qquad r = 10 ~\rm{m} \qquad V_{\rm{Tank}}=4000 ~\rm{m^3}
\qquad \rm{height}_{\rm{of~the~tank}} = h = ~?
$}} \\\\\\
\begin{array}{rcl}
h &=& \dfrac{ V_{\rm{Tank}} } {\pi \cdot r^2}\\\\
h &=& \dfrac{ 4000 ~\rm{m^3} } {\pi \cdot (10 ~\rm{m})^2}\\\\
h &=& \dfrac{ 4000 ~\rm{m^3} } {\pi \cdot 100 ~\rm{m}^2}\\\\
h &=& \dfrac{ 40\not{0}\not{0} ~\rm{m^3} } {\pi \cdot 1\not{0}\not{0} ~\rm{m}^2}\\\\
h &=& \dfrac{ 40 ~\rm{m^3} } {\pi ~\rm{m}^2}\\\\
h &=& \dfrac{ 40 } {\pi} ~\rm{m}\\\\
h &=& 12.7323954474 ~\rm{m}\\\\
h &\approx& 13 ~\rm{m}
\end{array}$$

Wieviel km² sind 4250000000 cm² ?

$$\small{\text{$
\begin{array}{rcl}
4~250~000~000~ \rm{cm}^2
\cdot \left( \dfrac{ 1~\rm{dm} } { 10~\rm{cm} } \right)
\cdot \left( \dfrac{ 1~\rm{dm} } { 10~\rm{cm} } \right)
&=& \dfrac{4~250~000~000}{100}~ \rm{dm}^2\\\\
\dfrac{4~250~000~000}{100}~ \rm{dm}^2
\cdot \left( \dfrac{ 1~\rm{m} } { 10~\rm{dm} } \right)
\cdot \left( \dfrac{ 1~\rm{m} } { 10~\rm{dm} } \right)
&=& \dfrac{4~250~000~000}{10~000}~ \rm{m}^2\\\\
\dfrac{4~250~000~000}{10~000}~ \rm{m}^2
\cdot \left( \dfrac{ 1~\rm{km} } { 1~000~\rm{m} } \right)
\cdot \left( \dfrac{ 1~\rm{km} } { 1~000~\rm{m} } \right)
&=& \dfrac{4~250~000~000}{10~000~000~000}~ \rm{km}^2\\\\
\dfrac{4~250~000~000}{10~000~000~000}~ \rm{km}^2
&=& \dfrac{4~250}{10~000} ~ \rm{km}^2
= 0,425~ \rm{km}^2
\end{array}
$}}\\\\\\
\boxed{\mathbf{4~250~000~000~ \rm{cm}^2 =0,425~ \rm{km}^2 }}$$

how do you convert degrees to radians ?

Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that the four chords form the sides of a convex quadrilateral ?

Let the six points 1, 2, 3, 4, 5, 6

All convex quadrilateral  are all selection(4 Numbers)  in ascending order from the set 1, 2, 3, 4, 5, 6

$$\rm{convex~quadrilateral } = \bordermatrix{Points& 1 & 2 & 3 & 4 & 5 & 6 \cr
1. &1 &2 &3 &4 & & \cr
2. &1 &2 &3 & &5 & \cr
3. &1 &2 &3 & & &6 \cr
4. &1 &2 & &4 &5 & \cr
5. &1 &2 & &4 & &6 \cr
6. &1 &2 & & &5 &6 \cr
7. &1 & &3 &4 &5 & \cr
8. &1 & &3 &4 & &6 \cr
9. &1 & &3 & &5 &6 \cr
10.&1 & & &4 &5 &6 \cr
11.& &2 &3 &4 &5 & \cr
12.& &2 &3 &4 & &6 \cr
13.& &2 &3 & &5 &6 \cr
14.& &2 & &4 &5 &6 \cr
15.& & &3 &4 &5 &6 \cr} = \binom64$$

Point-connections:

$$\small{\text{$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
&1&2&3&4&5&6 \\
\hline
1&x&1\Rightarrow2&1\Rightarrow3&1\Rightarrow4&1\Rightarrow5&1\Rightarrow6 \\
2& &x&2\Rightarrow3&2\Rightarrow4&2\Rightarrow5&2\Rightarrow6 \\
3& & &x&3\Rightarrow4&3\Rightarrow5&3\Rightarrow6 \\
4& & & &x&4\Rightarrow5&4\Rightarrow6 \\
5& & & & &x&5\Rightarrow6 \\
6& & & & & &x \\
\hline
\end{array}
$}} = \binom62=15$$

Four  chords from 15 pont-connections

  $$=\binom{\binom62}{4}=\binom{15}{4}$$

the probability that the four chords form the sides of a convex quadrilateral

 $$\small{\text{$
\begin{array}{rcl}
&=&\dfrac{\binom64}{\binom{15}{4}}
=\dfrac{ \dfrac{6!}{4!\cdot(6-4)!} }{ \dfrac{15!}{4!\cdot(15-4)!} }
=\dfrac{6!}{2!} \cdot \dfrac{11!}{15!}
=\dfrac{3\cdot 4\cdot 5\cdot 6}{ 12\cdot 13 \cdot 14 \cdot 15}
=\dfrac{36}{3276}
=0.01098901099
\end{array}
$}}$$

I have a triangle with C = 90 degrees and a c = 166 cm and last but not least an a = 130,96 cm 

which formulas and calculations do i have to do to find out the missing lengths (A,b, B) ?

$$\\\small{\text{$
\begin{array}{rcl}
\sin{(A)}&=&\dfrac{a}{c}\\\\
\sin{(A)}&=&\dfrac{ 130.96~\rm{cm} }{ 166.00~\rm{cm} }\\\\
\sin{(A)}&=&0.78891566265\\\\
A&=&52.0842936445\ensurement{^{\circ}}
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
B&=& 90\ensurement{^{\circ}} - A\\\\
B&=& 90\ensurement{^{\circ}} - 52.0842936445\ensurement{^{\circ}}\\\\
B&=& 37.9157063555\ensurement{^{\circ}}\\\\
\end{array}
$}}\\\\
\\\small{\text{$
\begin{array}{rcl}
\sin{(B)}&=&\dfrac{b}{c}\\\\
b &=& c\cdot \sin{(B)}\\\\
b &=& 166.00~\rm{cm} \cdot \sin{(37.9157063555\ensurement{^{\circ}})}\\\\
b &=& 166.00~\rm{cm} \cdot 0.61450148676 \\\\
b &=& 102.01~\rm{cm} \\\\
\end{array}
$}}$$