All the numbers listed are rational, since they can all be represented as a fraction, EXCEPT "2 in the house thingy", which I suppose is $\sqrt{2}$ .
This last one is irrational, because it cannot be represented as a finite fraction.

The following is a check to make sure:
1/3 is already a fraction
2/7 is already a fraction
2 in the house thingy = $\sqrt{2}$ cannot be converted into a finite fraction.
3.624 = 3624/1000 is a fraction
0.6 repeat = 2/3 is a fraction
0.87 repeat = 87/99 = 29/33 is a fraction
-5 = -5/1 is a fraction.

8 0

3 years ago

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Figure 6 shows a semicircle PTS with center O and radius 8cm. QST is a sector of a circle with center S and R is the midpoint of

sveticcg [70]

(a) <TOR=pi/3 radians

To determine <TOR we use the fact that in the right-angled triangle ORT we know two sides:

|OT|=radius=8cm and |OR|=radius/2=4cm

and can use the sine:

$\sin \angle OTR=\frac{r/2}{r}=\frac{1}{2}\implies \angle OTR =\frac{\pi}{6}$

and since <TRO=pi/2, it must be that

$\angle TOR =\pi-\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$

(b) The arc length is approximately 7.255 cm

In order to calculate the arc length QT, we need to first determine the length |ST| and the angle <OST.

Towards determining angle <OST:

$\angle SOT = \pi - \angle TOR = \pi - \frac{\pi}{3} = \frac{2}{3}\pi$

Next, draw a line connecting P and T. Realize that triangle PTS is right-angled with <PTS=pi/2. This follows from the Thales theorem. Since R is a midpoint between P and O, it follows that the triangles ORT and PRT are congruent. So the angles <PTR and <OTR are congruent. Knowing <PTS we can determine angle <OTS:

$\angle OTR \cong \angle PTR=\frac{\pi}{6}\implies\angle OTS=\angle PTS -\angle PTR -\angle OTR\\\angle OTS = \frac{\pi}{2}-\frac{\pi}{6}-\frac{\pi}{6}=\frac{\pi}{6}$

and so the angle <OST is

$\angle OST = \pi - \angle TOS - \angle OTS = \pi -\frac{2}{3}\pi - \frac{1}{6}\pi=\frac{\pi}{6}$

Towards determining |TS|:

Use cosine:

$\cos \angle OST =\frac{|RS|}{|ST|}\implies |ST|=\frac{\frac{3}{2}r}{\cos \frac{\pi}{6}}=\frac{12\cdot 2}{\sqrt{3}}=8\sqrt{3}cm$

Finally, we can determine the arc length QT:

$QT = {\angle OST}\cdot |ST|=\frac{\pi}{6}\cdot 8 \sqrt{3}=\frac{4\pi}{\sqrt{3}}\approx 7.255cm$

3 0

3 years ago