Posts tagged with: circles

60° is 1/6 of the circle (which has 360° of arc in total), so the length of the minor arc will be 1/6 of the circumference. 1/6 of 12π is 2π.

from Tumblr https://ift.tt/2JsMj92

Start by drawing it!

Note that OC = 5 and OD = 5 because both of those are also radii. Note also that because chord CD is perpendicular to OB, it’s bisected by OB. In other words, it’s split into 2 segments each measuring 4.

 

Things are really coming together! Because we know our Pythagorean triples when we see them, we know that we’ve got 3-4-5 triangles here. Therefore, the sine of angle OCD is 3/5.

from Tumblr https://ift.tt/2zgoxIQ

First, a circle inscribed in a square looks like this:

image

If that square has an area of 2, that means each of its sides has a length of √2. And THAT means that the radius of the circle is √2/2.

image

The area of a circle is πr^2. Plug √2/2 in for r and you’ve got your answer:

image

from Tumblr https://ift.tt/2vOO5Ll

Test 7 Section 4 #34

It’s important to know this about circles: many circle questionsespecially ones that deal with sectors like this one, are really just proportion questions. In this case, we know that every circle has 360° of arc around its center. The shaded portion represents 100°, so the proportion of the circle that is shaded is \dfrac{100}{360}=\dfrac{5}{18}.

Test 4 Section 4 Number 24

Because the question tells you that segment AB is a diameter, you know that arc ADB is exactly half the circumference of the circle. If the length of the arc is 8π, therefore, the full circumference of the circle must be 16π.

How do we solve for the radius if we know the circumference? We use the formula for circumference (provided at the beginning of the section if you don’t have it memorized).

    \begin{align*}C&=2\pi r\\16\pi&=2\pi r\\16&=2r\\8&=r\end{align*}

Test 7 Section 3 #18 please

To convert between radians and degrees, you just need to remember that a full circle, 360°, is the same as 2π radians. You can use that fact to solve any degree/radian conversion question with a simple ratio:

    \begin{align*}\dfrac{\text{angle in degrees}}{360^\circ}&=\dfrac{\text{angle in radians}}{2\pi}\end{align*}

In this case, we take the 720° we’re given and solve thusly:

    \begin{align*}\dfrac{720^\circ}{360^\circ}&=\dfrac{\text{angle in radians}}{2\pi}\\\\2&=\dfrac{\text{angle in radians}}{2\pi}\\\\4\pi&=\text{angle in radians}\end{align*}

That tells you that the a in  is equal to 4.

Practice question for Circles, Radians, and a Little More Trigonometry, #5, p. 272, 4th Ed.

Here’s the question:

So the first thing you’ll want to do is draw in a strategically placed radius (or two) and label the lengths you know.

Because we know that M is the midpoint of \overline{AB}, we can actually cut that 5\sqrt{3} in half: AM=\dfrac{5\sqrt{3}}{2} andBM=\dfrac{5\sqrt{3}}{2}.

We also know that we can always draw a perpendicular bisector from the center of a circle to the midpoint of a chord, so you know that by drawing in \overline{OM}, you’re creating two right triangles: AOM and BOM.

Let’s just look at triangle AOM.

At this point, we can certainly use the Pythagorean Theorem to solve, but the presence of the \sqrt{3} should have activated our Spidey-senses from the beginning: this is a 30°-60°-90° triangle! So we can also just see how the numbers we know fit into the known 1:\sqrt{3}:2 ratio. The hypotenuse in a 30°-60°-90° is double the length of the short leg. Our hypotenuse is 5, so the short leg we seek must be half that: \dfrac{5}{2}. We know we’re right because the long leg in a30°-60°-90° is the short leg times \sqrt{3}, and sure enough: \dfrac{5}{2}\times\sqrt{3}=\dfrac{5\sqrt{3}}{2}. So we have our answer. OM=\dfrac{5}{2}.

Gettin’ Pythaggy wit it:

    \begin{align*}OM^2+\left(\dfrac{5\sqrt{3}}{2}\right)^2&=5^2\\\\OM^2+\dfrac{25(3)}{4}&=25\\\\OM^2+\dfrac{75}{4}&=\dfrac{100}{4}\\\\OM^2&=\dfrac{25}{4}\\\\OM&=\dfrac{5}{2}\end{align*}

Test 8 Section 3 #20

This question is asking you to figure out the measure of a circle’s central angle from information about an arc length. This is a piece of cake if you remember that arc length, sector area (a sector is like a pizza slice shape), and central angle are all tied together via proportion:

    \begin{align*}\dfrac{\text{central angle measure}}{360^\circ}=\dfrac{\text{arc length}}{\text{circumference}}=\dfrac{\text{sector area}}{\text{circle area}}\end{align*}

Lots of circle questions (see some examples in the style of the old SAT here) will simply ask you to solve variations on that proportion.

In this case, when the question tells you that the length of arc BC is \dfrac{2}{5} of the circle’s circumference, it’s giving you the middle fraction! So we can solve for the value of x easily:

    \begin{align*}\dfrac{\text{central angle measure}}{360^\circ}&=\dfrac{\text{arc length}}{\text{circumference}}\\\\\dfrac{x^\circ}{360^\circ}&=\dfrac{2}{5}\\\\x^\circ&=144^\circ\end{align*}

Can you explain #9 on pg 273 in the PWN book please?

Sure. Here it is:

Break this into a few steps. First, note that when a wheel rolls without slipping, it travels a distance of one circumference when it makes one complete revolution. Since this wheel has a radius of 12 cm, it would travel 24π cm if it made one complete revolution. Of course, it’s not going to go quite as far.

Now, note that the wheel needs only to turn 30° for the line to be parallel to the ground. (Not obvious? Draw a line parallel to the ground through the wheel’s center.)

Once you turn that first 30°, you only need to turn 90° more to get the line to be perpendicular to the ground. So in total, the wheel will turn 30° + 90° = 120°.

120° is only one third of 360°, so the wheel only makes one third of a complete revolution. Therefore, it only travels one third of its complete circumference of 24π. One third of 24π is 8π, so the answer you’d grid in would be 8.

Hi Mike,

Can you please do practise test 7, section 4, question 29 (Whats the fastest way to do it which doesn’t involve using the distance formula for every answer option)

Yeah, NO distance formula! The way to get this one is to know the standard formula for a circle and to recognize that it’s basically given to you here. The standard formula for a circle of radius r centered at (hk) is (x-h)^2+(y-k)^2=r^2.

Therefore, (x+3)^2+(y-1)^2=5^2 means you have a circle of radius 5 centered at (–3, 1). Once you know the center, you shouldn’t have to do the distance formula to recognize that (3, 2) is more than 5 units away from (–3, 1).

Hi Mike,

Can you tell me how to do #27 (Section 4, Practice Test 6). I know how to do it by completing the square but was wondering if there is an easier way.

thanks!
Rachel

If you have a calculator that can graph the given equation without rearranging it (like an Nspire) then the easiest way is just to do that.

       

If you don’t have one of those, though, then completing the square is 100% the way to go.

    \begin{align*}2x^2-6x+2y^2+2y&=45\\\\x^2-3x+y^2+y&=\dfrac{45}{2}\\\\x^2-3x+\dfrac{9}{4}+y^2+y+\dfrac{1}{4}&=\dfrac{45}{2}+\dfrac{9}{4}+\dfrac{1}{4}\\\\\left(x-\dfrac{3}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2&=25\end{align*}

Of course, since the formula for a circle is (x-h)^2+(y-k)^2=r^2, that 25 on the right side of the equation above tells you the radius is 5.

Test 4, Section 4, Number 36 (Calculator Section)

Great question. Remember that arc length and central angle are related thusly:

    \begin{align*}\dfrac{\text{central angle}}{360^\circ}=\dfrac{\text{arc length}}{\text{circumference}}\end{align*}

You have a circle with radius 10, so its circumference is 2\pi r=20\pi, or about 62.83. If the arc formed by the central angle has a length between 5 and 6, that means it’s between \dfrac{5}{62.83} and \dfrac{6}{62.83} of the full circumference.

    \begin{align*}x_\text{smallest}=360\left(\dfrac{5}{62.83}\right)=28.65\end{align*}

    \begin{align*}x_\text{biggest}=360\left(\dfrac{6}{62.83}\right)=34.38\end{align*}

Therefore, x can be any integer between 28.65 and 34.38: 29, 30, 31, 32, 33, or 34.

Test 3 Section 4 #34

Remember that circle problems are often ratio problems, and this is definitely one of those cases. Generally, we can say that \dfrac{A_{\text{sector}}}{A_{\text{circle}}}=\dfrac{\text{central angle measure}}{360^\circ\text{ or }2\pi}.

Since we’re given the central angle in radians and no other information, we just need to figure out the ratio of the central angle to 2\pi.

    \begin{align*}\dfrac{\left(\dfrac{5\pi}{4}\right)}{2\pi}=\dfrac{5\pi}{8\pi}=\dfrac{5}{8}\end{align*}

You can either grid in that fraction or its decimal equivalent: .625.

Test 6 #20 from the no calculator section

The circle, you’re told, has a radius of 1, so that means its circumference is 2\pi (1) =2\pi. The arc between A and B has a length of \dfrac{\pi}{3}. How much of the circumference is that?

\dfrac{\left(\dfrac{\pi}{3}\right)}{2\pi}=\dfrac{\pi}{3}\times\dfrac{1}{2\pi}=\dfrac{1}{6}

 

Can you please explain College Board Test 5 Section 4 #36?

I understand that angle A is half of x- the only part of the college board explanation I’m not understanding is why we do (360-x) while setting up the equation. Thanks!

Sure. Without looking at CB’s solution, here’s how I do it.

First, as you pointed out, the measure of angle A is half of . That’ll be useful. The other thing I think is useful is to draw segment BC and mark the angles you create (I’ll say their measures are  and ). Now you’ve got two triangles, ABC and PBC.

Each triangle, of course, has angles that add up to 180°. You can write two equations that show this:

Triangle ABC: \dfrac{1}{2}x+20+20+b+c=180

Triangle PBC: x+b+c=180

All you need to do to solve for x is set those left sides equal to each other since they both equal 180 (b and c will cancel right out!).

    \begin{align*}\dfrac{1}{2}x+20+20+b+c&=x+b+c\\\dfrac{1}{2}x+40&=x\\40&=\dfrac{1}{2}x\\80&=x\end{align*}