Diameter and Chords - PSAT Math

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Question

The circle above has a radius of , and the measure of $\angle AOB$ is $120 ^\circ$ . What is the length of chord ?

Answer

To solve a chord problem, draw right triangles using the chord, the radii, and a line connecting the center of the circle to the chord at a right angle.

Now, the chord is split into two equal pieces, and angle AOB is bisected. Instead of one 120 degree angle, you now have two 30-60-90 triangles. 30-60-90 triangles are characterized by having sides in the following ratio:

So, to find the length of the chord, first find the length of each half. Because the triangles in your circle are similar to the 30-60-90 triangle above, you can set up a proportion. The hypotenuse of our triangle is 6 (the radius of the circle) so it is set over 2 (the hypotenuse of our model 30-60-90 triangle). Half of the chord of the circle is the leg of the triangle that is across from the 60 degree angle (120/2), so it corresponds to the $\sqrt{3}$ side of the model triangle.

$\frac{x}{6}=\frac{\sqrt{3}}{2}$

Therefore, $x=3\sqrt{3}$

Because x is equal to half of the chord, the answer is $6\sqrt{3}$ .

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Question

If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?

Answer

Set the area of the circle equal to four times the circumference πr_2 = 4(2_πr).

Cross out both π symbols and one r on each side leaves you with r = 4(2) so r = 8 and therefore d = 16.

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Question

Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?

Answer

For the circle to contain all 3 vertices, the hypotenuse must be the diameter of the circle. The hypotenuse, and therefore the diameter, is 5, since this must be a 3-4-5 right triangle.

The equation for the area of a circle is A = πr2.

$A=\pi (5/2)^2=6.25\pi$

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Question

The perimeter of a circle is 36 π. What is the diameter of the circle?

Answer

The perimeter of a circle = 2 πr = πd

Therefore d = 36

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Question

If the area of the circle touching the square in the picture above is $81\pi$ , what is the closest value to the area of the square?

Answer

Obtain the radius of the circle from the area.

$A=\pi r^2=81\pi$

Split the square up into 4 triangles by connecting opposite corners. These triangles will have a right angle at the center of the square, formed by two radii of the circle, and two 45-degree angles at the square's corners. Because you have a 45-45-90 triangle, you can calculate the sides of the triangles to be , , and $x\sqrt{2}$ . The radii of the circle (from the center to the corners of the square) will be 9. The hypotenuse (side of the square) must be $9\sqrt{2}$ .

The area of the square is then $(9\sqrt{2})^2=162$ .

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Question

Note: Figure NOT drawn to scale.

In the above circle, the length of arc $\widehat{MYN}$ is $42 \pi$ , and . What is the diameter of the circle?

Answer

Call the diameter . Since , $\widehat{MN}$ is $\frac{60}{360} = \frac{1}{6}$ of the circle, and $\widehat{MYN}$ is $\frac{5}{6 }$ of a circle with circumference $\pi d$ .

$\widehat{MYN}$ is $42 \pi$ in length, so

$\frac{5}{6} \pi d = 42 \pi$

$d = 42 \pi \div \frac{5}{6} \pi =\frac{ 42 \pi }{1}\cdot \frac{6}{5 \pi} = \frac{252}{5 } = 50 \frac{2}{5 }$

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Question

Note: Figure NOT drawn to scale.

In the above circle, the length of arc $\widehat{M N}$ is 10, and . Give the diameter of the circle. (Nearest tenth).

Answer

Call the diameter . Since , $\widehat{MN}$ is $\frac{72}{360} = \frac{1}{5}$ of a circle with circumference $\pi d$ . Since it is of length 10, the circumference of the circle is 5 times this, or 50. Therefore, set in the circumference formula:

$\pi d = C$

$\pi d = 50$

$d = \frac{50 }{\pi } \approx \frac{50 }{3.14159} \approx 15.9$

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Question

Let represent the area of a circle and represent its circumference. Which of the following equations expresses in terms of ?

Answer

The formula for the area of a circle is $A=\pi r^2$ , and the formula for circumference is $C=2\pi r$ . If we solve for C in terms of r, we get
$r=C/2\pi$ .

We can then substitute this value of r into the formula for the area:

$A=\pi r^2$

$=\pi (C/2\pi )^2$

$=C^2\pi /4\pi ^2$

$=C^2/4\pi$

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