Diameter and Chords - PSAT Math
Card 1 of 56
If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?
If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?
Tap to reveal 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.
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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Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?
Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?
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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.
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.
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The perimeter of a circle is 36 π. What is the diameter of the circle?
The perimeter of a circle is 36 π. What is the diameter of the circle?
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The perimeter of a circle = 2 πr = πd
Therefore d = 36
The perimeter of a circle = 2 πr = πd
Therefore d = 36
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If the area of the circle touching the square in the picture above is 
, what is the closest value to the area of the square?
If the area of the circle touching the square in the picture above is
Tap to reveal answer
Obtain the radius of the circle from the area.
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 
. 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 
.
The area of the square is then 
.
Obtain the radius of the circle from the area.
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
The area of the square is then
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Note: Figure NOT drawn to scale.
In the above circle, the length of arc 
is 
, and 
. What is the diameter of the circle?
Note: Figure NOT drawn to scale.
In the above circle, the length of arc
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Call the diameter 
. Since 
, 
is 
of the circle, and 
is 
of a circle with circumference 
.
is 
in length, so
Call the diameter
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Note: Figure NOT drawn to scale.
In the above circle, the length of arc 
is 10, and 
. Give the diameter of the circle. (Nearest tenth).
Note: Figure NOT drawn to scale.
In the above circle, the length of arc
Tap to reveal answer
Call the diameter 
. Since 
, 
is 
of a circle with circumference 
. Since it is of length 10, the circumference of the circle is 5 times this, or 50. Therefore, set 
in the circumference formula:
Call the diameter
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The circle above has a radius of 
, and the measure of 
is 
. What is the length of chord 
?
The circle above has a radius of
Tap to reveal 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 
side of the model triangle.
Therefore, 
Because x is equal to half of the chord, the answer is 
.
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
Therefore,
Because x is equal to half of the chord, the answer is
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Let 
represent the area of a circle and 
represent its circumference. Which of the following equations expresses 
in terms of 
?
Let
Tap to reveal answer
The formula for the area of a circle is 
, and the formula for circumference is 
. If we solve for C in terms of r, we get
.
We can then substitute this value of r into the formula for the area:
The formula for the area of a circle is
We can then substitute this value of r into the formula for the area:
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The circle above has a radius of , and the measure of is . What is the length of chord ?
The circle above has a radius of
Tap to reveal 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 side of the model triangle.
Therefore,
Because x is equal to half of the chord, the answer is .
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
Therefore,
Because x is equal to half of the chord, the answer is
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The circle above has a radius of , and the measure of is . What is the length of chord ?
The circle above has a radius of
Tap to reveal 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 side of the model triangle.
Therefore,
Because x is equal to half of the chord, the answer is .
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
Therefore,
Because x is equal to half of the chord, the answer is
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The circle above has a radius of , and the measure of is . What is the length of chord ?
The circle above has a radius of
Tap to reveal 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 side of the model triangle.
Therefore,
Because x is equal to half of the chord, the answer is .
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
Therefore,
Because x is equal to half of the chord, the answer is
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If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?
If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?
Tap to reveal 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.
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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Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?
Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?
Tap to reveal 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.
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.
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The perimeter of a circle is 36 π. What is the diameter of the circle?
The perimeter of a circle is 36 π. What is the diameter of the circle?
Tap to reveal answer
The perimeter of a circle = 2 πr = πd
Therefore d = 36
The perimeter of a circle = 2 πr = πd
Therefore d = 36
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If the area of the circle touching the square in the picture above is , what is the closest value to the area of the square?
If the area of the circle touching the square in the picture above is
Tap to reveal answer
Obtain the radius of the circle from the area.
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 . 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 .
The area of the square is then .
Obtain the radius of the circle from the area.
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
The area of the square is then
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Note: Figure NOT drawn to scale.
In the above circle, the length of arc is , and . What is the diameter of the circle?
Note: Figure NOT drawn to scale.
In the above circle, the length of arc
Tap to reveal answer
Call the diameter . Since , is of the circle, and is of a circle with circumference .
is in length, so
Call the diameter
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Note: Figure NOT drawn to scale.
In the above circle, the length of arc is 10, and . Give the diameter of the circle. (Nearest tenth).
Note: Figure NOT drawn to scale.
In the above circle, the length of arc
Tap to reveal answer
Call the diameter . Since , is of a circle with circumference . Since it is of length 10, the circumference of the circle is 5 times this, or 50. Therefore, set in the circumference formula:
Call the diameter
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The circle above has a radius of , and the measure of is . What is the length of chord ?
The circle above has a radius of
Tap to reveal 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 side of the model triangle.
Therefore,
Because x is equal to half of the chord, the answer is .
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
Therefore,
Because x is equal to half of the chord, the answer is
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Let represent the area of a circle and represent its circumference. Which of the following equations expresses in terms of ?
Let
Tap to reveal answer
The formula for the area of a circle is , and the formula for circumference is . If we solve for C in terms of r, we get
.
We can then substitute this value of r into the formula for the area:
The formula for the area of a circle is
We can then substitute this value of r into the formula for the area:
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If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?
If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?
Tap to reveal 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.
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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