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06 Dec 2021, 03:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
49% (02:16) correct
51%
(02:43)
wrong
based on 37
sessions
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In a circle, two parallel chords on the same side of a diameter have lengths 6 cm and 10 cm. If the distance between these chords is 2 cm, then the radius of the circle, in cm, is
A. \(5\)
B. \(4\sqrt{2}\)
C. \(\sqrt{34}\)
D. \(\sqrt{41}\)
E. \(6\sqrt{2}\)
Are You Up For the Challenge: 700 Level Questions
A. \(5\)
B. \(4\sqrt{2}\)
C. \(\sqrt{34}\)
D. \(\sqrt{41}\)
E. \(6\sqrt{2}\)
Are You Up For the Challenge: 700 Level Questions
16 Dec 2021, 21:34
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Concept: a chord drawn from the center of a circle, perpendicular to another chord, will bisect that other chord
Chord of length 10 must be closer to the diameter than chord of length 6
Let chord of length 6 = AB
Let chord length 10 = XY
Draw center O beneath both chords
Since chords AB and XY are parallel, we can draw a perpendicular line from center O that intersects each chord’s midpoint
This line starting from O drawn perpendicular to each chord bisects AB at point D and bisects XY at point Z
The distance between the chords = perpendicular distance between the 2 parallel lines
Therefore, on this line OZD ——> the portion ZD = 3
The portion OZ is unknown.
Finally, we can draw two radii in order to create two right triangles. One radius drawn from the center to the circumference at point X. And another radius drawn from the center to the circumference at point A.
This gives us two right triangles and we can apply the Pythagorean Theorem.
(1st) right triangle OAD
Since AD is the Bisected part of the Chord of length 6 ——-> AD = 3
Length from D to the center O = OZD = (OZ) + (2)
*where 2 is the perpendicular distance between the two chords
And finally the radius = R
(R)^2 = (OZ + 2)^2 + (3)^2 —— (eq 1)
(2nd) right triangle OXZ can be created
XZ is the bisected portion of the chord of length 10. Thus, XZ = 5
OZ is unknown
And the radius = OX = R
(R)^2 = (OD)^2 + (5)^2 —— (eq 2)
Set (eq 1) = (eq 2) in order to find the value of OZ
Let OZ’s length = U
(U + 2)^2 + (3)^2 = (U)^2 + (5)^2
U^2 + 4U + 4 + 9 = U^2 + 25
4U = 12 ——-> U = 3 = length of OZ
We can plug this into either Pythagorean Theorem formula to find the length of R - radius
(3)^2 + (5)^2 = (R)^2
R = sqrt(34)
Answer C
Posted from my mobile device
Chord of length 10 must be closer to the diameter than chord of length 6
Let chord of length 6 = AB
Let chord length 10 = XY
Draw center O beneath both chords
Since chords AB and XY are parallel, we can draw a perpendicular line from center O that intersects each chord’s midpoint
This line starting from O drawn perpendicular to each chord bisects AB at point D and bisects XY at point Z
The distance between the chords = perpendicular distance between the 2 parallel lines
Therefore, on this line OZD ——> the portion ZD = 3
The portion OZ is unknown.
Finally, we can draw two radii in order to create two right triangles. One radius drawn from the center to the circumference at point X. And another radius drawn from the center to the circumference at point A.
This gives us two right triangles and we can apply the Pythagorean Theorem.
(1st) right triangle OAD
Since AD is the Bisected part of the Chord of length 6 ——-> AD = 3
Length from D to the center O = OZD = (OZ) + (2)
*where 2 is the perpendicular distance between the two chords
And finally the radius = R
(R)^2 = (OZ + 2)^2 + (3)^2 —— (eq 1)
(2nd) right triangle OXZ can be created
XZ is the bisected portion of the chord of length 10. Thus, XZ = 5
OZ is unknown
And the radius = OX = R
(R)^2 = (OD)^2 + (5)^2 —— (eq 2)
Set (eq 1) = (eq 2) in order to find the value of OZ
Let OZ’s length = U
(U + 2)^2 + (3)^2 = (U)^2 + (5)^2
U^2 + 4U + 4 + 9 = U^2 + 25
4U = 12 ——-> U = 3 = length of OZ
We can plug this into either Pythagorean Theorem formula to find the length of R - radius
(3)^2 + (5)^2 = (R)^2
R = sqrt(34)
Answer C
Posted from my mobile device
17 Sep 2024, 02:31
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Here's an easier approach:
You can easily condense and organise the information as shown in the image attached. You get:
\((a + 2)^2 + 3^2 = r^2\) -- (1)
\(a^2 + 5^2 = r^2\) -- (2)
Subtract equation 1 from 2 to get value of \(a = 3\), then plug this value in eq. 2 to get \(r = \sqrt{34}\)
You can easily condense and organise the information as shown in the image attached. You get:
\((a + 2)^2 + 3^2 = r^2\) -- (1)
\(a^2 + 5^2 = r^2\) -- (2)
Subtract equation 1 from 2 to get value of \(a = 3\), then plug this value in eq. 2 to get \(r = \sqrt{34}\)
Attachments
Screenshot 2024-09-17 at 14.55.51.png [ 128.79 KiB | Viewed 1280 times ]
