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In the figure above, if BC is an arc in the circle with center O, then
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08 Dec 2022, 11:33
08 Dec 2022, 11:33
Expert Reply
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Question Stats:
61% (01:55) correct
38% (01:33) wrong based on 18 sessions
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GRE In the figure above, if BC is an arc in the circle with center O.jpg [ 15.75 KiB | Viewed 1934 times ]
In the figure above, if BC is an arc in the circle with center O, then AB − DC =
−10
10
\(2x \)
\(x^2 − 25\)
\(\sqrt{2x^2+50}\)
Re: In the figure above, if BC is an arc in the circle with center O, then
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11 Dec 2022, 04:00
11 Dec 2022, 04:00
Expert Reply
OE
To find AB, find the radius of the circle and then subtract OA. If the radius is r, then AB = r − (x − 5) = r − x + 5. Similarly, DC = r − (x + 5) = r − x − 5. So AB − DC = [r − x + 5] − [r − x − 5] = 10, so the answer is choice (B). If you use POE, you can eliminate choice (A) because you know the answer has to be positive. If you selected choice (E), you selected the radius.
To find AB, find the radius of the circle and then subtract OA. If the radius is r, then AB = r − (x − 5) = r − x + 5. Similarly, DC = r − (x + 5) = r − x − 5. So AB − DC = [r − x + 5] − [r − x − 5] = 10, so the answer is choice (B). If you use POE, you can eliminate choice (A) because you know the answer has to be positive. If you selected choice (E), you selected the radius.
In the figure above, if BC is an arc in the circle with center O, then
[#permalink]
15 Oct 2024, 12:06
15 Oct 2024, 12:06
1
Quote:
In the figure above, if BC is an arc in the circle with center O, then AB − DC =
−10
10
\(2x \)
\(x^2 − 25\)
\(\sqrt{2x^2+50}\)
−10
10
\(2x \)
\(x^2 − 25\)
\(\sqrt{2x^2+50}\)
Attachment:
Screen Shot 2024-10-15 at 3.52.15 PM.png [ 25.9 KiB | Viewed 229 times ]
In the given figure, the base of the rectangle = x+5 and the height of the rectangle = x-5.
(x+5) - (x-5) = 10
Implication:
The base and the height differ by 10.
Since 3:4:5 = 30:40:50, the base and the height will differ by 10 if they are 40 and 30, respectively, as shown in the figure above.
Since the radius in this case = 50, we get:
AB = (radius OB) - (height OA) = 50-30 = 20
DC = (radius OC) - (base OD) = 50-40 = 10
AB - DC = 20-10 = 10
Show: ::
B
