GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If point A is at (0, 8), point C is at (6, 0), and the distance from p
[#permalink]
09 Dec 2022, 05:59
09 Dec 2022, 05:59
Expert Reply
1
Bookmarks
00:00
Question Stats:
66% (03:00) correct
33% (03:28) wrong based on 3 sessions
Hide Show timer Statistics
Attachment:
GRe If point A is at (0, 8), point C is at (6, 0).jpg [ 14.83 KiB | Viewed 1339 times ]
If point A is at (0, 8), point C is at (6, 0), and the distance from point B to point C is x, what is \(\frac{x}{ \sqrt{7} }\) ?
Show: ::
2
Re: If point A is at (0, 8), point C is at (6, 0), and the distance from p
[#permalink]
11 Dec 2022, 04:00
11 Dec 2022, 04:00
Expert Reply
OE
You need to find the length of a leg of a triangle. By finding the lengths of the two other sides, you can use the Pythagorean theorem to find the third side. The hypotenuse of the triangle is the radius of the (quarter-) circle, which, since point A is at (0, 8), is 8. Since point C is (6, 0), the other leg is 6. From the Pythagorean theorem, \(6^2 + x^2 = 8^2\), so \(x^2 = 28\),\( x= \sqrt{28} = \sqrt{4 * 7} = \sqrt{4} * \sqrt{7} = 2 * \sqrt{7}\), so \(\frac{x}{ \sqrt{7} }\)=2 .
You need to find the length of a leg of a triangle. By finding the lengths of the two other sides, you can use the Pythagorean theorem to find the third side. The hypotenuse of the triangle is the radius of the (quarter-) circle, which, since point A is at (0, 8), is 8. Since point C is (6, 0), the other leg is 6. From the Pythagorean theorem, \(6^2 + x^2 = 8^2\), so \(x^2 = 28\),\( x= \sqrt{28} = \sqrt{4 * 7} = \sqrt{4} * \sqrt{7} = 2 * \sqrt{7}\), so \(\frac{x}{ \sqrt{7} }\)=2 .
Re: If point A is at (0, 8), point C is at (6, 0), and the distance from p
[#permalink]
19 Dec 2022, 19:00
19 Dec 2022, 19:00
Carcass wrote:
OE
You need to find the length of a leg of a triangle. By finding the lengths of the two other sides, you can use the Pythagorean theorem to find the third side. The hypotenuse of the triangle is the radius of the (quarter-) circle, which, since point A is at (0, 8), is 8. Since point C is (6, 0), the other leg is 6. From the Pythagorean theorem, \(6^2 + x^2 = 8^2\), so \(x^2 = 28\),\( x= \sqrt{28} = \sqrt{4 * 7} = \sqrt{4} * \sqrt{7} = 2 * \sqrt{7}\), so \(\frac{x}{ \sqrt{7} }\)=2 .
You need to find the length of a leg of a triangle. By finding the lengths of the two other sides, you can use the Pythagorean theorem to find the third side. The hypotenuse of the triangle is the radius of the (quarter-) circle, which, since point A is at (0, 8), is 8. Since point C is (6, 0), the other leg is 6. From the Pythagorean theorem, \(6^2 + x^2 = 8^2\), so \(x^2 = 28\),\( x= \sqrt{28} = \sqrt{4 * 7} = \sqrt{4} * \sqrt{7} = 2 * \sqrt{7}\), so \(\frac{x}{ \sqrt{7} }\)=2 .
What grants us permission to assume the circle is exactly a quarter circle?
