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.
On a rectangular coordinate plane, a circle centered at (0,
[#permalink]
06 Oct 2020, 10:39
06 Oct 2020, 10:39
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
46% (02:48) correct
53% (02:38) wrong based on 39 sessions
Hide Show timer Statistics
On a rectangular coordinate plane, a circle centered at (0, 0) is inscribed within a square with adjacent vertices at (\(0, −2 \sqrt{2}\)) and (\(2 \sqrt{2}, 0\)). What is the area of the region, rounded to the nearest tenth, that is inside the square but outside the circle?
Show: :: OA
3.4
Re: On a rectangular coordinate plane, a circle centered at (0,
[#permalink]
06 Oct 2020, 20:33
06 Oct 2020, 20:33
2
1
Bookmarks
side of sqaure =4=diameter of circle
area of square =16
area of circle=4π
area of region that is inside square and outside circle=16-4π=3.42
answer is 3.4
area of square =16
area of circle=4π
area of region that is inside square and outside circle=16-4π=3.42
answer is 3.4
Re: On a rectangular coordinate plane, a circle centered at (0,
[#permalink]
06 Jul 2023, 04:30
06 Jul 2023, 04:30
1
Since the vertices of the square containing the circle are on 0,-2\sqrt{2} root 2 and 2\sqrt{2},0, it means that if a line is drawn to the two points we will have one edge of the square.
The length between each of the two vertices to the origin are both equal and are 2\sqrt{2}.
That means that they are two equal sides of an isosceles 1,1,\sqrt{2} triangle where the length from the vertices to the origin are the 1,1, equal sides.
Hence the \sqrt{2} side of the 1,1, \sqrt{2} triangle is 2\sqrt{2} * \sqrt{2} which is 4.
Hence all the sides of the square have length, L = 4.
Since the circle is inscribed in the square, then the diameter, D = 4 and the radius, r = 2.
Area of the square = L raised to power 2 =
L * L = 4*4
Area of the circle is pi * r raised to power 2 =
pi * r * r = 3.142 * 2 * 2 = 3.142 * 4
Area of the region outside the circle but in the square =
4*4 - 3.142 * 4 = (4 - 3.142) * 4 = 0.858*4
= 3.432 approximately 3.4 to the nearest tenth.
Please give me a kudos if my answer helps.
The length between each of the two vertices to the origin are both equal and are 2\sqrt{2}.
That means that they are two equal sides of an isosceles 1,1,\sqrt{2} triangle where the length from the vertices to the origin are the 1,1, equal sides.
Hence the \sqrt{2} side of the 1,1, \sqrt{2} triangle is 2\sqrt{2} * \sqrt{2} which is 4.
Hence all the sides of the square have length, L = 4.
Since the circle is inscribed in the square, then the diameter, D = 4 and the radius, r = 2.
Area of the square = L raised to power 2 =
L * L = 4*4
Area of the circle is pi * r raised to power 2 =
pi * r * r = 3.142 * 2 * 2 = 3.142 * 4
Area of the region outside the circle but in the square =
4*4 - 3.142 * 4 = (4 - 3.142) * 4 = 0.858*4
= 3.432 approximately 3.4 to the nearest tenth.
Please give me a kudos if my answer helps.
