area
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lê đức anh 11/01/2020 at 04:54
nửa chu vi là 60: 2 = 30cm
độ dài mỗi cạnh là 30:(2+1)x2=20cm;20:2=10cm
diện tích là 20x10=200cm2
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Nguyễn Mạnh Hùng 28/08/2018 at 07:53
60 x 50 = 300 (cm2)
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lê đức anh 11/01/2020 at 04:58
độ dài cạnh hình vuông đậm là
30:2=15cm
S hình vuông đó là
15x15=225cm\(^2\)
đ/s:225cm\(^2\)
Huỳnh Anh Phương
16/08/2018 at 14:13
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Phan Minh Anh 17/08/2018 at 00:06
The area of the shaded region in the picture is:
24x48=1152 ( m2 )
Huỳnh Anh Phương selected this answer. -
jimin bts cute 17/08/2018 at 08:33
The area of the shaded region is:
\(48\times24=1152\left(m^2\right)\)
Answer: \(1152m^2\)
mk ko nhanh như bn khác nhưng hãy k mk! *_<
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Phan Minh Anh 17/08/2018 at 00:04
Diện tích của khu vực bóng mờ là:
24x48=1152 (m2)
Đáp số: 1152 m2.
Quoc Tran Anh Le Coordinator
05/08/2018 at 03:47
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Tôn Thất Khắc Trịnh 24/07/2018 at 13:22
OwO, it's a typo: it's suppossed to be width, okay?
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If you use the AMGM theorem, you'll know that any rectangle achieves the maximum possible area when both of their measurements are equal, making it a square (a square is technically a rectangle too)
So we have p=q
Therefore, the equation will be q2+q=13+q2
<=>q=13(cm)
<=>p=q=13(cm)
<=>S=pq=132=169(cm2)
(P.S: don't do this here, cuz you'll get to a dead end, oof)
\(pq+q=13+q^2\)
\(\Leftrightarrow S=q^2-q+13\)
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Uchiha Sasuke 24/07/2018 at 13:34
OMG OMG THANK YOU SO VERY MUCH!!!!!!!!!!!!!!!!!!!!! :DDDDDDDDDDDD
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Fixida 18/04/2018 at 13:19
The area of the square is:
38 x 38 = 1444(m2)
Answer: 1444 m2
huynh anh phuong selected this answer. -
The area of the square is: \(38\cdot38=38^2=1444\left(m^2\right)\)
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¤« 18/04/2018 at 13:03
The area of the square is: 38⋅38=382=1444(m2)
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¤« 16/04/2018 at 14:05
But i also have the answer
The area of the rectangle is :
8 x 9 = 72 ( cm2 )
Answer : 72 cm2
huynh anh phuong selected this answer. -
Fixida 14/04/2018 at 13:25
But i also have the answer
The area of the rectangle is :
8 x 9 = 72 ( cm2 )
Answer : 72 cm2
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Fixida 14/04/2018 at 13:24
I'm the length must be a 9 cm, width is 8 cm
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The area of the square is : 9 x 9 = 81 (cm2)
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Vũ Mạnh Hùng 03/12/2017 at 20:24
9*9=81(cm2)
it is correct can you check and for me
huynh anh phuong selected this answer. -
Trần Quỳnh Anh 27/12/2017 at 19:39
The edge of square is 9cm
\(\Rightarrow\) The area : 9.9=81 (cm2)
So the area is 81 cm2
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Chibi 11/04/2017 at 11:31
Center of the circle: ABC
=> AB = BC = AC = 2R = 4
=> ABC is a equilateral triangle
The area for the shaded region: S
The area for a sector definition by A and 2 tangential points: SA
S = SABC - 3SA
SABC = \(\dfrac{1}{2}\).4.4.\(\dfrac{\sqrt{3}}{2}\) = 4\(\sqrt{3}\)
SA = \(\dfrac{60}{360}\)Scircles = \(\dfrac{1}{6}\)\(\pi\)22 = \(\dfrac{2\pi}{3}\)
=> S = 4\(\sqrt{3}\) - 3\(\dfrac{2\pi}{3}\) = 4\(\sqrt{3}\) - 2\(\pi\)
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FA KAKALOTS 28/01/2018 at 22:09
Center of the circle: ABC
=> AB = BC = AC = 2R = 4
=> ABC is a equilateral triangle
The area for the shaded region: S
The area for a sector definition by A and 2 tangential points: SA
S = SABC - 3SA
SABC = 12
.4.4.√32 = 4√3
SA = 60360
Scircles = 16π22 = 2π3
=> S = 4√3
- 32π3 = 4√3 - 2π
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tth 05/11/2017 at 19:11
Center of the circle: ABC=> AB = BC = AC = 2R = 4
=> ABC is a equilateral triangle
The area for the shaded region: S
The area for a sector definition by A and 2 tangential points: SA
S = SABC - 3SA
SABC = 12
.4.4.√32 = 4√3
SA = 60360
Scircles = 16π22 = 2π3=> S = 4√3
- 32π3 = 4√3 - 2π
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An Duong 09/04/2017 at 07:31
We have \(\dfrac{SP}{PB}=\dfrac{area\left(APS\right)}{area\left(ABP\right)}=\dfrac{5}{6}\)
Call the area of PSR be x, the area of CSR be y, we have:
\(\dfrac{area\left(PSR\right)}{area\left(PBR\right)}=\dfrac{SP}{PB}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{5}{6}\) \(\Rightarrow x=\dfrac{35}{6}\)
\(\dfrac{BR}{CR}=\dfrac{area\left(BSR\right)}{area\left(CRS\right)}=\dfrac{7+x}{y}\) (1)
\(\dfrac{BR}{CR}=\dfrac{area\left(ABR\right)}{area\left(ACR\right)}=\dfrac{13}{x+y+5}\) (2)
(1), (2) => \(\dfrac{7+x}{y}=\dfrac{13}{x+y+5}\)
\(\Rightarrow y=\dfrac{\left(7+x\right)\left(5+x\right)}{6-x}=\dfrac{5005}{6}\)
So, \(area\left(ABC\right)=5+6+7+x+y\)
\(=5+6+7+\dfrac{35}{6}+\dfrac{5005}{6}=858\)
John selected this answer. -
FA KAKALOTS 28/01/2018 at 22:09
A B C R S P 5 6 7 M N x y
We have SPPB=area(APS)area(ABP)=56
Call the area of PSR be x, the area of CSR be y, we have:
area(PSR)area(PBR)=SPPB=56
⇒x7=56
⇒x=356
BRCR=area(BSR)area(CRS)=7+xy
(1)
BRCR=area(ABR)area(ACR)=13x+y+5
(2)
(1), (2) => 7+xy=13x+y+5
⇒y=(7+x)(5+x)6−x=50056
So, area(ABC)=5+6+7+x+y
=5+6+7+356+50056=858
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An Duong 25/03/2017 at 21:31
Because a line through a center of a rectangle (or a circle) divide it into two part with equivalent area.
So you should cut the paper by the line connecting two centers of the rectangle and the circle (see following figure)
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Min Hoang moderators
16/03/2017 at 11:05
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mathlove 16/03/2017 at 18:29
The circle of radius 3 have an area \(9\pi\). We sign r as the radius of the pictured quadrant of the done cỉcle, then
\(r=3\sqrt{2}+3\). Put x is the area to calculate, we have
\(\dfrac{1}{4}\pi r^2=2x+\pi.3^2+\left(3^2-\dfrac{1}{4}.\pi.3^2\right)=2x+9\left(1+\dfrac{3\pi}{4}\right)\)
\(\Leftrightarrow\dfrac{\pi}{4}\left(3\sqrt{2}+3\right)^2=2x+9\left(1+\dfrac{3\pi}{4}\right)\Leftrightarrow2x=\dfrac{\left(27+18\sqrt{2}\right)\pi}{4}-\dfrac{36+27\pi}{4}\)
\(\Leftrightarrow x=\dfrac{9\sqrt{2}\pi-36}{4}\)
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mathlove 17/03/2017 at 10:45
We have \(y=3^2-\left(\dfrac{1}{4}\pi3^2\right)\) and \(r-3=3\sqrt{2}\Rightarrow r=3+3\sqrt{2}\).
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FA KAKALOTS 28/01/2018 at 22:09
The circle of radius 3 have an area 9π
. We sign r as the radius of the pictured quadrant of the done cỉcle, then
r=3√2+3
. Put x is the area to calculate, we have
14πr2=2x+π.32+(32−14.π.32)=2x+9(1+3π4)
⇔π4(3√2+3)2=2x+9(1+3π4)⇔2x=(27+18√2)π4−36+27π4
⇔x=9√2π−364
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mathlove 13/03/2017 at 18:12
Setting x is the are to find. Easy to see that \(\left(1\right)+\left(2\right)+\left(1\right)=1-\dfrac{\pi}{4}\).
According to the previous post: \(\left(1\right)=1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\). So that\(\left(2\right)=\left(1-\dfrac{\pi}{4}\right)-2\left(1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\right)=-1+\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{2}\).
Therefore \(x=\dfrac{\pi}{4}-\left[3.\left(2\right)+2.\left(1\right)\right]=\dfrac{\pi}{4}-\left[\left(-3+\dfrac{\pi}{4}+\dfrac{3\sqrt{3}}{2}\right)+\left(2-\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{3}\right)\right]\)
\(=1+\dfrac{\pi}{3}-\sqrt{3}\) .
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FA KAKALOTS 28/01/2018 at 22:10
Setting x is the are to find. Easy to see that (1)+(2)+(1)=1−π4.
According to the previous post: (1)=1−√34−π6. So that
(2)=(1−π4)−2(1−√34−π6)=−1+π12+√32
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Therefore x=π4−[3.(2)+2.(1)]=π4−[(−3+π4+3√32)+(2−√32−π3)]
=1+π3−√3
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mathlove 11/03/2017 at 18:27
Let x is the area to calculate. We see that EAD is equilateral triangle with the edge equal to 1, the equilateral line equal to \(\dfrac{\sqrt{3}}{2}\) . So \(EF=1-\dfrac{\sqrt{3}}{2}\) .
We have the angle EDC is \(30^0\) , so that \(\dfrac{1}{2}.\dfrac{1}{2}\left(1-\dfrac{\sqrt{3}}{2}\right)-\dfrac{x}{2}=\dfrac{\pi}{12}-\dfrac{1}{2}.1.1.\sin30^0=\dfrac{\pi}{12}-\dfrac{1}{4}\)
So \(x=1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\) .
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FA KAKALOTS 28/01/2018 at 22:10
Let x is the area to calculate. We see that EAD is equilateral triangle with the edge equal to 1, the equilateral line equal to √32 . So EF=1−√32
.
We have the angle EDC is 300
, so that 12.12(1−√32)−x2=π12−12.1.1.sin300=π12−14
So x=1−√34−π6
.