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area

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huynh anh phuong
29/06/2020 at 11:12
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A rectangle has its length is 8cm longer than its width. If we double the width, its new length is still 8cm longer its new width. What is the smallest possible area of that rectangle ?

area


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Huỳnh Anh Phương
23/11/2018 at 09:47
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In the rectangle ABCD, segement AB is twice segement BC. The perimeter of the rectangle  ABCD  is 60cm. Find the area of the rectangle ABCD?

areaperimetergeometry

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    FacuFeri 24/11/2018 at 09:45

    My answer is :

     \(200cm^2\)

    :D

    Huỳnh Anh Phương selected this answer.
  • ...
    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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Huỳnh Anh Phương
24/08/2018 at 03:42
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Find the area of the shaded region in the shape below:

2m 50cm 60cm

areageometry

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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 04:56

    60x50=3000cm\(^2\)

    hk tốt


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Huỳnh Anh Phương
17/08/2018 at 13:33
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After Lina used the box, she ranks the box as shown. If each edge is 30cm long, find the area on the shadow region? Express your answer in 2 ways. 

30cm

areageometry

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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\(^2\)


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Huỳnh Anh Phương
16/08/2018 at 14:13
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What is the area of the shaded region in the picture below?

48m 24m

area

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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! *_<

  • ...
    Phan Minh Anh 17/08/2018 at 00:04

    Diện tích của khu vực bóng mờ là:

            24x48=1152 (m2)

                      Đáp số: 1152 m2.


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Quoc Tran Anh Le Coordinator
05/08/2018 at 03:47
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One circular base of a cylinder with radius 2 inches and height 4 inches is glued flush to the center of each face of a 4-inch cube. What is the surface area of the resulting solid? Express your answer as a decimal to the nearest tenth.

Mathcountsarea


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LOL
24/07/2018 at 13:48
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The trapezoid ABCD has \(\widehat{A}=\widehat{B}=90^o\), \(AB^2=AD\times BC\) and \(AC^2+BD^2=64\). Find the greatest possible area of the trapezoid ABCD.

area


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Uchiha Sasuke
24/07/2018 at 02:12
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A rectangle has length p cm and breadth q cm, where p and q are integers and p and q satisfy the equation pq + q = 13 + q2 then the maximum possible area of the rectangle is...

area

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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?
    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)
    \(pq+q=13+q^2\)
    \(\Leftrightarrow S=q^2-q+13\)
     

    Selected by MathYouLike
  • ...
    Uchiha Sasuke 24/07/2018 at 13:34

    OMG OMG THANK YOU SO VERY MUCH!!!!!!!!!!!!!!!!!!!!! :DDDDDDDDDDDD


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huynh anh phuong
18/04/2018 at 10:46
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What is area of the square?

38m

area

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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.
  • ...
    Lê Quốc Trần Anh Coordinator 18/04/2018 at 10:50

    The area of the square is: \(38\cdot38=38^2=1444\left(m^2\right)\)

  • ...
    ¤« 18/04/2018 at 13:03

    The area of the square is: 38⋅38=382=1444(m2)


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huynh anh phuong
14/04/2018 at 02:52
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The area of the rectangle is ... cm2.

8cm 9cm

area

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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

  • ...
    Fixida 14/04/2018 at 13:24

    I'm the length must be a 9 cm, width is 8 cm


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huynh anh phuong
03/12/2017 at 17:35
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The edge of square is 9cm. Find the area?

area

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    Phan Thanh Tinh Coordinator 03/12/2017 at 17:48

    The area of the square is : 9 x 9 = 81 (cm2)

    Selected by MathYouLike
  • ...
    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 (cm​2)

    So the area is 81 cm​2


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london
11/04/2017 at 10:44
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 Given three circles of radius 2, tangent to each other as shown in the following diagram, what is the area for the shaded region? 

undefined

area

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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\(\pi\)

    Selected by MathYouLike
  • ...
    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

  • ...
    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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John
08/04/2017 at 19:59
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Let R and S be points on the sides BC and AC , respectively, of ΔABC , and let P be the intersection of AR and BS . Determine the area of ΔABC if the areas of ΔAPS , ΔAPB , and ΔBPR are 5, 6, and 7, respectively

area

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    An Duong 09/04/2017 at 07:31

    A B C R S P 5 6 7 M N x y

    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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John
25/03/2017 at 16:44
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Given a rectangle paper with a circle hole as the figure below. How to cut the paper with a line so that we have two parts with equal area.

area

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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)

    Selected by MathYouLike

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Min Hoang moderators
16/03/2017 at 11:05
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A circle of radius 3 is inscribed in the pictured quadrant of a circle. Find the area of the shaded section. 

area

  • ...
    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  

                   \(\dfrac{1}{4}\pi r^2=2x+\pi.3^2+\left(3^2-\dfrac{1}{4}.\pi.3^2\right)=2x+9\left(1+\dfrac{3\pi}{4}\right)\)

          ​ ​\(\Leftrightarrow\dfrac{\pi}{4}\left(3\sqrt{2}+3\right)^2=2x+9\left(1+\dfrac{3\pi}{4}\right)\Leftrightarrow2x=\dfrac{\left(27+18\sqrt{2}\right)\pi}{4}-\dfrac{36+27\pi}{4}\)

           \(\Leftrightarrow x=\dfrac{9\sqrt{2}\pi-36}{4}\)

    Selected by MathYouLike
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    mathlove 17/03/2017 at 10:45

    x x y

    We have    \(y=3^2-\left(\dfrac{1}{4}\pi3^2\right)\) and  \(r-3=3\sqrt{2}\Rightarrow r=3+3\sqrt{2}\).

    Selected by MathYouLike
  • ...
    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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John
12/03/2017 at 18:39
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Continuing the previous post:

Question by John - Discuss with MathYouLike

I have another problem: Calculate the area of the curved square below (crossed area):

1

area

  • ...
    mathlove 13/03/2017 at 18:12

    undefined


    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       \(x=\dfrac{\pi}{4}-\left[3.\left(2\right)+2.\left(1\right)\right]=\dfrac{\pi}{4}-\left[\left(-3+\dfrac{\pi}{4}+\dfrac{3\sqrt{3}}{2}\right)+\left(2-\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{3}\right)\right]\)

                                                                       \(=1+\dfrac{\pi}{3}-\sqrt{3}\) .

    Selected by MathYouLike
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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

    .

    Therefore       x=π4−[3.(2)+2.(1)]=π4−[(−3+π4+3√32)+(2−√32−π3)]

                                                                       =1+π3−√3

     .


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John
10/03/2017 at 15:29
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Calculate the crossed area in the figure below:

1

area

  • ...
    mathlove 11/03/2017 at 18:27

    undefined

    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   \(x=1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\) .

                                           

    John selected this answer.
  • ...
    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

     .


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