My answer is :

 \(200cm^2\)

:D

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

60 x 50 = 300 (cm²)

60x50=3000cm\(^2\)

hk tốt

độ dài cạnh hình vuông đậm là

     30:2=15cm

S hình vuông đó là

       15x15=225cm\(^2\)

                 đ/s:225cm\(^2\)

The area of the shaded region in the picture is:

                   24x48=1152 ( m2 )

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! *_<

Diện tích của khu vực bóng mờ là:

        24x48=1152 (m2)

                  Đáp số: 1152 m2.

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

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

The area of the square is:

         38 x 38 = 1444(m2)

                  Answer: 1444 m2

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

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

But i also have the answer

           The area of the rectangle is :

                    8 x 9 = 72 ( cm2 )

                              Answer : 72 cm2

But i also have the answer

           The area of the rectangle is :

                    8 x 9 = 72 ( cm2 )

                              Answer : 72 cm2

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

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

9*9=81(cm2)

it is correct can you check and for me

The edge of square is 9cm

\(\Rightarrow\) The area : 9.9=81 (cm^​2)

So the area is 81 cm^​2

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: S_A

S = S_ABC - 3S_A

S_ABC = \(\dfrac{1}{2}\).4.4.\(\dfrac{\sqrt{3}}{2}\) = 4\(\sqrt{3}\)

S_A = \(\dfrac{60}{360}\)S_circles = \(\dfrac{1}{6}\)\(\pi\)2² = \(\dfrac{2\pi}{3}\)

^{=> S = 4\(\sqrt{3}\)} - 3\(\dfrac{2\pi}{3}\) = 4\(\sqrt{3}\) - 2\(\pi\)

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

    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

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, \(area\left(ABC\right)=5+6+7+x+y\)

                             \(=5+6+7+\dfrac{35}{6}+\dfrac{5005}{6}=858\)

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

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)

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  

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

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

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

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

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

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

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

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

 .

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

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

 .