All categories

3 years ago

NO LINKS OR REPORT Please help! Find the area of the figure

Mathematics

2 answers:

Oduvanchick [21]3 years ago

4 0

Answer:

$Aear=2450cm^{2}$

Step-by-step explanation:

The figure is made up of a rectangle, parallelogram, and square.

Area of the rectangle is 50(15)=750 cm²

Area of the parallelogram is 40 (40)=1600cm²

Area of the square is 10(10)=100 cm²

The total area is 750+1600+100= 2450 cm²

$-------------------$

hope it helps..

have a great day!!!

dalvyx [7]3 years ago

3 0

Answer:

Report this if u do i wont care all it does is gives me a warning and if i do it some more it will delete my account and i dont care if it gets deleted because it wont effect me in real life so report it

Step-by-step explanation:

im just spitting facts my guy

You might be interested in

If the circle has the same diameter as the edge length of the square, then the area of this circle is ___________the area of the

WINSTONCH [101]

Answer:

The area of this circle is $(\frac{\pi}{2} )$ the area of the square.

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Therefore, Φsquare is $(\frac{2}{\pi} )$ ϕcircle

Step-by-step explanation:

Area of the circle is given by;

$A_c = \frac{\pi d^2}{4}$

Area of the square is given by;

$A_s = L^2$

relationship between the edge length of the square, d, and length of its side, L,

$d = \sqrt{L^2 + L^2} \\\\d = \sqrt{2L^2}$

But area of the square , $A_s = L^2$

$d = \sqrt{2A_s}$

Then, the area of the square in terms of the edge length is given by;

$A_s = \frac{d^2}{2}$

Area of the circle in terms of area of the square is given by;

$A_c = \frac{\pi d^2}{4} = \frac{\pi}{2}(\frac{d^2}{2} )\\\\But \ A_s = \frac{d^2}{2} \\\\A_c = \frac{\pi}{2}(\frac{d^2}{2} )\\\\A_c = \frac{\pi}{2}(A_s )$

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Ф = E.A

Flux through the surface of the circle is given by;

$\phi _{circle} = E.(\frac{\pi d^2}{4})$

Flux through the surface of the square is given by;

$\phi _{square} = E.(\frac{d^2}{2} )\\\\\phi _{square} =E.(\frac{d^2}{2} ).(\frac{\pi}{2} ).(\frac{2}{\pi} )\\\\\phi _{square} =E.(\frac{\pi d^2}{4} ).(\frac{2}{\pi} )\\\\\phi _{square} =(\phi _{circle}).(\frac{2}{\pi} )$