Question

Answer that please! Thanks

Answer 1

Hey here is the answer to ur question....!
Given:
AC=2
BC=2
BE=2
DE=2
DF=4
Angle DCB=angleCBE=angleBED
To find: area of the given figure
Solution: in BEDC,
Since BE=2
=>CD=2 (Opposite sides are equal in a square)
Therefore ,area of square BEDC=s*s=2*2=4 units^2
In triangle ABC,
AC=CB=2
=>area of triangle =1/2*b*h
=>area of triangle ABC=1/2*2*2
=>area of triangle =2 units^2
In triangle DEF,
DF=4
DE=2
=>area of triangle DEF=1/2*4*2
=>area of triangle DEF=4units ^2
Therefore ,area of the given figure ABEF=area of triangle ABC+area of triangle DEF+area of square BEDC
=>2+4+4
=>area of the given figure ABEF=10 units^2
Hope this helps u.........!!!!!!!

Answer 2

One way to find the area of this shape is to add up the areas of all of the interior shapes. To find the area of a triangle, we use the formula A=1/2bh. To find the area of a square, we use the formula A = s^2.

We will begin by finding the area of the triangle on the left side of the figure.

A = 1/2bh= 1/2(2)(2)= 1/2(4)= 2 units^2

Then we can do the same for the triangle on the right side of the figure.

A = 1/2bh = (1/2)(4)(2) = (1/2)(8) = 4 units^2

Finally, we can find the area of the inner square.

A = s^2 = 2^2 = 4 units^2

Now, we must add up the areas of these smaller shapes to find the total area.

A = 2 + 4 + 4 = 10 units^2

Therefore, your answer is 10 units^2.

Hope this helps!