The answer is 37. Hope this helps.
Answer:
Part 1) The area of the shaded region is
Part 2) The area of the shaded region is
Step-by-step explanation:
Part 1) Figure N 1
I assume that the figure ABCD is a square
we know that
The area of the shaded region is equal to the area of the square ABCD minus the area of semicircle BC plus the area of semicircle AD
therefore
The area of the shaded region is equal to the area of the square ABCD
The area of the square is
Part 2) Figure N 2
I assume that the triangle ABC is a right isosceles triangle
so
AB=BC
AB ⊥ BC
The area of the shaded region is equal to the area of triangle plus the area of semicircle
A) <em>Find the area of the triangle ABC</em>
The area of triangle is
substitute
B) Find the area of semicircle
The area of semicircle is equal to
we have
-----> the radius is half the diameter
substitute
therefore
The area of the figure is equal to
The lateral surface area is the surface area of the cube excluding the base and the top, which would make 4 square planes.
First we can find the area of a square plane:
4 1/2 x 4 1//2
=20 1/4 ft²
We can then multiply the area of plane by 4:
20 1/4 x 4
=81 ft²
Therefore the answer is 81 ft².
Hope it helps!
Answer:
Width = 2
Length = 16
Step-by-step explanation:
lets start by writing everything we know
Area = 32 sqft
l = 2w + 12
l * w = 32
we can substitute l for 2w + 21 to figure what w equals
(2w + 12)w = 32
distribute the w
2w^2 + 12w = 32
2w^2 + 12w - 32 = 0
use quadratic formula to solve
w = -8, or 2
it can't be -8, so we will check with 2
2(2)^2 + 12(2) - 32 = 0
8 + 24 -32 = 0
32-32 = 0
l = 2(2) + 12
l = 4 + 12
l = 16
2*16 =32
Width = 2
Length = 16