What would be the surface area of the following object if it we dilate it by a factor of 2?
The surface area of a cube is:
S=6 A
Where A is the area of each face. Since each face of the cube is a square, we have:
A=a^2
Where a is the length of an edge. Dilating the cube by a factor of 2, the length of each edge of the new cube is 6. Therefore:
- a=6
- A=6^2=36
Therefore the surface area of the new cube is:
S= 6 \times 36
S=216
What would be the volume of the following object if it we dilate it by a factor of \dfrac{1}{2} ?
The surface area of a cuboid is with length a width b and height c equals:
S= a \times b \times c
Dilating the cube by a factor of \dfrac{1}{2}, we have:
- a=1
- b=2
- c=\dfrac{5}{2}
Therefore the volume of the new cube is:
V= 1 \times 2 \times \dfrac{2}{5} = 5
S=5
What would be the surface area of the following sphere if it we dilate it by a factor of 3?
The surface area of a cube is:
S=4 \pi r^2
Where r is the radius of the sphere. Dilating the sphere by a factor of 3, the radius of the new sphere is 15. Therefore the surface area of the new sphere is:
S=4 \pi \times 15^2 = 900 \pi
S=900 \pi
What would be the perimeter of the following object if it we dilate it by a factor of \dfrac{1}{3} ?
The perimeter of the rectangle is:
P=2\left(a+b\right)
Where a and b are the length and width of the rectangle. Dilating the cube by a factor of \dfrac{1}{3}, the length and width of the new triangle are 5 and 2, respectively:
- a=5
- b=2
Therefore the perimeter of the new rectangle is:
P= 2\left(5+2\right) = 14
P=14
What would be the surface area of the following cylinder if it we dilate it by a factor of 2?
The volume of a cylinder is:
S=A \times h
Where A is the area of the base and h is the height. Since the base of the cylinder is a circle, we have:
A=\pi r^2
Where r is the radius of the cylinder. Dilating the cylinder by a factor of 2, we have:
- r=4
- h=10
- A=\pi \left(4\right)^2=16 \pi
Therefore the surface area of the new cylinder is:
S= 16 \pi \times 10 = 160 \pi
V= 160 \pi
What would be the area of the following object if it we dilate it by a factor of \dfrac{1}{4} ?
The surface area of a triangle is:
S=\dfrac{1}{2}\times b \times h
Where b is the base and h is the height. Dilating the triangle by a factor of \dfrac{1}{4}, the base and height of the new triangle is:
- b=4
- h=\dfrac{3}{2}
Therefore the surface area of the new triangle is:
S= \dfrac{1}{2} \times 4 \times \dfrac{3}{2}=3
S=3
What would be the surface area of the following object if it we dilate it by a factor of 3?
The volume of a cone is:
S=\dfrac{1}{3}.A.h
Where A is the area of the base. Since the area of the base is a square, we have:
A=\pi r^2
Where r is the radius of the cone. Dilating the cone by a factor of 3, we have:
- r=6
- h=12
- A=36\pi
Therefore the volume of the new cone is:
V= \dfrac{1}{3} \times 36\pi \times 12=144\pi
V=144\pi