In a right triangle the length of a hypotenuse is c and the length of one leg is a, and the length of the other leg is b, what i
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Answer:
The radius of circle B is 6 times greater than the radius of circle A
The area of circle B is 36 times greater than the area of circle A
Step-by-step explanation:
we have
<em>Circle A</em>
The radius of circle A is
-----> the radius is half the diameter
<em>Circle B</em>
Compare the radius of both circles
The radius of circle B is six times greater than the radius of circle A
Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared
All circles are similar
In this problem the scale factor is 6
so
therefore
The area of circle B is 36 times greater than the area of circle A
The surface area of the first figure is 160.
The net of the second figure is attached, and the surface area is 408.
To find the surface area of the first figure, we find the area of the base:
10*10 = 100
Now we find the area of each of the 4 lateral faces (triangles):
A=1/2bh = 1/2(3)(10) = 15
The total is 100+15(4) = 160
To find the surface area of the second figure, find the area of the bases (triangles):
A=1/2bh = 1/2(6)(8)= 24
The area of the bottom lateral face is 15*6 = 90.
The area of the slanted lateral face is 15*10 = 150.
The area of the vertical lateral face is 15*8 = 120.
Together we have 24(2) + 90 + 150 + 120 = 408.
The net of the second figure is attached, and the surface area is 408.
To find the surface area of the first figure, we find the area of the base:
10*10 = 100
Now we find the area of each of the 4 lateral faces (triangles):
A=1/2bh = 1/2(3)(10) = 15
The total is 100+15(4) = 160
To find the surface area of the second figure, find the area of the bases (triangles):
A=1/2bh = 1/2(6)(8)= 24
The area of the bottom lateral face is 15*6 = 90.
The area of the slanted lateral face is 15*10 = 150.
The area of the vertical lateral face is 15*8 = 120.
Together we have 24(2) + 90 + 150 + 120 = 408.