Acircle with a radius of 1/2ft is dilated by a scale factor of 8. which statements about the new circle are true? check all that apply. the length of the new radius will be 4 feet. the length of the new radius will be 32 feet. the new circumference will be 8 times the original circumference. the new circumference will be 64 times the original circumference. the new area will be 8 times the original area. the new area will be 64 times the original area. the new circumference will be feet. the new area will be square feet.

A) The length of the new radius will be 4 feet

C) The new circumference will be 8 times the original circumference

F) The new area will be 64 times the original area

H) The new area will be 16π square feet

Step-by-step explanation:

Verify each statement

see the attached figure to better understand the problem

Part A) The length of the new radius will be 4 feet

The statement is true

Because

we know that

The radius of the original circle is r=1/2 ft

The scale factor is 8

so

To find out the length of the new radius multiply the radius of the original circle by the scale factor

(1/2)(8)=4 ft

Part B)  The length of the new radius will be 32 feet.

The statement is false

Because, the length of the new radius is 4 ft (see part A)

Part C) The new circumference will be 8 times the original circumference

The statement is true

Because

The original circumference is equal to

The new circumference is equal to

Divide

so

the new circumference will be the original circumference multiplied by 8

Part D) The new circumference will be 64 times the original circumference

The statement is false

Because, the new circumference will be 8 times the original circumference (see part C)

Part E) The new area will be 8 times the original area

The statement is false

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

The scale factor is 8

so

The scale factor squared is 8²=64

therefore

The new area will be 64 times the original area

Part F) The new area will be 64 times the original area

The statement is true

(see Part E)

Part G) The new circumference will be 8π feet

The statement is false

Because, the new circumference will be 4π feet (see part C)

Part H) The new area will be 16π square feet

The statement is true

Because

The area of a circle is

we have

substitute

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.