Mathematics, 30.07.2019 01:00 leslie122372

Question 2: a circular picture is 8 inches in diameter. part a what is the area of the picture in square inches? a. 4π square inches b. 8π square inches c. 16π square inches d. 32π square inches part b a frame that is 2 inches wide surrounds the picture. what is the total area of the picture and the frame in square inches? a. 4π square inches b. 12π square inches c. 36π square inches d. 40π square inches really important answer correctly

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Answers:
_______________________________________________________
Part A) The correct answer is: [C]: " 16 $\pi$ square inches " .
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Part B) The correct answer is: [C]: " 36 $\pi$ square inches " .
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Note:
_______________________________________________________
Part A)
_______________________________________________________
The formula for the area, "A", of a circle:

A = $\pi * r^2$ ;

in which:

" A = area of the circle" ;
" r = radius of the circle" ;

Note: radius, "r" = d / 2 ; in which: "d" is the diameter of the circle ;

Given: " d = 8 in " ; " r = (8 in) / 2 = 4 in " ;
_________________________________________________
A = $\pi$ * (4 in)² = $\pi$ * 4² * (in)² ;

= $\pi$ * (4*4) * (in²) ;

= $\pi$ * 16 * in² ;
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→ which is: Answer choice: [C]: " 16 $\pi$ square inches " .
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Part B)
________________________________________________________
We now have a circular frame with a thickness of "2 inches"
surrounding the original circle. This adds "4 inches" to the diameter of that of the original circle in "Part A" —
{Note: "2 inches on one side of the circle" ; PLUS "2 inches on the other side of the circle" ; equals "4 inches".}.

The diameter, "d", of this "new circle" ;
= "8 inches (that of the previous circle)" , PLUS "4 inches" , equals "12 inches" .

→ d = 12 in. ;

The radius , "r" :

→ r = d/2 = 12 in / 2 = 6 in.

  → r = 6 in.

The area, "A" of a circle:
______________________________________________________
A = $\pi$ * r² ;

= $\pi$ * (6 in)² ;

= $\pi$ * (6²) * (in²) ;

= $\pi$ * (6*6) * (in²) ;

  = $\pi$ * (36) * (in²) ;
_______________________________________________________
→ which is: Answer choice: [C]:  " 36 $\pi$ square inches " .
_______________________________________________________

Answer from: Quest

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.

Find the value of the variable and the length of each chord for #5 and #6.

Answer from: Quest

answer: $34.31

step-by-step explanation:

$11.99 x 3 = $35.97

$35.97 - 10% = $32.37

$32.37 + 6% = $34.31

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