The figure below is a net for a rectangular prism. side a = 21 inches, side b = 17 inches, and side c = 11 inches. what is the surface area of this figure?

a. 1,550 square inches
b. 775 square inches
c. 3,927 square inches
d. 1,319 square inches

The answer is 1,550

Step-by-step explanation:

First you find the area of the rectangle 1

Area = ab

=(21 in) (17in)

=357 sq in.

this will be the same for rectangle 3 = 357 sq. in.

next find the area of rectangle 2

area=ac

=(21 in) (11 in)

=231 sq in.

this will be the same for rectangle 4 = 231 sq. in.

next find the area of rectangle 5

area = bc

=(17 in) (11 in.)

=187 sq. in

this will be the same for rectangle 6 = 187 sq. in.

Next add all areas together

Total surface area = 2(ac) + 2(ab) + 2(bc)

= 2(231) sq in. + 2(357) sq in. + 2(187) sq. in

= 462 + 174+ 374

= 1550 sq. in

  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.

i dont know

step-by-step explanation: