Quadrilateral ABCD in the figure below represents a scaled-down model of a walkway around a historic site. Quadrilateral EFGH represents the actual walkway. ABCD is similar to EFGH. Two irregular similar quadrilaterals ABCD and EFGH are drawn. AB measures 5 inches, BC measures 4 inches, CD measures 4 inches and AD measures 3 inches. EF measures 45 feet. What is the total length, in feet, of the actual walkway?
when polygons are similar, their sides are proportional. can you observe that AB and EF are 'corresponding' sides ?
yes! i was thinking the ration would be 9:1..?
not exactly, but "numerically" yes!
like 5in became 45 ft *not 45in*
*shocked* YYYAAAYY! i was 1/2 right :))
oh. you're right..
lol, yes! so you can find all other sides now ?
would i just multiply 9 to each side.?
exactly.
okay. great! even if the units are different.?
if you wish to use a mathematical equation for 'corresponding sides being proportional' then, \(\large \dfrac{EF}{AB}=\dfrac{45}{5}=\dfrac{EH}{AD}=\dfrac{HG}{DC}=\dfrac{GF}{CB}=9\)
once i have all the sides, i then find the area.?
maintain the units, like for GF = 9*CB = 9*4 = 36 write that as 36ft and not 36in
alright. do you mind if i get back to you with my answer to make sure it's all good.?
sure! i mean i don't mind :)
thank youu! and would i have to use the area of a trapezoid for this one.? @hartnn
"What is the total length, in feet, of the actual walkway? " you don't need area :P
ok. so add it all up..?
total length of walkaway = perimeter of walkaway.
yes, perimeter is just sum of sides :)
would it be 144.? @hartnn
144 " ft " is correct! :) i just got a shorter way to do it :P
since the sides are 9 times larger, just add up sides of smaller triangle and multiply by 9 ! 9 (3+4+4+5) = 144
**smaller polygon
and why didn't you tell me this earlier -_-..?
LOL!!
however i feel victorious, so I don't know XD
lol, because i "just" got that idea. anyways, the whole point of this problem was to practice, "corresponding sides of similar polygons are proportional", so we are good with this method :)
si!! thanks youu!! Your aweosme!
you are awesome learner too! :D most welcome ^_^
Join our real-time social learning platform and learn together with your friends!