SOLID MENSURATION QUESTION (Need help ASAP)
I've been trying different kinds of method but it's not even helping
A piece of wire length 52m is cut into two parts. Each part is then bent to form a square. It is found that the combined area of the two squares is 109 sq. m. Find the measures of the sides of the two squares.
Need help ASAP :(
tA = ((52 - Ps1)/4 )^2 + ((52 - Ps2)/4)^2
i tried this but it didnt work
Stacey Warren - Expert brainly.com
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if you cut the wire into two lengths x & y
then you know that x+y = 52
the squares have sides x/4 and y/4
so the area of the squares are (x/4)^2 and (y/4)^2
so (x/4)^2 +(y/4)^2 = 109
I have solved by graphing (plot both equations and find intersection points), but not yet solved analytically
@MrNood Soo how can i solve the sides? the x and y's... it seems impossible if thats the given equation.
you have 2 equations
(x/4)^2 +(y/4)^2 = 109
from the first y= (52-x)
substitiute that into the second equation
you will get a quadratic in x
solve for x
then solve for y in the first equation.
The answer is x/4 and y/4 for th esquare side lenghts
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Oh i see i get it now
let me try it first
note - the quadratic has TWO real solutions
you can choose EITHER of them since we don't care whch 'end' of the wire is which
Thank you so much @MrNood XD
and oyur answer is?....
first square is 10m per sides. with the perimeter of 40m
second square is 3m per sides with the perimeter of 12m
then total perimeter is 40 + 12 = 52
and area is 10^2 + 3^2 = 100 + 9 = 109m^2
i finally get it now thanks a lot
good to see someone stick it out through some complicated calculation!
took me 1 week to understand the question properly... i used this site as my last resort. glad i finally understand this. lol thanks again