Question

. The sum of the areas of two squates is 41
cm. If the sum of their perimeters is 36 cm, find the
sides of the two squares.​

Answer 1

Step-by-step explanation:

given x square + y square=41

4x+4y=36

x+y=9

x=9-y

(9-y)^2+y^2=41

81+2y^2-18y=41

2y^2-18y+40=0

y^2-9y+20=0

y^2-4y-5y+20=0

y(y-4)-5(y-4)=0

(y-5)(y-4)=0

y=5 or 4

if y= 5 then x=4

if y=4 then x=5

so sides are 4 and 5 or 5 and 4

Answer 2

Answer:

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Step-by-step explanation:

Let the sides are a and b

Now condition 1

a^2+b^2=41. and 4(a+b)=36

(a+b) =9

(a+b) ^2-2ab=41

=>9^2-2ab=41

=>2ab=81-41

=>ab=40/2=20

Now,

(a-b) ^2=(a+b)^2-4ab

=81-4(20)

=1

(a-b) =1

a+b=9

a-b=1

(-) (-) (-)

_________

2b=8

b=4

And. ab=20

a=20/b

a=20/4

a=5

Therefore the values of a=5 and b=4