Sum of areas of two squares is 468 metres square. if the difference of their perimeter is 24, find the sides of two squares.?

Question

Answer

Let the side of the 1st square be 'a' m and that of the 2nd be 'A' m.
#Area of the 1st square = a2 sq m.
#Area of the 2nd square =A2 sq m.
#Their perimeters would be 4a and 4A respectively.

¶Given 4A - 4a = 24
A - a = 6-(1)
A2 + a2 = 468-(2)
*From (1), A=a+6.
#Substituting for A in (2) , we get...

(a+6)2+a2 = 468a2+12a+36+a2=468a2+12a+36=468a2+6a+18=234a2+6a-216=0a2+18a-12a-216=0a(a+18)-12(a+18)=0(a-12) (a+18)=0a=12,-18.

¶So , the side of the 1st square is 12m, and the side of the 2nd square is 18m.

Answer

Answer

Let the side of one square be a and other one is b.
area of square =a×a
and b×b .
a.a+b.b=468(given) (1) ,
perimeter of square =4a
and 4b ,
4a-4b= 24 ,
a-b=6 (2) ,
a=6+b (3),
from (1) and (2)
a square +b square =(a-b) square + 2a.b ,
(6)square + 2 a.b = 468 ,
36+ 2 a.b = 468 ,
2 a.b= 468-36 ,
a.b= 216 (4),
from (3) and (4) ,
b×(6+b)= 216 ,
b square + 6b - 216= 0 ,
b square +18b - 12b -216 =0 ,
b (18 +b) - 12(18+b) =0 ,
(b-12)(18+ b)=0 ,
b= 12 or - 18 ,
-18 is not possible. so side of 1 square is 12 .
from (2) ,
a- 12 = 6 ,
a = 18 ,
so the side of these 2 squares are 12 and 18.

Answer

Solution:

Given:

Sum of areas of two squares = 468 m²

Difference of their perimeter = 24

Let the length of sides of two squares be x and y.

Area of first square = x²

Area of second square = y²

Sum of the areas of the squares = x² + y² = 468

Difference of their perimeters = 4x - 4y = 24

Simplifying the equations:

x² + y² = 468

x - y = 6

Solving the equations:

y = x - 6

Substituting y in the first equation:

x² + (x - 6)² = 468

2x² - 12x + 180 = 0

x² - 6x + 90 = 0

x = 10 or x = 6

Checking the solutions:

If x = 10, then y = 4

If x = 6, then y = 0 (Not possible as side cannot be negative)

Therefore, the sides of the squares are:

First square: 10 m

Second square: 4 m

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