Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .
Answers
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Let the length of the side of the smaller square be x .
Perimeter of smaller square = 4x
Let the length of the side of the larger square be y m.
Perimeter of larger square = 4y
Now, 4y−4x=24
or, y−x=6 (Dividing both sides by 4)
or, y=x+6
Now, area of larger square = y 2=(x+6)2
= x2 +12x+36
Area of smaller square = x2
Now, x2+12x+36+x2=468 (Sum of areas)
or, 2 x2+12x+36=468
or, x2 +6 x +18=234 (Dividing both sides by 2)
or, x2 +6x+18−234=0
or, x2 +6x−216=0
or, (x+18)(x−12)=0 (Factoring the equation)
or, x=−18,12
Since x cannot be negative,therefore,
x=12m
Therefore, x + 6 = 18m
So the sides are 18m and 12m respectively.
Hope that helps.
Let the two squares be A and B. A has sides of a, while B has sides of b,
Now for the two equations.
a^2 + b^2 = 468 …(1) [Sum of the areas]
4a - 4b = 24 …(2) (perimeter of the two squares), which can be written as
a - b = 6, or
a = b+6. Put that value in (1) to get
(b+6)^2 + b^2 = 468, or
b^2 + 12b + 36 + b^2 = 468, or
2b^2 + 12b + 36 = 468, or
b^2 + 6b + 18 = 234, or
b^2 + 6b + 18 - 234 = 0, or
b^2 + 6b -216 = 0, or
b^2 + 18b - 12b - 216 = 0, or
b(b + 18) - 12(b + 18) = 0, or
(b-12)(b+18) = 0
b = 12 m or -18m (inadmissible)
Then a = b+6 = 12 + 6 = 18m
So A is a square of 18 m and B is a square of 12 m
Check: Area of A = 18^2 = 324 sq m. area of B = 12^2 = 144 sq m and their sum is 324 + 144 = 468 sq m.
Let the side of the first square be 'a'm and that of the second be
Let the side of the first square be 'a'm and that of the second be ′
Let the side of the first square be 'a'm and that of the second be ′
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =468
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =4682a 2
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =4682a 2 +12a+36=4682+6a+18=234+6a−216=18a−12a−216=0
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =4682a 2 +12a+36=4682+6a+18=234+6a−216=18a−12a−216=0a(a+18)−12(a+18)=0
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =4682a 2 +12a+36=4682+6a+18=234+6a−216=18a−12a−216=0a(a+18)−12(a+18)=0(a−12)(a+18)=0
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =4682a 2 +12a+36=4682+6a+18=234+6a−216=18a−12a−216=0a(a+18)−12(a+18)=0(a−12)(a+18)=0a=12,−18
Let the side of the first square be 'a'm and that of the second be ′ Given 4A−4a=24A−a=6 --(1From (1), A=a+6Substituting for A in (2), we get(a+6) =468 12a+36+a 2 =4682a 2 +12a+36=4682+6a+18=234+6a−216=18a−12a−216=0a(a+18)−12(a+18)=0(a−12)(a+18)=0a=12,−18So, the side of the first square is 12 m. and the side of the second square is 18 m.