Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of the two squares.

# Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of the two squares.

1. A
6m, 12m
2. B
11m, 18m
3. C
18m, 12m
4. D
None of these

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

Let the side of the first square be a.
Let the side of Second Square be b.
Area of 1st square= a2
Area of 2nd square= b2
Perimeter of 1st square= 4a
Perimeter of 2nd square= 4b
Difference of perimeters of the square is 24m.
4a-4b=24
4(a-b) =24
a - b= 6         --------- (1)
Sum of the areas of the square is 468m2. ------------ (2)
From eq (1), we write a as
a=b+6                  ------------ (3)
Put the value of a in the (2) equation.    Side of the square is never negative. So we consider b=12.
Put the value of b in eq (3).
We get,      a=18 and b=12
So, the two sides of the square are 18m and 12m.
Hence, Option (3) is the correct Ans:.

## Related content

 Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formula Perimeter of Triangle Formula Area Formulae Volume of Cone Formula Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics  +91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)