Question

Sum of the areas of two square is 260m^2. If the difference of their perimeters is 24m then find the sides of the two squares.

Answer 1

Answer:

8 m

Step-by-step explanation:

Let side of first square = x

side of second square = y

sum of area of two square = 260 sq m

i.e     x^2 + y^2 = 260    (i)

now the difference of their perimeter = 24 m

i.e     4x - 4y = 24

=> x - y = 6     (ii)   (taken 4 common)

squarring equation (ii)

(x - y)^2 = 6^2

x^2 + y^2 - 2 xy = 36

260 - 2 xy =36           (value from equation (i)

224 - 2 (6+y) y =0       (value from equation (ii)

2y^2 + 12 y - 224 = 0    (multiplying both side with  -1)

y^2 +6y - 112 =0

y^2 + 14 y - 8y - 112 = 0

y (y+14) - 8 (y+14) =0

y= 8 or -14 m

here we take positive number 

so y= 8

therefore  x-y = 6

=> x= 6+8 = 14 m

thus , the side of small square is 8m.

thanks for such brilliant question

Answer 2

SOLUTION

Let a metres & b metres be the side of two squares.Then, (a^2)m & (b^2)m are the areas of the sides and their perimeters are (4a)m &(4b)m respectively.

Therefore,

=) 4a-4b= 24

=) 4(a-b)= 24

=) a-b= 6

=) b= (a-6)...............(1)

Given that,

Sum of their areas = 260m^2

=) a^2 +b^2 = 260

=) a^2 +(a-6)^2= 260 [from (1)]

=) 2a^2 -12a-224= 0

=) a^2 -6a-112= 0

=) a^2 -14a+8a-112=0

=) a(a-14) +8(a-14)=0

=) (a-14) (a+8)= 0

=) a-14=0 or a+8=0

=) a= 14 or a= -8

Sides of a square can be never be negative.

So,

=) a= 14

=) b= (14-6) = 8

Thus, the sides are 14m & 8m

hope it helps ☺️