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

Solution :

Let x and y be the side length of squares.

x2 + y2 = 468 (1)

4x - 4y = 24

x - y = 6

x = 6 + y (2)

(6+y)2 + y2 = 468

36+y2+12y+y2-468 = 0

2y2 + 12y - 432 = 0

y2 + 6y - 216 = 0

(y - 12)(y + 18) = 0

y = 12 and y = -18 (not admissible)

If y = 12, then x = 18

So, the side length of required squares are 12 and 18 respectively.

Step-by-step explanation:

#itzuryesul

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5

a - base

b - exponent

so,