Midsquare

The diagonal of square $EFGH$ is the length of square $ABCD$ , which is $\sqrt {120}$ . Knowing that the length diagonal of square is $\sqrt 2$ times its length, the length of square $EFGH$ is $\frac {\sqrt{120}}{\sqrt 2} = \sqrt {60}$ . Thus it's area is simply the square of its length which gives $60$ as the answer.

As " ABCD " is a square with area 120 , and E ,F,G,H are the mid points AE =EB=EF=FC= > > = √30 as C is 90° , then √FG = (√30)^2 + (√30)^2 , then FG=2√15 same with ( GH , GE , EF ) then the area of the Square [EFGH] = 2√15 *2√15 = 60

Answer = 60

ABCD is a square, therefore, sides AB, BC, CB & DA = root120 (as 120 is the area of ABCD). Root120 can be simplified to 2root30; therefore, AB, BC, CD & DA = 2root30. Thus, EA, AF, FB, BG, GC, CH, HD & DE must equal root30. By using basic phythagoras' theorem, we can calculate the sides EF, FG, GH & HE; (root30)^2+(root30)^2 = ((30)+(30))^2 = root60. If sides EF, FG, GH & HE equal root60, [EFGH] = 60 as (root60)*(root60) = 60.

PS, I'm not sure how to employ mathematical signs (such as the sqrt sign) yet, so I apologise if this looks scruffy/unclear.