The side of square is x meters.The midpoints of its side are joined to form another square whose area is 16 sq. meters. Find the are of the portion of the bigger square outside the smaller square

Question

anonymous · Answer

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BE^2+BF^2 = EF^2

i.e (x/2)^2+(x/2)^2 = 4^2                  [since area of inner square is 16,side is 4]
Now from x, you can get area of bigger square.

jfruscianterhcp · Answer

show solution how?

jfruscianterhcp · Answer

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anonymous · Answer

Yeah exactly, now apply pythagoras  theorem to one of those triangles.

jfruscianterhcp · Answer

how? you do it haha show soulution

anonymous · Answer

$$BE ^{2}+BF ^{2} = EF ^{2}$$
$$(X/2)^{2} + (X/2)^{2} = 4^{2}$$
$$X ^{2}/2 = 16$$
$$X ^{2}= 32$$
$$X = \sqrt{32}$$
$$X = 4\sqrt{2}$$

jfruscianterhcp · Answer

but that is not the answer

jfruscianterhcp · Answer

the answer should be 16

anonymous · Answer

Yeah, 16 is the answer.
I only gave you the value of x.
Now find the area. Then subtract the area of inner square from it.