Many a times I came across to questions in which a square is formed by joining the mid-points of an enclosing figure. So everytime I came across to some common result in all such question which I will be sharing with you in this post.
1) Figure formed by joining the mid points of a Square: 
Remember: Figure formed will be Square. And the area of this Square is half the area of the enclosing square. For eg. If the side of enclosing square is : a cms, then
side of the Square formed by joining mid points = b= a/(sqrt2).
We got this result using Pythagoras Theorum.
Area of this Square : b^2 = a^2/2 = Half of the enclosing Square.
2. When enclosing figure is a Rectangle.
Remember: Figure formed by joinging the mid points of this figure won’t be rectangle or square. It will be a Rhombus (whose diagnols bisect each other at 90deg and all sides are equal)
Area of this rhombus too will be half of the area of the enclosing rectangle. Let the sides of rectangle be p cms and q cms (p<q)
Then Length of Rhombus = sqrt(p^2 + q^2)/2 { Using Pythagoras Theorum}
Area of Rhombus = half the product of diagnols.
= (1/2)* pq {As length of diagnols of rhombus will be same as sides of rectangle enclosing it}
= Half of area of rectangle
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