NIMCET 2009 Mathematics PYQ — A square with side is revolved about its centre through 45°. … | Mathem Solvex | Mathem Solvex
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NIMCET 2009 — Mathematics PYQ
NIMCET | Mathematics | 2009
A square with side a is revolved about its centre through 45°. What is the area common to both the squares?
Choose the correct answer:
A.
2(2−1)a2 sq units
(Correct Answer)
B.
2(2+1)a2 sq units
C.
(3−1)a2 sq units
D.
(5−1)a2 sq units
Correct Answer:
2(2−1)a2 sq units
Explanation
. Understanding the Geometry
When a square is rotated by 45° about its center, the intersection of the original square and the rotated square forms a regular octagon.
2. Calculating the Side of the Octagon
Let the side of the original square be a. When we rotate it, small right-angled triangles are formed at each corner of the square that are not part of the intersection. Let the length of the segment removed from each corner of the square side be x.
The side of the resulting octagon (s) can be expressed in two ways:
As the remaining part of the square side: s=a−2x
As the hypotenuse of the corner triangle (which is an isosceles right triangle): s=x2+x2=x2
Equating these two:
a−2x=x2
a=x(2+2)
x=2+2a
3. Finding the Area of the Overlap (Octagon)
The area of the common region is the Area of the Square minus the areas of the 4 corner triangles.
Area of one corner triangle: 21⋅x⋅x=2x2
Area of 4 corner triangles: 4⋅2x2=2x2
Substitute the value of x:
Common Area=a2−2(2+2a)2
Common Area=a2−4+2+422a2
Common Area=a2−6+422a2
Common Area=a2−3+22a2
To simplify, rationalize the denominator:
3+221⋅3−223−22=9−83−22=3−22
Substitute back:
Common Area=a2−a2(3−22)
Common Area=a2(1−3+22)
Common Area=a2(22−2)
Common Area=2(2−1)a2
Final Answer:
The correct option is (a).
Explanation
. Understanding the Geometry
When a square is rotated by 45° about its center, the intersection of the original square and the rotated square forms a regular octagon.
2. Calculating the Side of the Octagon
Let the side of the original square be a. When we rotate it, small right-angled triangles are formed at each corner of the square that are not part of the intersection. Let the length of the segment removed from each corner of the square side be x.
The side of the resulting octagon (s) can be expressed in two ways:
As the remaining part of the square side: s=a−2x
As the hypotenuse of the corner triangle (which is an isosceles right triangle): s=x2+x2=x2
Equating these two:
a−2x=x2
a=x(2+2)
x=2+2a
3. Finding the Area of the Overlap (Octagon)
The area of the common region is the Area of the Square minus the areas of the 4 corner triangles.
