a rhombus can be created by dragging the four sides of the square to a certain angle. so how does the area of the new formed rhombus differ from the square.it can be proven mathematically but how do they become different??
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$\begingroup$Each frame of the following animation features a rhombus of side $1$. The area reaches a maximum of $1$ when the rombus is a square. This shows how "stretching" the rhombus decreases its area.
Think about what happens when you "tilt" a square into a rhombus, keeping the horizontal lines of the square horizontal, but rotating the vertical lines a little bit clockwise. Compare the new area to the original area. Doing this has added a triangular area to the right of the square, removed an identical triangular area from the left, but also removed a rectangular area from the top of the square because the height has decreased. The triangular areas cancel out, but the rectangular area removed due to the decrease in height means that the rhombus has a smaller area.
$\endgroup$ $\begingroup$One way to think about it is this: the area of a rhombus with side length $s$ is the sum of the areas of $4$ congruent right triangles with hypotenuse $s.$ (The legs of these triangles are the segments from the rhombus's center to its vertices.)
If we consider any right triangle of hypotenuse $s,$ and let the hypotenuse be a diameter of a circle, then we find that exactly $3$ points of the triangle lie on the circle: the vertices.
So, dragging the corners of the rhombus around has the same effect as dragging the right-angle corner of a right triangle around the edge of its circumscribed circle. As the right-angle corner gets closer to the hypotenuse-diameter, the triangle's area gets smaller (the base is the same, but the height is decreasing). The triangle's area is maximized when the right-angle corner is at its greatest distance from the hypotenuse-diameter (when the height is maximized). This happens exactly when the right triangle is isosceles. The right-triangle sections of a rhombus are isosceles exactly when it is a square.
$\endgroup$ $\begingroup$Since the area of rhombus is " base* height".If we consider a rhombus as a square then heigt is increasing. Consequently the area is increased. So the area of rhombus is smaller than square using same sides.
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