Square vs Rhombus

There is a nice, geometrical solution to the question.

Let the side length for both the figures be $a$ units.

Area of rhombus = $a^{2}\sin\theta$

Area of square = $a^{2}$

One can clearly see that in case of the square, $\theta = 90^{o}$ , so rhombus has to have the smaller area.

The area of a square is $(Side)^2$ and the area of rhombus is $Base\times\space Height$ .

Consider the set-ups below:

Notice that $AE$ is perpendicular to $BC$ so, $AE^2=AB^2-BE^2=a^2-BE^2.$ So, definitely, $EF>AE.$

The rest is left to the reader.

Consider an extreme case - a rhombus that is incredibly skinny yet still has the same side length as a square. In other words, the acute angle between two adjacent sides of the rhombus is very small. Clearly, the area of the rhombus is much smaller than the area of the square, so that is our answer.

Split the Rhombus into two triangles. The area is base * height. The new base is the length of the dividing line and is always less than the length of any side. and the new height is always always less than the length of any side therefore the product of two reduced numbers cannot sum up to side*side

• Let's recall that─ a square is an special case of rhombus. A rhombus whose one angle (consequently, each angle) is a right angle is a square.

• The area of a rhombus with side length $s$ and one angle $\theta$ is $s^2\sin{\theta}$ . How? Let me provide a hint: the area of a triangle, whose one angle is $\theta$ , and $a$ and $b$ are the two adjacent sides of the angle, is $\frac{1}{2}ab\sin{\theta}$ .

• Now, as $0^\circ{} <\theta< 180^\circ{}$ , the maximum area of a rhombus with side length $s$ occurs at $\theta = 90^\circ{}$ , that is when the rhombus is also a $square$ .

• So, if the square and the rhombus are not congruent, but have equal side length, then the rhombus has smaller area .

Solve this by looking at an extreme case. If the rhombus is completely flattened it will have zero volume. So the rhombus has less area than the square.

consider a square made of frames put that frame on ground

now give some bearings to lower two vertices and give it a push

after some time the whole frame collapses and has zero area

Let the rhombus area is S and d1, d2 are its diagonal. We have: And: a^2 is the quare area. So the rhombus has to have the smaller area.

The area of a square or rhimbus is given by a² sinA. Where a is it's an arm and A is angle between two adjacent arms. For square A=90° and for rhombus A= lesser or greater than 90°. Since sin90°=1 and for other angles between 0° to 180° it is less than 1. That's why rhombus has smaller area.

Extend the folding of the square to the limit and it reaches zero area. So the area diminishes the moment the square starts to fold!

Intuitively, if you allow the rhombus to deform further it becomes flatter until the area must become zero. Therefore, the area of any rhombus that is not a square must be less than that of the square.

The rhombus can be made be squeezing opposite corners of the square. As it is squeezed its area decreases to zero - it never increases.

Let the length be 2 cm:

Then the area of the square is $4\ cm^2$

The area of rhombus is $\dfrac{4\sqrt{3}}{2}$ .

$\large \color{#20A900}\therefore \text{The area of rhombus is less than the area of the square}$

I found the area of the rhombus by dividing it into 4 right-angled triangles as we know that a rhombus is made of 2 equilateral triangles

Imagine "pulling" from two corners of the square in order to get a rhombus, and you continued until it folds into a straight line which has zero area. So the area has to go from l^2 to 0 and the area of the rhombus is in between

A Rhombus always has four equal sides; a square always has four equal sides. A square always has four 90 degree angles. The angles of a rhombus can be any measure (each pair of opposite angles will be equal, and all four will always add up to 360). A square can also be defined as a parallelogram with equal diagonals that bisect the angles. If a figure is both a rectangle (right angles) and a rhombus (equal edge lengths), then it is a square. ... A square has a larger area than any other quadrilateral with the same perimeter.

The area of a square is (side)^2 or a^2 (Considering side length = a) and the area of rhombus is Base*Height or a^2 sinϴ. the acute angle between two adjacent sides of the rhombus is very small. Clearly, the area of the rhombus is much smaller than the area of the square, so that is our answer.

Base times the height.