Question

prove area of rectangle

Answer 1

Proof

Pick two rectangles ABCD and ABC′D′ that share a side:



I claim that their areas are in the ratio of their sides:

S(ABCD)S(ABC′D′)=BCBC′.

Proving that takes three steps (see the derivation of the equation of the linear function, it is very similar.) First, if BC′=nBCthen clearly ABC′D′ consists of n identical copies of ABCD such that, by Axiom 2,

S(ABCD)S(ABC′D′)=1n=BCBC′.

On the second step, be turn what we just arrived at on its head: Assume that mBC′=BC. Then as before but switching the roles of the two rectangles,

S(ABC′D′)S(ABCD)=1m=B′C′BC.

Combining the two discoveries, if BC′=nmBC then also

S(ABC′D′)S(ABCD)=nm=B′C′BC.

This settles the question where the side lengths are in a rational ratio. For the third step we have to assume that the area function is 

continuous. If that is really so, then whatever holds for the rational factors also hold for the real ones.

Thus we see that the area S(ABCD) is bound to be proportional to BC. Now, observe that instead of fixing AB we could have fixed, say BC. The result would be that the area S(ABCD) is bound to be proportional to AB.It follows that the area S(ABCD) ought to be proportional to the product AB⋅BC.However, to be consistent with the requirement that for the unit square U, S(U)=1, it is necessary that S(ABCD) be equal not just proportional AB⋅BC.

Answer 2

Area of a rectangle = length×breadth
Example: If length of a rectangle is 4cm and breadth is 5cm . Find the area of the rectangle?
Sol) Area of a rectangle=length×breadth
=4cm. × 5cm
= 20 sq cm