Question

A square and a rectangle are of equal area. it the perimeter of rectangle is p of square is q, then :
(i) P= Q
(ii) P<Q
(iii) P>Q
(iv) P--< Q ​

Answer 1

Answer:

Suppose the side of a square is a

And length and breadth of rectangle are l and b

a^2=lb

p=2(l+b)

q=4a

(l+b)^2=l^2+b^2+2lb

=l^2+b^2+2a^2

So (l+b)^2>a^2

or (l+b)>a

Since (l+b)>a

So P>Q

Answer 2

Answer:

Option(iii)

Step-by-step explanation:

Area of rectangle = length * breadth

A square and a rectangle are of equal area.

Area of Square = Area of rectangle

=> Area of square = l * b

So, each side of square = √lb

Given,

Perimeter of rectangle = 2(l + b) = p  ----- (1)

Perimeter of square = 4√lb = q   ---- (2)

On Squaring both the equations, we get

p = 4(l + b)²   ----- (3)

q = 16 lb  ----- (4)

Subtract 8lb from each of the above equations.

p = 4(l + b)² - 8ab

  = 4(l² + b² + 2ab) - 8ab

 = l² + b²

q = 16lb - 8lb

  = 8lb

Now,

Divide by 4 both the equations.

p = l² + b²

q = 2 lb

Now,

p - q = l² + b² - 2lb

p - q = (l - b)²

But (l - b) is positive.

Hence, p - q is also positive.

Therefore,

P > Q

Hope it helps!