Aptitude - Area - Discussion

The value of diagonal(rectangle)=root of 41
Removing the square here,
Root(l^2+b^2)=root(41)
Cancel the root , so we get l^2+b^2=41
Here we have to find the perimeter of rectangle (i.e)=2(l+b)
We all know that,
(l+b)^2=l^2+b^2+2lb.

The area of the rectangle= l*b is nothing but 20 (it is given in the question)
now apply in the above formula,
(l+b)^2=41+2*20,
=41+40,
(l+b)^2 =81,
we get l+b=9.
To find: perimeter of the rectangle= 2(l+b),
=2*9,
=18.

As u know hypotenuse^2 = base^2+ alt^2
Since √41 is the diagonal
l^2 + b^2 = 41
Also lb = 20 (Since area = 20)
Now we have to find 2(l+b) (Perimeter)
So remember the formula (a+b)^2 = a^2 + 2ab + b^2
So (l+b)^2 = l^2 + b^2 + 2lb
= 41+(2*20)
=81
Therefore, l+b = 9
Now perimeter = 2(l+b)
= 2*9
=18 cm
:)

h^2 = b^2+a^2.

l^2 + b^2 = 41.

Let put l^2 + b^2 - 2lb + 2lb = 41.

Area is given in question than lb = 20.

So (l^2 + b^2 + 2lb) - 2lb = 41.

= (l+b)^2 - 2(20) = 41.

= (l+b)^2 = 41 + 40.

= (l+b)^2 = 81.

= (l+b) = 9.

Put in formula of perimeter = 2(l+b)

= 2*9 = 18.

Solution.

As we know the perimeter of a rectangle is = 2 (l+b).
So, As we have. Area = 20m and by applying Pythagoras (l^2+b^2=41).
So, the parameters of rectangle= 2 (l+b) ^2 =4 (l^2+b^2+2lb)
= 4 (41+40)
= (81×4)
={324}.
P^2= 324.
Then P= 18m.

How +40 came?

l^2 + b^2 + 2(lb) = 41.
l^2 + b^2 + 2(20) =41.

Then it should be like
l^2 + b^2 = 41 - 40,
l^2 + b^2 = 1.

How 81 will come?
I am not able to understand this. Please, anyone, help me to get it.

By PYTHAGORAS THEOREM;

L^2+B^2=D^2,
=5^2+4^2 =41^2,
=5X4=20.
Then;
The perimeter of Rectangle =2(5+4)=40.

What is the need to write the 40 there though that 2lb came from the formula (l+b)2?

I can't understand this. Please someone explain this clearly?

L^2+b^2=41.

So how (l+b) ^2=41 become. Explain this please?

How +40 came?

It should be -40 isn't it?