# When an error of 1% is made in the length and breadth of a rectangle, what is the percentage error (%) in the area of rectangle?

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An error of 1% is made in calculating the length and breadth of a rectangle. Let the error in measuring the length be L be dL and the error in measuring the breadth W be dW.

The error in the area A which is the product A = W*L is `(dA)/A = sqrt(((dL)/L)^2 + ((dW)/W)^2)`

=> `(dA)/A = sqrt(0.01^2 + 0.01^2)`

=> `(dA)/A = sqrt(2*10^-4)`

=> `(dA)/A = sqrt 2*10^-2`

=> `(dA)/A = sqrt 2 %`

**The error in the area is `sqrt 2` % of the area estimated.**

I appreciate your effort. But this ques is MCQ. Options are-

a. 0

b. 2

c. 1

d. 4

And correct answer is b. 2

I needed solution.