Question

7. When measuring the side of the square, 10% less error measurement was made. percentage of calculated area error. ? A. 20% B. 22% C. 21% D. 19%
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Answer 1

Answer:

Calculus Method

Δss=0.05

=> Area, A= s2

=> lnA=2lns

=> ΔAA=2Δss

=> ΔAA=0.1

On partially differentiating,

=> ΔAA

Arithmetic Method

New side, s'= 1.05s

So new area, A'= 1.1025s^2

Now, change in area, ∆A= (1.1025–1)s^2

So, percentage error,

∆A%/A= 0.1025×100= 10.25%

Now,

Why there are changes in values and which one is correct?

The arithmetic one is correct.

Change is because we approximate the change in function in calculus

So we say,

f(X+∆x)≈ y + dy/dx(∆x) ; y= f(X)

At X= X1,

f(x+∆x)≈ f(X1)+ (dy/dx)x=x1(∆x)

So, we get A(1)+ 2×0.5; where A(s)= s^2

Thus you get….

Thank you

hope it helps you✌

Answer 2

Answer:

Question - When measuring the side of the square, 10% less error measurement was made. percentage of calculated area error. ? A. 20% B. 22% C. 21% D. 19%

Answer .

So, first we have to suppose a number

Let the supposed number be 10

So, 10%of 10 is equal to 1

Now the side is 10+1=11

Now we multiply 10×10 because the area of square is side× side

=100sq unit

Then we find the area of new side which is11

11×11=121 sq unit

Now we have to find the difference of area

So,

121-100

=21

21 is 21% of 100

Now the answer is 21%

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