Aptitude - Area - Discussion

Suppose,

The Original Length of one side is 10m, Then area for this is 100m² &
Length when 2% error is excess is 10+0.2, it means now the length of a side is 10.2m and the area for this is 104.04m².
Now,
The area due to error excess - Original area = 104.04 - 100.
= 4.04.
Also, we can assume the length of a side is 100m or whatever we want.

Here, it is mentioned that the error excess of 2% so for that reason we add the value of the error on the original side instead of subtracting.

Suppose side = 10, Error excess = 2%.
i.e. 10 + 2% = 10.2,
A 1 = 10 * 10 = 100,
A2 = 10.2 *10.2 = 104.4,
A2 - A1 = 104.4 - 100 = 4.04%.

It is simple;

Error is 2% so, take it 1.02.
Now one side of a square is 'a'.
Area = (a*1.02)*(a*1.02),
= (a*a*1.0404).

So, the total error in area = 4.04%.

Use the formula {A+B+(AB/100)}.
Here A=2 for side.
For area A=B.
2+2+2*2/100=4.04.

We can also solve this is in 404/10.

In order to find the percentage error, the formula is [ (x*x) /100+2*x]%.

How? Explain the answer please.

You are right, Thanks @Harsh.

Percentage error=
( Experimental value - theoretical value )/theoretical value *100.

Therefore,
Here percentage error= [(102)^2 - (100)^2]/(100)^2,
= 404*100/(100)^2.

Let the actual side of the square be "a" units.

then Actual area=a*a sq units
2% excess in side=102*a/100 (lets call this side "a1").

Excess Area =a1*a1=(102*a/100)*(102*a/100)
=(2601*a^2)/2500 sq units.

Now, error in area=Excess Area-Actual area
=(2601*a^2)/2500-(a^2) sq units
= (101a^2)/2500 sq units.

Now percentage error=(error in area/actual area)*100
=(((101a^2)/2500)/a^2)*100
=101/25
= 4.04%.