Question

The length and breadth of a rectangle are 25.0 cm and 16.7 cm respectively. These have been measured to an accuracy of 0.1 cm. Determine the percentage error in the area of the rectangle.

Answer 1

Given

• length of rectangle , l = 25.0 cm
• breadth of rectangle , b = 16.7 cm
• l and b are measured to an accuracy of 0.1

To find

• Percentage error in the area of rectangle

Knowledge required

• Error of a product

Suppose , Z = AB

and measured values of A and B are A ± Δ A and B ± Δ B respectively. then,

$\bigstar\boxed{\sf{\dfrac{\triangle Z}{Z}=\dfrac{\triangle A}{A}+ \dfrac{\triangle B}{B}}}$

hence, the rule is : When two quantities are multiplied , the relative error in the result is the sum of the relative errors in the multipliers.

Solution

given that ,

Δ l = Δ b =  0.1

Let, Area of rectangle be A

then,

$\implies\sf{A=l\times b}\\\\\\ \implies\sf{\dfrac{\triangle A}{A}=\dfrac{\triangle l }{l} + \dfrac{\triangle b}{b}}\\\\\\\implies \sf{\dfrac{\triangle A}{A}=\dfrac{0.1}{25.0}+\dfrac{0.1}{16.7}}\\\\\\\implies\sf{\dfrac{\triangle A}{A}=0.004\;\;+0.006}\\\\\\\implies\sf{\dfrac{\triangle A}{A} = 0.01 }$

$\sf{converting\:into\:percentage\:error}\\\\\implies\sf{\dfrac{\triangle A}{A}\times 100 \% =0.01\times 100\%}\\\\ \\\implies\sf{\dfrac{\triangle A}{A}\times 100\% = 1 .0\; \%}$

$\implies\underline{\boxed{\sf{percentage\:error\:in\:A=1.0\;\%}}}$

Hence, the percentage error in Area of rectangle is 1.0 % .