Question

The length and breadth of rectangle are measured as 10 cm and 5 cm respectively with 0.1 cm accuracy.Calculate area of rectangle in terms of error limits.​

Answer 1

Answer:

• Area of Rectangle in terms of error limit = 50 ± 1.5 cm sq.

Explanation:

Given:

• Length of Rectangle with accuracy = 10 ± 0.1 cm
• Breadth of Rectangle with accuracy = 5 ± 0.1 cm

To Find:

• Area of Rectangle in terms of error limits.

Now, we know that,

=> Area of Rectangle = length × breadth

=> Area of Rectangle = 10 × 5

=> Area of Rectangle = 50 cm sq.

Now, we will find area of Rectangle by error limit.

=> ∆A = A × (∆L/L + ∆B/B)

Now, put the values

=> ∆A = 50 × (0.1/10 + 0.1/5)

=> ∆A = 50 × (0.01 + 0.02)

=> ∆A = 50 × (0.03)

=> ∆A = 1.5

So, Area of Rectangle in terms of error limit are 50 ± 1.5 cm sq.

Answer 2

★ GiveN :

Length of rectangle with accuracy = 10 ± 0.1 cm

Breadth of rectangle with accuracy = 5 ± 0.1 cm

$\rule{200}{1}$

★ To FinD :

We have to find the area of rectangle in terms of error limits.

$\rule{200}{1}$

★ SolutioN :

Firstly, we will calculate the area of rectangle.

So, We know the formula to find the area of rectangle.

$\Large{\implies{\boxed{\boxed{\sf{Area = Length \times Breadth}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow Area = 10 \times 5} \\ \\ \sf{\dashrightarrow Area = 50} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Area = 50 \: cm^2}}}}}$

$\rule{150}{2}$

Now, we will calculate the area of rectangle in terms of error limits by using error limit method.

$\Large{\implies{\boxed{\boxed{\sf{\Delta A = A \times \bigg( \frac{\Delta L}{L} + \frac{\Delta B}{B} \bigg)}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow \Delta A = 50 \times \bigg(\frac{0.1}{10} + \frac{0.1}{5} \bigg)} \\ \\ \sf{\dashrightarrow \Delta A = 50 \times \frac{0.1 + 0.2}{10}} \\ \\ \sf{\dashrightarrow \Delta A = 50 \times \frac{0.3}{10}} \\ \\ \sf{\dashrightarrow \Delta A = 50 \times 0.03} \\ \\ \sf{\dashrightarrow \Delta A = 1.5}$

$\therefore$ Area of rectangle in terms of error limiting is 50 ± 1.5 cm².