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Question

Find the circumference (in m) of the largest circle that can be inscribed in a rectangle whose dimensions are given as 114 m and 63 m.
Take $\pi = \frac{22}{7}$

The correct answer is
198

To find the circumference of the largest circle that can be inscribed within a rectangle, we need to determine the circle's diameter. The largest possible circle that fits inside a rectangle will have a diameter equal to the length of the shorter side of the rectangle.

Rectangle Dimensions Analysis

The given dimensions of the rectangle are 114 m and 63 m.

  • Length = 114 m
  • Width = 63 m

Comparing the two dimensions, the shorter side is 63 m.

Diameter of the Inscribed Circle

Therefore, the diameter (d) of the largest circle that can be inscribed in this rectangle is equal to the shorter side:

d = 63 m

Calculating the Circumference

The formula for the circumference (C) of a circle is:

$$ C = \pi d $$

We are given that $\pi = \frac{22}{7}$.

Substituting the values:

$$ C = \frac{22}{7} \times 63 $$

Now, perform the calculation:

$$ C = 22 \times \frac{63}{7} $$

$$ C = 22 \times 9 $$

$$ C = 198 $$

The circumference is in meters (m).

Conclusion

The circumference of the largest circle that can be inscribed in the rectangle is 198 m.

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Important Questions from 2-D Mensuration

  1. The length of a rectangular plot is $(x^2 + xy + y^2)$ m and its breadth is $(x^2 - 5xy - y^2)$ m.
    Find its perimeter, when $x = 1$ and $y = -1$.
  2. The ratio between the perimeter and breadth of a rectangle is 3: 1. If the area of the rectangle is $98 \text{ cm}^2$, find the perimeter (in cm) of the rectangle.
  3. The shorter side of a rectangle is 15 cm less than the longer side. The numerical value of its area is equal to 5 times the numerical value of its perimeter. What is the length (in cm) of its longer side?

  4. In a circle of radius 10.5 cm, if the angle of a sector is $\frac{2\pi}{3}$, then the perimeter of the sector is (in cm):
    (Take $\pi = \frac{22}{7}$)
  5. The length of a rectangular pitch is 30 m more than its breadth. Its area is $18,271$ m$^{2}$. Its breadth (in m) is:
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