NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 11: Area Related To Circles

Exercise 11.2

Step 1: A square has all sides equal. Here each side = a cm.

Step 2: An inscribed circle means the circle touches all four sides of the square from inside.

Step 3: The diameter of this circle = side of the square = a cm.

Step 4: Radius of the circle \(r = \dfrac{\text{diameter}}{2} = \dfrac{a}{2}\) cm.

Step 5: Formula for area of a circle = \(\pi r^2\). Substituting radius: \(\pi \left(\dfrac{a}{2}\right)^2 = \pi \dfrac{a^2}{4} = \dfrac{\pi a^2}{4}\) cm².

Step 6: The given claim is \(\pi a^2\) cm². But the actual area is only \(\dfrac{\pi a^2}{4}\) cm², which is one-fourth of the claim.

Final Answer: The statement is False.

Step 1: The circle has radius \(a\) cm. The diameter of the circle is twice the radius:

\( \text{Diameter} = 2 \times a = 2a \; \text{cm} \)

Step 2: A square circumscribing the circle means the circle just touches the middle of all four sides of the square. Therefore, the side of the square is equal to the diameter of the circle.

So, side of square = \(2a\) cm.

Step 3: The perimeter of a square is the total length around it. Formula: \( \text{Perimeter} = 4 \times \text{side} \)

Substitute the value of side:

\( \text{Perimeter} = 4 \times 2a = 8a \; \text{cm} \)

Step 4: Therefore, the statement is correct. Final Answer: True.

Step 1: Let the diameter of the circle be \(d\). Then the radius is half of the diameter:

\(r = \dfrac{d}{2}\).

Step 2: For the inner square (inscribed in the circle):

• The diagonal of this square is equal to the diameter of the circle (because all 4 vertices touch the circle).
• So, diagonal of inner square \( = 2r \).
• If the diagonal of a square is \(d_{sq}\), then side of square = \(\dfrac{d_{sq}}{\sqrt{2}}\).
• Therefore, side of inner square \(s_i = \dfrac{2r}{\sqrt{2}} = r\sqrt{2}\).
• Area of inner square \(A_i = (s_i)^2 = (r\sqrt{2})^2 = 2r^2\).

Step 3: For the outer square (circumscribed around the circle):

• The side of this square is equal to the diameter of the circle, because the circle touches the middle of each side of the square.
• So, side of outer square \(s_o = 2r\).
• Area of outer square \(A_o = (s_o)^2 = (2r)^2 = 4r^2\).

Step 4: Compare the two areas:

\(\dfrac{A_o}{A_i} = \dfrac{4r^2}{2r^2} = 2\).

Step 5: The outer square’s area is only 2 times the inner square’s area, not 4 times.

Final Answer: The statement is false.

Step 1: Understand the terms.

• A sector of a circle looks like a slice of pizza. Its area is given by:
  \( A_{\text{sector}} = \tfrac{1}{2} r^2 \theta \), where:
  • \(r\) = radius (in metres)
  • \(\theta\) = angle at the centre (in radians)
• A segment is part of the circle cut off by a chord (a straight line inside the circle). It is smaller than the whole sector because it does not include the triangular part.

Step 2: Relation between sector and segment.

Area of segment = Area of sector − Area of triangle (formed by the two radii and the chord).

Step 3: Area of the triangle.

The triangle is isosceles, and its area is:

\( A_{\text{triangle}} = \tfrac{1}{2} r^2 \sin \theta \), which is always positive (for \(0 < \theta < 2\pi\)).

Step 4: Substitution.

So, area of the segment:

\( A_{\text{segment}} = \tfrac{1}{2} r^2 \theta - \tfrac{1}{2} r^2 \sin \theta \).

Step 5: Compare.

Since we subtract a positive quantity (triangle area), the segment’s area is always smaller than the sector’s area.

Final Answer: Yes, it is true. The area of a segment is less than the area of its corresponding sector.

Step 1: When a circular wheel makes one complete revolution, the distance it covers is equal to the circumference of the wheel.

Step 2: The formula for circumference is:

\( C = 2\pi r \)   or   \( C = \pi d \)

where \(r\) = radius of the wheel and \(d\) = diameter of the wheel.

Step 3: In SI units, if the diameter is given as \(d\) cm, we must first convert to metres (since SI unit of length is metre):

\( d\,\text{cm} = \dfrac{d}{100}\,\text{m} \).

Step 4: Substituting into the circumference formula:

\( C = \pi d \) (in cm)   or   \( C = \pi \times \dfrac{d}{100} \) m (in SI).

Step 5: The statement says distance = \(2\pi d\). But the correct formula is \(\pi d\).

Conclusion: Therefore, the given statement is False.

Step 1: A wheel is circular. When it makes one complete turn (one revolution), the distance it moves forward is equal to the circumference of the wheel.

Step 2: The circumference of a circle is given by the formula:

\( C = 2\pi r \)

where \(r\) is the radius of the wheel (in metres, SI unit of length).

Step 3: So, in one revolution, the wheel covers a distance of \(2\pi r\) metres.

Step 4: If the wheel covers a total distance of \(s\) metres, then the number of revolutions is:

\( \text{Number of revolutions} = \dfrac{\text{Total distance}}{\text{Distance per revolution}} \)

\( = \dfrac{s}{2\pi r} \)

Step 5: This is exactly the same as given in the question. Therefore, the statement is True.

Step 1: Recall the formulas

For a circle of radius \(r\) metres:

• Area = \(\pi r^2\) square metres (\(m^2\))
• Circumference = \(2\pi r\) metres (\(m\))

Step 2: Compare their numerical values

We are not comparing the units (since area and length are different quantities), but only the numbers that come in front of the units.

So, compare: \(\pi r^2\) (number from area) and \(2\pi r\) (number from circumference).

Step 3: Cancel common factor

Both expressions have \(\pi\). Divide both sides by \(\pi\):

Compare \(r^2\) with \(2r\).

Step 4: Simplify

Divide both sides by \(r\) (assuming \(r > 0\)):

Compare \(r\) with 2.

Step 5: Interpret

• If \(r > 2\) m → Area number > Circumference number
• If \(r = 2\) m → Area number = Circumference number
• If \(r < 2\) m → Area number < Circumference number

Step 6: Conclusion

The statement says “the area is greater than the circumference” as if it is always true. But it depends on the radius:

Only for radii greater than 2 m is the area number larger. For smaller radii, it is not.

Therefore, the statement is False.

Step 1: Recall the formula for the length of an arc of a circle:

\( l = r \times \theta \)

Here:

• \(l\) = arc length (in metres, SI unit)
• \(r\) = radius (in metres)
• \(\theta\) = angle at the centre (in radians, SI unit)

Step 2: For the first circle (radius \(r\), angle \(\alpha\)):

Arc length = \( l_1 = r \alpha \).

Step 3: For the second circle (radius \(2r\), angle \(\beta\)):

Arc length = \( l_2 = 2r \beta \).

Step 4: It is given that these two arc lengths are equal:

\( l_1 = l_2 \Rightarrow r \alpha = 2r \beta \).

Step 5: Divide both sides by \(r\) (since \(r > 0\)):

\( \alpha = 2 \beta \).

Step 6: This shows that the angle of the sector of the first circle (\(\alpha\)) is double the angle of the sector of the second circle (\(\beta\)).

Step 7: The given statement claims exactly this relationship. So the statement is true.

Step 8: But the question is asking whether this statement is false. Since we proved it is true, the correct answer is that saying it is false is itself false.

Step 1: Recall the formula for the area of a sector:

\( A = \tfrac{1}{2} r^2 \theta \)

Here, \(r\) is the radius (in SI unit: metre), and \(\theta\) is the angle at the centre (in radians).

Step 2: We also know that the arc length of a sector is:

\( l = r \theta \)

Step 3: Substitute \(\theta = \tfrac{l}{r}\) into the area formula:

\( A = \tfrac{1}{2} r^2 \cdot \tfrac{l}{r} \)

Step 4: Simplify the expression:

\( A = \tfrac{1}{2} r \cdot l \)

Step 5: This shows that the area of a sector depends on both the radius \(r\) (metres) and the arc length \(l\) (metres).

Step 6: If two sectors have the same arc length \(l\), but the radii of the circles are different, then their areas will be different because the formula has \(r\) multiplied with \(l\).

Step 7: Therefore, the statement that their areas are equal is False, unless the radii are also the same.

Step 1: Recall the formula for the area of a sector in terms of its arc length.

For any circle, the area of a sector is given by:

\( A = \tfrac{1}{2} r \times l \)

where:

• \(A\) = area of the sector (measured in square metres, \(\text{m}^2\))
• \(r\) = radius of the circle (measured in metres, \(\text{m}\))
• \(l\) = length of the arc (measured in metres, \(\text{m}\))

Step 2: Write the condition given in the problem.

The areas of the two sectors are equal. That means:

\( A_1 = A_2 \)

So,

\( \tfrac{1}{2} r_1 l_1 = \tfrac{1}{2} r_2 l_2 \)

which simplifies to:

\( r_1 l_1 = r_2 l_2 \)

Step 3: Check if this condition forces \(l_1 = l_2\).

Notice that the equality only says the product of radius and arc length is the same for both sectors. This does not mean the arc lengths must be equal.

Step 4: Give an example to make it clear.

Suppose:

• Sector 1: radius \(r_1 = 2\,\text{m}\), arc length \(l_1 = 6\,\text{m}\)
• Sector 2: radius \(r_2 = 3\,\text{m}\), arc length \(l_2 = 4\,\text{m}\)

Now, \( r_1 l_1 = 2 \times 6 = 12 \) and \( r_2 l_2 = 3 \times 4 = 12 \).

So the areas of the sectors are equal. But \(l_1 = 6\,\text{m}\) and \(l_2 = 4\,\text{m}\), which are not equal.

Final Step: Therefore, it is not necessary that the arc lengths are equal if the areas are equal.

Answer: False.

Step 1: We are given a rectangle of length \(a\) cm and breadth \(b\) cm, where \(a > b\). This means the rectangle is longer than it is wide.

Step 2: We want to fit the largest possible circle inside this rectangle. A circle that fits inside a rectangle is called an inscribed circle.

Step 3: The circle cannot be larger than the smaller side of the rectangle, because it has to fit completely inside. The smaller side here is the breadth \(b\).

Step 4: So, the diameter of the largest circle will be exactly equal to the breadth \(b\). That means:

\[ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, D = b \, (\text{in cm}) \]

Step 5: The radius of a circle is half the diameter. Therefore:

\[ r = \dfrac{b}{2} \, (\text{in cm}) \]

Step 6: The formula for the area of a circle is:

\[ A = \pi r^2 \]

Step 7: Substituting \(r = \dfrac{b}{2}\):

\[ A = \pi \left( \dfrac{b}{2} \right)^2 \]

Step 8: Simplify the expression:

\[ A = \pi \cdot \dfrac{b^2}{4} = \dfrac{\pi b^2}{4} \, (\text{in cm}^2) \]

Step 9: The question says the area is \(\pi b^2\) cm². But our calculation shows it is actually \(\dfrac{\pi b^2}{4}\) cm².

Conclusion: Since \(\dfrac{\pi b^2}{4} \neq \pi b^2\), the statement is False.

Step 1: Recall the formula for the circumference of a circle.

Circumference \(C = 2 \pi r\), where \(r\) is the radius (measured in metres in SI).

Step 2: Suppose the circumferences of two circles are equal:

\(2 \pi r_1 = 2 \pi r_2\)

Step 3: Cancel \(2\pi\) from both sides:

\(r_1 = r_2\)

Step 4: If the radii are equal, then both circles are of the same size.

Step 5: Recall the formula for the area of a circle:

Area \(A = \pi r^2\), measured in square metres (m²).

Step 6: Since \(r_1 = r_2\), their areas will also be equal:

\(\pi r_1^2 = \pi r_2^2\)

Final Answer: Yes, if two circles have equal circumferences, their radii are equal, so their areas must also be equal. True.

Step 1: Formula for the area of a circle is:

\( A = \pi r^2 \) (where \(r\) is the radius in metres).

Step 2: If the areas of two circles are equal, then:

\( \pi r_1^2 = \pi r_2^2 \)

Step 3: Cancel \(\pi\) from both sides:

\( r_1^2 = r_2^2 \)

Step 4: Taking square root on both sides (and radius is always positive in SI units, so we ignore the negative value):

\( r_1 = r_2 \)

Step 5: Formula for circumference of a circle is:

\( C = 2 \pi r \) (measured in metres).

Step 6: Since \( r_1 = r_2 \), we get:

\( 2 \pi r_1 = 2 \pi r_2 \)

Final Answer: The circumferences of both circles are equal. So, the statement is True.

Step 1: A square is inscribed in a circle. This means all four vertices (corners) of the square touch the circle.

Step 2: In such a case, the diagonal of the square is equal to the diameter of the circle.

So, diagonal of the square = \(p\) cm.

Step 3: Formula for the area of a square in terms of its diagonal:

\[ \text{Area of square} = \dfrac{(\text{diagonal})^2}{2} \]

Step 4: Substitute diagonal = \(p\):

\[ \text{Area} = \dfrac{p^2}{2} \;\text{cm}^2 \]

Step 5: The statement given says the area is \(p^2\) cm². But we calculated it as \(\tfrac{p^2}{2}\) cm².

Final Answer: The statement is false. The correct area is \(\tfrac{p^2}{2}\) cm².