NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles — Exercise 11.2

NCERT Exemplar Problems, Class 10 Mathematics — Areas Related to Circles, Exercise 11.2. All 14 questions solved.

Fourteen statements to judge, and nine of them are false — an unusually high proportion, and a hint about how they are built. Most are a correct formula with one factor wrong: $\pi a^2$ where $\frac{\pi a^2}{4}$ belongs, $2\pi d$ where $\pi d$ belongs, $p^2$ where $\frac{p^2}{2}$ belongs.

So the working method is to derive the quantity yourself and compare, rather than to look at the printed expression and ask whether it feels right. A factor of $4$ is invisible to intuition.

Three results carry most of the exercise:

$$\text{arc length } \ell = \frac{\theta}{360^\circ} \times 2\pi r, \qquad \text{sector area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{\ell r}{2}$$

That last form, sector area $= \frac{\ell r}{2}$, is the one worth memorising: it settles Questions 9 and 10 immediately, because it shows that equal arcs on different circles give different areas.

Two of the printed answers are wrong, and this page says which and why — Question 11’s verdict is simply the opposite of the truth, and Question 8’s answers a question other than the one printed. Both are worked in full below.

A note on the current syllabus. This chapter is now a single numbered section, “Areas of Sector and Segment of a Circle”, plus a Summary — seven pages in total, with “Areas of Combinations of Plane Figures” removed entirely. One consequence matters for Question 4. The textbook’s Summary states, without qualification:

Area of segment of a circle = Area of the corresponding sector − Area of the corresponding triangle.

That is precisely the statement this exercise asks you to judge, and the answer is that it holds only for the minor segment. The textbook is not wrong — its body carries a Remark that “segment” and “sector” mean the minor ones unless stated otherwise, and a Note giving the major-segment formula. But a student revising from the Summary alone will walk into Question 4.

Key insight. When a statement compares two circles, ask what is being held fixed. Equal arc lengths on different circles force different angles and different sector areas; equal areas force different arcs. Only within one circle do arc, angle and sector area rise and fall together. Questions 8, 9 and 10 are that one observation, asked three ways.

Write ‘True’ or ‘False’ and justify your answer in each of the following:

Question 1

Is the area of the circle inscribed in a square of side $a$ cm, $\pi a^2$ cm²? Give reasons for your answer.

Solution. A circle inscribed in a square touches all four sides, so it spans the full width of the square:

$$\text{diameter} = a \implies r = \frac{a}{2}$$

Therefore

$$\text{Area} = \pi r^2 = \pi\left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{4} \text{ cm}^2$$

The claimed $\pi a^2$ is four times too large. No.

A check that needs no algebra: the square itself has area $a^2$, and the inscribed circle must be smaller. But $\pi a^2 \approx 3.14a^2$ is more than three times the square’s own area, which is impossible for a figure drawn inside it. Whenever a claimed area exceeds the region containing it, the error is a missing division.

Question 2

Will it be true to say that the perimeter of a square circumscribing a circle of radius $a$ cm is $8a$ cm? Give reasons for your answer.

Solution. The square is now on the outside, with the circle touching all four of its sides. The circle spans the square’s width, so

$$\text{side of square} = \text{diameter of circle} = 2a \text{ cm}$$

and the perimeter is

$$4 \times 2a = 8a \text{ cm}$$

True.

Compare with Question 1, which is the same configuration with the roles reversed. There the circle was inside and the halving was needed; here the square is outside and the doubling is needed. The two are placed together because the same relation — side $=$ diameter — is applied in opposite directions.

Question 3

In Fig 11.3, a square is inscribed in a circle of diameter $d$ and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer.

Solution. Name both squares in terms of $d$.

The inner square has its four corners on the circle, so its diagonal is the diameter:

$$\text{diagonal} = d \implies \text{side} = \frac{d}{\sqrt2} \implies \text{area} = \frac{d^2}{2}$$

The outer square has its sides touching the circle, so its side is the diameter:

$$\text{side} = d \implies \text{area} = d^2$$

The ratio is

$$\frac{d^2}{\frac{d^2}{2}} = 2$$

So the outer square is twice the inner one, not four times. No.

Where the plausible “four times” comes from. The outer square’s side is $\sqrt2$ times the inner square’s side, and people reach for “double the side means four times the area”. But the side is not doubled — it is multiplied by $\sqrt2$, and $(\sqrt2)^2 = 2$.

The key’s own phrasing puts it neatly: the side of the outer square equals the diagonal of the inner square. Since a square’s diagonal is $\sqrt2$ times its side, the areas are in the ratio $2 : 1$.

Question 4

Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?

Solution. It depends which segment, and that is the whole question.

For a minor segment. The minor segment is what remains of the minor sector after the triangle formed by the two radii and the chord is removed:

$$\text{minor segment} = \text{minor sector} – \text{triangle}$$

Since the triangle has positive area, the segment is indeed smaller than its sector.

For a major segment. The corresponding sector is now the major sector, and the triangle lies outside it, on the far side of the chord. So the triangle must be added:

$$\text{major segment} = \text{major sector} + \text{triangle}$$

and the segment is larger than its sector.

So the statement is false in general — it holds only for the minor segment.

A warning about revising from the Summary. The rationalised Class 10 textbook’s Summary states, with no qualification, that “Area of segment of a circle = Area of the corresponding sector − Area of the corresponding triangle”. Read on its own that is the claim this question declares false. The textbook is saved by a Remark in its body — “when we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise” — and by a Note giving the major-segment formula. This question is exactly the case where things are stated otherwise.

Question 5

Is it true that the distance travelled by a circular wheel of diameter $d$ cm in one revolution is $2\pi d$ cm? Why?

Solution. In one revolution a wheel lays down exactly its own circumference along the ground. So the distance is the circumference, and for a wheel of diameter $d$:

$$\text{circumference} = \pi d \text{ cm}$$

The claim of $2\pi d$ is twice the truth. No.

The error is a familiar one: $2\pi r$ and $\pi d$ are the same thing, because $d = 2r$. Writing $2\pi d$ applies the doubling twice over.

A number makes it concrete. A wheel of diameter $70$ cm travels $\pi \times 70 \approx 220$ cm per revolution, not $440$ cm. If $440$ were right, a bicycle wheel would cover twice the ground it actually does.

Question 6

In covering a distance $s$ metres, a circular wheel of radius $r$ metres makes $\dfrac{s}{2\pi r}$ revolutions. Is this statement true? Why?

Solution. Each revolution covers one circumference, which for radius $r$ is $2\pi r$ metres. So the number of revolutions needed to cover $s$ metres is

$$\frac{\text{total distance}}{\text{distance per revolution}} = \frac{s}{2\pi r}$$

True.

This is Question 5 done correctly, which is why the two sit together. There the circumference was given in terms of the diameter and wrongly doubled; here it is given in terms of the radius and is right.

A units check confirms the structure: $s$ is in metres and $2\pi r$ is in metres, so the quotient is a pure number — which is what a count of revolutions must be. An expression like $s \times 2\pi r$ would have units of square metres and could not be a count.

Question 7

The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?

Solution. Compare the two expressions directly:

$$\text{area} = \pi r^2, \qquad \text{circumference} = 2\pi r$$

$$\pi r^2 > 2\pi r \iff r > 2$$

dividing both sides by the positive quantity $\pi r$.

So the statement holds only when $r > 2$, and fails otherwise. It is therefore false as a general claim.

The three cases are worth naming:

• $r < 2$: the circumference is numerically larger. A circle of radius $1$ has area $\pi \approx 3.14$ and circumference $2\pi \approx 6.28$.
• $r = 2$: they are numerically equal, both $4\pi$.
• $r > 2$: the area is larger. At $r = 7$ the area is $154$ and the circumference $44$.

A word on “numerical value”. The question says numerical value precisely because the two quantities have different units — one is an area and the other a length, so comparing them is meaningless unless you agree to compare the bare numbers. That also means the comparison depends on the unit chosen: the same circle can have $r > 2$ in centimetres and $r < 2$ in inches.

Question 8

If the length of an arc of a circle of radius $r$ is equal to that of an arc of a circle of radius $2r$, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?

Solution. Work out whether the claim is true, then answer the question that was actually asked — which is not the same thing.

Let the first circle have radius $r$ and sector angle $\theta_1$, and the second have radius $2r$ and sector angle $\theta_2$. Their arc lengths are

$$\ell_1 = \frac{\theta_1}{360^\circ} \times 2\pi r, \qquad \ell_2 = \frac{\theta_2}{360^\circ} \times 2\pi(2r)$$

Setting $\ell_1 = \ell_2$ and cancelling the common $\dfrac{2\pi r}{360^\circ}$:

$$\theta_1 = 2\theta_2$$

So the first circle’s angle is double the second’s. The claim in the statement is true.

The question asks: “Is this statement false?” Since the statement is true, the answer is No — it is not false.

A concrete instance. Take $r = 7$ cm, so the circles have radii $7$ and $14$. A $60^\circ$ sector of the small circle has arc $\frac{60}{360} \times 2\pi(7) = \frac{22}{3}$ cm. The second circle needs $\frac{22}{3} = \frac{\theta_2}{360} \times 2\pi(14)$, giving $\theta_2 = 30^\circ$. And $60^\circ$ is indeed double $30^\circ$.

The printed answer answers a different question. The key gives “Yes”, with the reason “it will be true for the arcs of the same circle”. Two things are wrong with that.

First, “Yes” would mean the statement is false, and the calculation above shows it is not. The key appears to be answering “is this statement true?” — which is how Questions 7 and 9 are phrased — rather than the negatively-worded question actually printed here.

Second, the reason given belongs to Questions 9 and 10. Those are about a property that fails when the circles differ, so “it will be true for the arcs of the same circle” is the right explanation there. Here the relation $\theta_1 = 2\theta_2$ holds precisely because the circles are different; on a single circle, equal arcs would give equal angles, not doubled ones. The sentence cannot be this question’s reason under any reading.

Question 9

The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?

Solution. Use the compact form of the sector area, which is exactly built for this comparison:

$$\text{sector area} = \frac{\ell r}{2}$$

where $\ell$ is the arc length. (It follows from $\frac{\theta}{360}\pi r^2 = \frac{r}{2} \times \frac{\theta}{360} \times 2\pi r$.)

Now the two sectors have the same $\ell$ but belong to circles of different radii, so their areas $\frac{\ell r_1}{2}$ and $\frac{\ell r_2}{2}$ are in the ratio $r_1 : r_2$ — and that is not $1 : 1$. The areas are not equal, so the statement is false.

Numbers make it vivid. Take $\ell = \frac{44}{3}$ cm on circles of radii $7$ cm and $21$ cm:

$$\text{area}_1 = \frac{\frac{44}{3} \times 7}{2} = \frac{154}{3} \approx 51.3 \text{ cm}^2, \qquad \text{area}_2 = \frac{\frac{44}{3} \times 21}{2} = 154 \text{ cm}^2$$

Three times as large, from the same length of arc. This is exactly Question 16 of Exercise 11.4, where the same pair of circles is set as a calculation.

Within a single circle the statement would be true — there $r$ is fixed, so $\frac{\ell r}{2}$ depends only on $\ell$.

Question 10

The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?

Solution. The converse of Question 9, and it fails for the same reason. From

$$\text{area} = \frac{\ell r}{2}$$

equal areas give

$$\ell_1 r_1 = \ell_2 r_2 \implies \frac{\ell_1}{\ell_2} = \frac{r_2}{r_1}$$

So with different radii the arcs must be in the inverse ratio — the bigger circle needs the shorter arc to make the same area. They cannot be equal unless the radii are.

A worked instance. Take circles of radii $7$ cm and $14$ cm and ask for sectors of area $77$ cm² each:

• radius $7$: $\frac{\ell \times 7}{2} = 77$ gives $\ell = 22$ cm
• radius $14$: $\frac{\ell \times 14}{2} = 77$ gives $\ell = 11$ cm

Equal areas, and one arc is twice the other.

So it is not necessary — and in fact it is impossible, unless the two circles happen to be the same size.

Question 11

Is the area of the largest circle that can be drawn inside a rectangle of length $a$ cm and breadth $b$ cm $(a > b)$ is $\pi b^2$ cm²? Why?

Solution. The circle has to fit within both dimensions, so its diameter cannot exceed the smaller one. Since $a > b$, the breadth is the binding constraint:

$$\text{diameter} = b \implies r = \frac{b}{2}$$

Therefore

$$\text{Area} = \pi r^2 = \pi\left(\frac{b}{2}\right)^2 = \frac{\pi b^2}{4} \text{ cm}^2$$

The claimed $\pi b^2$ is four times too large. The answer is No.

This is Question 1 with a rectangle in place of the square, and the same factor of $4$ goes missing. The check is the same too: the rectangle has area $ab$, and $\pi b^2 \approx 3.14b^2$ would exceed it whenever $a < 3.14b$ — so for a rectangle only slightly longer than it is wide, the claimed circle would be bigger than the rectangle containing it.

The printed answer is wrong. The key gives “Yes”, with the reason “radius of the circle breadth of the rectangle” — a sentence with a word missing, as printed. The mathematics is not in doubt: a circle of radius $\frac{b}{2}$ has area $\frac{\pi b^2}{4}$, which is not $\pi b^2$, so the verdict must be No.

The likely explanation is visible in the key itself. The intended reason was presumably “radius of the circle $=$ half the breadth of the rectangle”, which supports a verdict of No; and the “Yes” looks to have been carried down from the two entries immediately following, Questions 12 and 13, which both begin “Yes, their radii are equal”.

Question 12

Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?

Solution. Yes, and the reason is that a circle is completely determined by its radius.

$$2\pi r_1 = 2\pi r_2 \implies r_1 = r_2$$

Once the radii are equal the circles are congruent, so every measurement of them agrees — areas included:

$$\pi r_1^2 = \pi r_2^2$$

Why this differs from Questions 9 and 10. There, two sectors were compared, and a sector has two independent parameters — its radius and its angle. Fixing one quantity (the arc length, or the area) leaves the other free, so the circles could differ. A whole circle has only one parameter, so fixing any one of its measurements fixes all of them.

That is the distinction the three questions are testing together: for whole circles, circumference, radius, area and diameter all determine one another; for sectors, they do not.

Question 13

Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?

Solution. The converse of Question 12, and true for the same reason.

$$\pi r_1^2 = \pi r_2^2 \implies r_1^2 = r_2^2 \implies r_1 = r_2$$

taking positive roots, since a radius is a length. Equal radii give equal circumferences:

$$2\pi r_1 = 2\pi r_2$$

The step worth pausing on is $r_1^2 = r_2^2 \implies r_1 = r_2$. In general $x^2 = y^2$ gives $x = \pm y$, so the negative possibility has to be dismissed — and here it is dismissed by the fact that radii are positive. Questions of this shape elsewhere are not always so safe; Exercise 8.4’s Questions 5 and 11 turn on exactly that point.

Question 14

Is it true to say that area of a square inscribed in a circle of diameter $p$ cm is $p^2$ cm²? Why?

Solution. A square inscribed in a circle has all four corners on the circle, so its longest internal line — the diagonal — is the diameter:

$$\text{diagonal} = p$$

For a square of side $s$ the diagonal is $s\sqrt2$, so

$$s\sqrt2 = p \implies s = \frac{p}{\sqrt2} \implies \text{area} = s^2 = \frac{p^2}{2} \text{ cm}^2$$

The claimed $p^2$ is twice the truth. No.

The quickest route avoids the surd altogether. The diagonals of a square are equal and perpendicular, and both are diameters here, so

$$\text{area} = \frac12 d_1 d_2 = \frac12 \times p \times p = \frac{p^2}{2}$$

Where $p^2$ would be right. It is the area of the square circumscribing the same circle, whose side — not diagonal — equals $p$. So the claim confuses the inscribed square with the circumscribed one, and the two differ by the factor of $2$ that Question 3 established.

Common mistakes

• Questions 1, 11 and 14 — losing a factor while converting between a diameter and a radius. Each claims an area that is $2$ or $4$ times too large, and the error is always the same step. A reliable check: an inscribed figure can never have more area than the figure containing it, so compare against that before deciding.
• Question 5 — writing $2\pi d$. The circumference is $2\pi r$ or $\pi d$, never both doublings at once. Choose whether you are working in radii or diameters and stay there.
• Question 4 — quoting the Summary’s segment formula. “Segment $=$ sector $-$ triangle” holds for the minor segment; for the major one the triangle is added. The textbook’s own Remark makes the convention explicit, and this question is the case where the convention is set aside.
• Questions 9 and 10 — expecting arc length to determine sector area. A sector has two free parameters, so fixing the arc leaves the radius free. Use sector area $= \frac{\ell r}{2}$ and the dependence on $r$ is impossible to miss.
• Questions 12 and 13 — treating whole circles like sectors. A circle has only one parameter, so any single measurement pins down all the others. That is exactly why the answers here are “yes” while Questions 9 and 10 are “no”.
• Question 3 — reasoning that $\sqrt2$ times the side means $4$ times the area. Doubling the side quadruples the area, but the side here is multiplied by $\sqrt2$, so the area doubles.
• Question 8 — answering the statement instead of the question. The statement is true, and the question asks whether it is false, so the answer is “No”. Negatively-worded questions repay reading twice — the printed key itself was caught by this one.

Practise next

• Exercise 11.3 — sixteen short-answer questions, where the sector and segment formulae of Questions 4, 9 and 10 are applied to composite shaded regions.
• Exercise 11.4 — twenty long-answer questions, including Question 16, which is Question 9 here set as a calculation on circles of radii $7$ cm and $21$ cm.
• Exercise 11.1 — worth a second pass, especially Questions 7 and 8, which are Questions 1 and 14 here with numbers instead of letters.