NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles — Exercise 11.1

NCERT Exemplar Problems, Class 10 Mathematics — Areas Related to Circles, Exercise 11.1. All 10 multiple-choice questions solved.

Ten questions, and every one of them runs on two formulae you have known since Class 7:

$$\text{Circumference} = 2\pi r, \qquad \text{Area} = \pi r^2$$

What the exercise actually tests is the squaring. Circumferences add like lengths, so a circle whose circumference is the sum of two others has $R = R_1 + R_2$. Areas add like squares, so a circle whose area is the sum of two others has $R^2 = R_1^2 + R_2^2$. Questions 1, 2, 6, 9 and 10 — half the exercise — are that single distinction, asked five different ways.

The other half is about inscribing and circumscribing, where the only thing to get right is which length is which:

• a circle inside a square of side $a$: diameter $= a$
• a square inside a circle of radius $r$: diagonal $= 2r$, so side $= r\sqrt2$

A note on the current syllabus. This chapter has been cut harder than any other in Class 10. The rationalised version has exactly one numbered section — “Areas of Sector and Segment of a Circle” — plus a Summary, and runs to seven pages. Both “Perimeter and Area of a Circle: A Review” and “Areas of Combinations of Plane Figures” have gone entirely.

So every question here uses a result the chapter no longer states, though the Summary’s three points silently assume it and it has been on the syllabus since Class 7. None of this exercise is hard; it is just no longer taught in the place you would look for it.

Key insight. Before touching a calculator, ask whether the quantity being added is a length or an area. Circumferences and radii add directly; areas add as squares of the radii. Getting that backwards is the single error this exercise is built to catch, and it accounts for one wrong option in almost every question.

Choose the correct answer from the given four options:

Question 1

If the sum of the areas of two circles with radii $R_1$ and $R_2$ is equal to the area of a circle of radius $R$, then

Solution. Write the condition out:

$$\pi R_1^2 + \pi R_2^2 = \pi R^2$$

Every term carries a factor of $\pi$, so divide it out:

$$R_1^2 + R_2^2 = R^2$$

That is option (B), and it is an equation, not an inequality — which disposes of (C) and (D) at once. The question says the areas are equal, so the relation between the radii must also be an equality.

Option (A) is the trap, and it is the wrong one for a good reason: $R_1 + R_2 = R$ would follow if the circumferences were being added, not the areas. Test it with numbers: $R_1 = 3$, $R_2 = 4$ gives $R = 5$ from (B), and indeed $9 + 16 = 25$. But $3 + 4 = 7 \ne 5$, so (A) fails.

Notice that (B) forces $R_1 + R_2 > R$ — because $(R_1+R_2)^2 = R_1^2 + R_2^2 + 2R_1R_2 > R^2$. So the sum of the two radii always overshoots.

Question 2

If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius $R$, then

Solution. The same manipulation with circumferences instead of areas:

$$2\pi R_1 + 2\pi R_2 = 2\pi R$$

Divide by $2\pi$:

$$R_1 + R_2 = R$$

Option (A).

Read this against Question 1 and the pair becomes the whole point of the exercise. Circumference is proportional to the radius, so adding circumferences adds radii. Area is proportional to the square of the radius, so adding areas adds squares. The two questions are deliberately placed side by side with nearly identical wording and different answers.

Option (D) is worth dismissing explicitly: the relation is completely determined, not indefinite. Whenever a proportionality holds — and circumference is exactly proportional to radius — a sum on one side gives a sum on the other.

Question 3

If the circumference of a circle and the perimeter of a square are equal, then

Solution. Let the circle have radius $r$ and the square have side $a$. The condition is

$$2\pi r = 4a \implies a = \frac{\pi r}{2}$$

Now compare the areas:

$$\text{circle} = \pi r^2, \qquad \text{square} = a^2 = \frac{\pi^2 r^2}{4}$$

Take the ratio, which is cleaner than subtracting:

$$\frac{\text{circle}}{\text{square}} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}} = \frac{4}{\pi}$$

Since $\pi \approx 3.14 < 4$, this ratio is greater than $1$ — about $1.27$. So the circle is larger, and option (B) is right.

The result is a real theorem, not an artefact of these two shapes. Among all plane figures of a given perimeter, the circle encloses the greatest area. Squares do better than rectangles, regular hexagons do better than squares, and the circle beats them all. So the answer here could have been predicted before any algebra.

Option (D) is wrong for the same reason as in Question 2: the ratio $\frac{4}{\pi}$ does not depend on $r$ at all, so the comparison is fixed for every size.

Question 4

Area of the largest triangle that can be inscribed in a semi-circle of radius $r$ units is

Solution. Take the base along the diameter and maximise base and height separately — they turn out to be maximised by the same triangle, which is why the answer is clean.

The base. Any chord of the semicircle is at most the diameter, so the longest possible base is

$$\text{base} = 2r$$

The height. With the base along the diameter, the third vertex lies on the arc. Its distance from the diameter is at most the radius, achieved at the top of the arc:

$$\text{height} = r$$

Both maxima are attained together by the isosceles triangle whose apex is the midpoint of the arc, so

$$\text{Area} = \frac12 \times 2r \times r = r^2$$

Option (C), $2r^2$, comes from forgetting the $\frac12$ in the triangle formula. Option (B) comes from taking the base as $r$ rather than $2r$.

A quick plausibility check: the semicircle has area $\frac{\pi r^2}{2} \approx 1.57r^2$, and the triangle sits inside it. So the answer must be below $1.57r^2$, which rules out (C) immediately. It must also be a decent fraction of that — $r^2$ is about $64\%$ of the semicircle, which is right for a triangle filling it as fully as a triangle can.

Question 5

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

Solution. Question 3 has already done the work: with equal perimeters,

$$\frac{\text{area of circle}}{\text{area of square}} = \frac{4}{\pi}$$

Now put in the chapter’s value $\pi = \dfrac{22}{7}$:

$$\frac{4}{\pi} = 4 \times \frac{7}{22} = \frac{28}{22} = \frac{14}{11}$$

So the ratio is $14 : 11$.

Check the order before choosing. The question says “the ratio of their areas” with the circle named first, so it is circle : square. Option (D), $11 : 14$, is the same ratio the other way round — and Question 3 has already told you which of the two is bigger, so the first number must be the larger. That alone settles it between (B) and (D).

Options (A) and (C) are $\pi$ and $\frac{1}{\pi}$ themselves, which is what you get by comparing the areas without squaring the relation between $a$ and $r$.

Question 6

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters $16$ m and $12$ m in a locality. The radius of the new park would be

Solution. The diameters are given, so halve them first: the radii are $8$ m and $6$ m.

Areas add, so squares of radii add (Question 1):

$$R^2 = 8^2 + 6^2 = 64 + 36 = 100 \implies R = 10 \text{ m}$$

the $6$–$8$–$10$ triangle.

Two traps are waiting here, and both appear among the options.

Forgetting to halve. Using the diameters as radii gives $R^2 = 256 + 144 = 400$, so $R = 20$ — option (C). That is in fact the diameter of the correct answer, which is what makes it so tempting.

Adding instead of squaring. $8 + 6 = 14$, which is not offered, but $16 + 12 – 4 = 24$ is — option (D) is the sum of the two diameters minus nothing sensible, and option (B) is close to the $14$ that the wrong method produces.

The reliable habit is to write down the radii on their own line before starting, and to note which of “add the radii” and “add the squares” the question calls for.

Question 7

The area of the circle that can be inscribed in a square of side $6$ cm is

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

$$\text{diameter} = \text{side} = 6 \text{ cm} \implies r = 3 \text{ cm}$$

$$\text{Area} = \pi r^2 = 9\pi \text{ cm}^2$$

Option (A), $36\pi$, is what you get by using the side as the radius — the same halving error as Question 6, and the commonest slip in the whole chapter.

A sanity check that catches it instantly: the square has area $36$ cm², and the inscribed circle must be smaller. Since $9\pi \approx 28.3$, the circle fills about $79\%$ of the square, which is right — an inscribed circle always occupies $\frac{\pi}{4} \approx 78.5\%$ of its square, whatever the size. If your answer exceeds the square’s own area, the halving has gone missing.

Question 8

The area of the square that can be inscribed in a circle of radius $8$ cm is

Solution. The square is inside the circle this time, with all four corners on the circle. The longest line inside the square is its diagonal, and that is what must equal the diameter:

$$\text{diagonal} = 2r = 16 \text{ cm}$$

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

$$a\sqrt2 = 16 \implies a = \frac{16}{\sqrt2} = 8\sqrt2$$

$$\text{Area} = a^2 = \left(8\sqrt2\right)^2 = 64 \times 2 = 128 \text{ cm}^2$$

There is a shortcut worth knowing, which avoids the surd entirely. The two diagonals of a square are equal and perpendicular, so

$$\text{Area} = \frac12 d_1 d_2 = \frac12 \times 16 \times 16 = 128 \text{ cm}^2$$

Option (A), $256$, is the square of the diameter — the square circumscribing this circle, not inscribed in it. Option (D), $64$, treats the radius as the side.

Compare with Question 7. There the circle was inside and the side was the diameter; here the square is inside and the diagonal is the diameter. Whichever figure is inside, its longest internal measurement is what the outer figure constrains.

Question 9

The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters $36$ cm and $20$ cm is

Solution. Circumferences add, so diameters add — that is the fastest route here, because the question gives diameters.

$$\pi D = \pi(36) + \pi(20) \implies D = 56 \text{ cm}$$

But the question asks for the radius:

$$R = \frac{56}{2} = 28 \text{ cm}$$

Option (A), $56$ cm, is the diameter — the number you find one line before the answer, and the reason this question is worth marking carefully. Whenever a question mixes diameters in the data with a radius in the answer, write down which you have at each stage.

The same by radii, for completeness: the given radii are $18$ and $10$, and $18 + 10 = 28$ ✓.

Question 10

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii $24$ cm and $7$ cm is

Solution. Areas add, so squares of radii add:

$$R^2 = 24^2 + 7^2 = 576 + 49 = 625 \implies R = 25 \text{ cm}$$

the $7$–$24$–$25$ triangle. The question asks for the diameter:

$$D = 2 \times 25 = 50 \text{ cm}$$

Every wrong option here is a real number from the working, which is what makes this a good final question:

• (B) $25$ cm is the radius — the answer to a question that was not asked.
• (A) $31$ cm is $24 + 7$, the result of adding radii when areas were meant.
• (C) $62$ cm is $2 \times 31$, the same error followed by the correct doubling.

Only (D) gets both steps right. The pattern to internalise: square, add, root, then check whether a radius or a diameter was requested.

Common mistakes

• Questions 1, 2, 9 and 10 — adding the wrong quantity. Circumferences and radii add directly; areas add as squares of the radii. Every one of these four questions offers the answer produced by the other rule, so the arithmetic will never warn you.
• Questions 6, 7 and 9 — forgetting to halve a diameter, or to double a radius. Question 6 gives diameters and wants a radius, Question 9 gives diameters and wants a radius, Question 10 gives radii and wants a diameter. Write down $r$ and $d$ separately at the start.
• Questions 7 and 8 — using the wrong internal measurement. A circle inside a square has its diameter equal to the side; a square inside a circle has its diagonal equal to the diameter. Whichever shape is inside, ask which of its measurements is longest and set that against the outer shape.
• Question 4 — dropping the $\frac12$. The largest triangle in a semicircle has base $2r$ and height $r$, giving $r^2$ — not $2r^2$. A useful check is that the triangle must fit inside the semicircle’s area of $\frac{\pi r^2}{2} \approx 1.57r^2$.
• Question 5 — giving the ratio the wrong way round. The question names the circle first, and Question 3 has already established that the circle is the larger, so the first number must exceed the second. That alone decides between $14:11$ and $11:14$.
• Questions 3 and 5 — treating the comparison as size-dependent. The ratio $\frac{4}{\pi}$ has no $r$ in it, so the circle beats the square at every scale. Options offering “nothing definite can be said” are wrong wherever a proportionality is involved.
• Throughout — computing decimals too early. Keeping $\pi$ symbolic until the last step makes the cancellations visible; in Question 3 the whole answer is the single ratio $\frac{4}{\pi}$, which a decimal would have hidden.

Practise next

• Exercise 11.2 — fourteen true-or-false statements on exactly these relations, plus the sector and segment formulae. Two of them are Questions 7 and 8 here restated as claims to be judged.
• Exercise 11.3 — sixteen short-answer questions, where these results are combined with sectors and segments to find shaded areas of composite figures.
• Exercise 9.1 — worth revisiting for the circle geometry that fixes the lengths these area formulae then use.