NCERT Solutions Class 9 Maths Exercise 6.3 — Measuring Space: Perimeter and Area

NCERT Class 9 Mathematics — Ganita Manjari Part I, Measuring Space: Perimeter and Area, Exercise Set 6.3. All 10 questions solved.

This exercise is the area counterpart of Exercise Set 6.1. Three formulas do the work:

$$\text{circle} = \pi r^2 \qquad \text{sector} = \frac{\theta}{360} \times \pi r^2 \qquad \text{segment} = \text{sector} – \text{triangle}$$

A sector is the pie slice between two radii; a segment is the smaller piece cut off by the chord alone. The segment is always the sector minus the triangle formed by the two radii and the chord — which is why every segment question needs two calculations, not one.

Watch the value of $\pi$: the exercise says $\frac{22}{7}$ unless told otherwise, and Questions 4 and 5 say otherwise.

Key insight. A sector is a fraction of the whole circle, and the fraction is $\frac{\theta}{360}$. Once that is written down, every sector and segment question is the circle’s area multiplied by a fraction, possibly with a triangle subtracted. Even the minute hand of Question 3 is just a sector, once you convert minutes into degrees.

Question 1

Find the area of a sector of a circle with radius $7$ cm if the angle of the sector is $60°$.

Solution. $60°$ is $\frac{60}{360} = \frac{1}{6}$ of a full turn, so the sector is one sixth of the circle:

$$\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 7^2 = \frac{1}{6} \times 154 = \frac{77}{3} \approx 25.67 \text{ cm}^2$$

Question 2

Find the area of a quadrant of a circle whose circumference is $44$ cm.

Solution. First find the radius from the circumference:

$$2 \times \frac{22}{7} \times r = 44 \quad\Longrightarrow\quad r = 7 \text{ cm}$$

A quadrant is a quarter of the circle:

$$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 49 = \frac{154}{4} = 38.5 \text{ cm}^2$$

Question 3

The length of the minute hand of a clock is $7$ cm. Find the area swept by the minute hand in $10$ minutes.

Solution. The minute hand sweeps a full circle in $60$ minutes, so in $10$ minutes it sweeps

$$\frac{10}{60} \times 360° = 60°$$

The area swept is the sector of radius $7$ cm and angle $60°$ — the same sector as Question 1:

$$\text{Area} = \frac{60}{360} \times \frac{22}{7} \times 49 = \frac{77}{3} \approx 25.67 \text{ cm}^2$$

Question 4

A chord of a circle of radius $10$ cm subtends $90°$ at the centre. Find the area of the corresponding: (i) minor sector, (ii) major sector. (Use $\pi \approx 3.14$.)

Solution. The whole circle has area

$$\pi r^2 = 3.14 \times 100 = 314 \text{ cm}^2$$

(i) Minor sector, angle $90°$, a quarter of the circle:

$$\frac{90}{360} \times 314 = \frac{314}{4} = 78.5 \text{ cm}^2$$

(ii) Major sector, the rest of the circle, angle $360° – 90° = 270°$:

$$\frac{270}{360} \times 314 = \frac{3}{4} \times 314 = 235.5 \text{ cm}^2$$

Check: $78.5 + 235.5 = 314$ ✓ — the two sectors make up the whole circle.

Question 5

A chord of a circle of radius $15$ cm subtends an angle of $60°$ at the centre. Find the areas of the corresponding minor and major segments. (Use $\pi \approx 3.14$ and $\sqrt3 \approx 1.73$.)

Solution. A segment is a sector minus a triangle, so both pieces are needed.

The minor sector, angle $60°$:

$$\frac{60}{360} \times 3.14 \times 225 = \frac{1}{6} \times 706.5 = 117.75 \text{ cm}^2$$

The triangle. Two radii with $60°$ between them make an equilateral triangle of side $15$ cm (the base angles are also $60°$). Its area is

$$\frac{\sqrt3}{4}(15)^2 = \frac{1.73}{4} \times 225 = 97.3125 \text{ cm}^2$$

The minor segment:

$$117.75 – 97.3125 = 20.4375 \approx 20.44 \text{ cm}^2$$

The major segment is everything else in the circle:

$$706.5 – 20.4375 = 686.0625 \approx 686.06 \text{ cm}^2$$

Question 6

A car has two wipers which do not overlap. Each wiper has a blade of length $28$ cm and sweeps through an angle of $120°$. Find the total area cleaned at each sweep of the blades.

Solution. Each blade sweeps a sector of radius $28$ cm through $120°$ — a third of a circle:

$$\frac{120}{360} \times \frac{22}{7} \times 28^2 = \frac{1}{3} \times \frac{22}{7} \times 784 = \frac{1}{3} \times 2464 = \frac{2464}{3} \text{ cm}^2$$

The wipers do not overlap, so the two areas simply add and nothing is double-counted:

$$\text{total} = 2 \times \frac{2464}{3} = \frac{4928}{3} \approx 1642.67 \text{ cm}^2$$

Question 7

*A chord of a circle of radius $r$ subtends an angle of $60°$ at the centre. Show that the area of the corresponding minor segment is $\pi r^2\left( \dfrac{1}{6} – \dfrac{\sqrt3}{4} \right)$.

Solution. The formula as printed cannot be right, and the page below derives the correct one. Look at the bracket:

$$\frac{1}{6} \approx 0.167 \qquad \frac{\sqrt3}{4} \approx 0.433$$

so the bracket is negative, about $-0.266$, and the whole expression comes out around $-0.84\,r^2$. An area can never be negative.

The correct derivation. Minor segment $=$ sector $-$ triangle.

The sector, angle $60°$, is one sixth of the circle:

$$\frac{60}{360} \times \pi r^2 = \frac{\pi r^2}{6}$$

The triangle. The two radii and the chord form a triangle with two sides $r$ and $60°$ between them — so the base angles are $60°$ too, and it is equilateral with side $r$:

$$\frac{\sqrt3}{4}r^2$$

Subtract:

$$\text{minor segment} = \frac{\pi r^2}{6} – \frac{\sqrt3}{4}r^2 = r^2\left( \frac{\pi}{6} – \frac{\sqrt3}{4} \right)$$

Check the sign: $\frac{\pi}{6} \approx 0.524$ and $\frac{\sqrt3}{4} \approx 0.433$, so the bracket is about $0.091$ — positive, as an area must be. With $r = 15$ this gives $20.4$, which matches Question 5 ✓

What the misprint is. The $\pi$ belongs to the sector term only, not to the whole bracket. Written correctly it is $r^2\left( \frac{\pi}{6} – \frac{\sqrt3}{4} \right)$; the book has moved the $\pi$ outside the bracket, where it wrongly multiplies the triangle term as well.

Question 8

*An equilateral triangle is inscribed in a circle of radius $r$. Show that the ratio of the area of the triangle to the area of the circle is $\dfrac{3\sqrt3}{4\pi} \approx 0.413$.

Solution. Find the side of the triangle. Join the centre to the three vertices. This makes three isosceles triangles, each with two sides $r$ and, by symmetry, an angle of $\frac{360°}{3} = 120°$ between them.

The side of the equilateral triangle is the chord subtending $120°$:

$$\text{side} = 2r\sin 60° = 2r \times \frac{\sqrt3}{2} = \sqrt3\,r$$

The triangle’s area. For an equilateral triangle of side $a$, the area is $\frac{\sqrt3}{4}a^2$:

$$\frac{\sqrt3}{4}\left( \sqrt3\,r \right)^2 = \frac{\sqrt3}{4} \times 3r^2 = \frac{3\sqrt3}{4}r^2$$

The ratio.

$$\frac{\frac{3\sqrt3}{4}r^2}{\pi r^2} = \frac{3\sqrt3}{4\pi} \approx \frac{5.196}{12.566} \approx 0.413 \qquad \blacksquare$$

The $r^2$ cancels, so the ratio is the same for every circle — which is what makes it worth stating.

Question 9

*A square is inscribed in a circle of radius $r$. Show that the ratio of the area of the square to the area of the circle is $\dfrac{2}{\pi} \approx 0.637$.

Solution. Find the side. The square’s diagonal is a diameter of the circle, since the angle in a semicircle is a right angle:

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

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

$$a\sqrt2 = 2r \quad\Longrightarrow\quad a = \frac{2r}{\sqrt2} = \sqrt2\,r$$

The square’s area.

$$a^2 = \left( \sqrt2\,r \right)^2 = 2r^2$$

(Equivalently, area $=$ half the square of the diagonal $= \frac{1}{2}(2r)^2 = 2r^2$.)

The ratio.

$$\frac{2r^2}{\pi r^2} = \frac{2}{\pi} \approx 0.637 \qquad \blacksquare$$

Question 10

*A hexagon is inscribed in a circle of radius $r$. Show that the ratio of the area of the hexagon to the area of the circle is $\dfrac{3\sqrt3}{2\pi} \approx 0.827$. Can you see why the answer is exactly twice the answer to Question 8?

Solution. Find the area. Join the centre to the six vertices. Each of the six triangles has two sides $r$ with $\frac{360°}{6} = 60°$ between them — so each is equilateral with side $r$, of area $\frac{\sqrt3}{4}r^2$.

$$\text{hexagon} = 6 \times \frac{\sqrt3}{4}r^2 = \frac{3\sqrt3}{2}r^2$$

The ratio.

$$\frac{\frac{3\sqrt3}{2}r^2}{\pi r^2} = \frac{3\sqrt3}{2\pi} \approx 0.827 \qquad \blacksquare$$

Why exactly twice Question 8. Compare the two answers:

$$\frac{3\sqrt3}{2\pi} = 2 \times \frac{3\sqrt3}{4\pi}$$

so the hexagon has exactly twice the area of the inscribed equilateral triangle.

The reason is that both are whole-number multiples of the same unit: an equilateral triangle of side $r$, of area $\frac{\sqrt3}{4}r^2$. The hexagon is $6$ of them, as just shown; and Question 8’s triangle has area $\frac{3\sqrt3}{4}r^2$, which is $3$ of them. Six against three is exactly double — a ratio of $2 : 1$.

Geometrically, joining alternate vertices of the regular hexagon gives precisely the inscribed equilateral triangle of Question 8, and it covers half the hexagon.

Notice also how the three ratios climb: $0.413$ for the triangle, $0.637$ for the square, $0.827$ for the hexagon. The more sides a regular inscribed polygon has, the closer it comes to filling the circle — and the ratio approaches $1$ as the number of sides grows.

Common mistakes

• Questions 1 to 3, using $2\pi r$ instead of $\pi r^2$. This exercise is about area, so the circle’s area $\pi r^2$ is what gets multiplied by $\frac{\theta}{360}$.
• Question 2, using $44$ as the radius or the diameter. $44$ cm is the circumference; the radius has to be found from it first.
• Question 3, using $10$ as the angle. Ten minutes is $60$ degrees. The conversion is $\frac{10}{60} \times 360°$, and skipping it makes the sector six times too small.
• Question 4, forgetting which $\pi$ to use. This question specifies $3.14$, not $\frac{22}{7}$. The two give slightly different answers, and following the instruction matters.
• Question 5, giving the sector as the segment. A segment is the sector minus the triangle. Here that is a difference of $117.75$ and $97.31$ — nearly a factor of six.
• Question 5, computing the triangle as $\frac{1}{2}(15)(15)$. That would be right for a $90°$ angle. For $60°$ the triangle is equilateral, with area $\frac{\sqrt3}{4}a^2$.
• Question 6, halving for the overlap. The question states the wipers do not overlap, so the two sectors are simply added.
• Question 7, trying to derive the printed formula. It cannot be derived, because it is negative. Checking the sign of a bracket before trusting it is the quickest way to catch this kind of error.
• Questions 8 to 10, taking the polygon’s side as $r$ in every case. Only for the hexagon is the side equal to the radius. The equilateral triangle has side $\sqrt3\,r$ and the square $\sqrt2\,r$.

Practise next

• The End-of-Chapter Exercises, Questions 21, 22, 24, 25 and 27 — five composite-area problems built from sectors and segments of exactly this kind.
• Exercise Set 6.1 — the perimeter versions of these same shapes, worth setting beside them to keep the two formulas apart.