NCERT Solutions Class 9 Maths Exercise 6.1 — Measuring Space: Perimeter and Area

NCERT Class 9 Mathematics — Ganita Manjari Part I, Measuring Space: Perimeter and Area, Exercise Set 6.1. All 8 questions solved.

This exercise is about the perimeter of curved shapes. Four formulas cover all of it, with $r$ the radius and $\theta$ the angle at the centre in degrees:

$$\text{circumference} = 2\pi r \qquad \text{arc length} = \frac{\theta}{360} \times 2\pi r$$ $$\text{perimeter of a sector} = \frac{\theta}{360} \times 2\pi r + 2r \qquad \text{(arc plus two radii)}$$

Throughout, use $\pi = \dfrac{22}{7}$ as the exercise instructs. That value is chosen to make the arithmetic come out neatly whenever the radius is a multiple of $7$ — and several questions here are built that way.

Key insight. For a composite shape, walk round the boundary once and add up what you actually travel along. A dashed construction line inside the figure is not part of the perimeter, and a straight edge that has been replaced by an arc contributes the arc, not the edge. Most errors in Question 5 come from counting a line that is not on the boundary.

Question 1

The perimeter of a circle is $44$ cm. What is its radius?

Solution.

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

Question 2

Calculate, correct to $3$ significant figures, the circumference of a circle with:

Solution. $C = 2\pi r$ with $\pi = \frac{22}{7}$ in each case.

(i) $r = 7$ cm: $\;C = 2 \times \frac{22}{7} \times 7 = 44$, so $C = 44.0$ cm.

(ii) $r = 10$ cm: $\;C = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} = 62.857\ldots$, so $C = 62.9$ cm.

(iii) $r = 12$ cm: $\;C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} = 75.428\ldots$, so $C = 75.4$ cm.

Question 3

Calculate the length of the arc of a circle if:

Solution. Arc length $= \dfrac{\theta}{360} \times 2\pi r$ — the fraction of the full circumference that the angle takes up.

(i) $r = 3.5$ cm, $\theta = 60°$. The angle is one sixth of a full turn:

$$\frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5 = \frac{1}{6} \times 22 = \frac{11}{3} \approx 3.67 \text{ cm}$$

(ii) $r = 6.3$ m, $\theta = 120°$. One third of a full turn:

$$\frac{120}{360} \times 2 \times \frac{22}{7} \times 6.3 = \frac{1}{3} \times 39.6 = 13.2 \text{ m}$$

Question 4

Find the perimeter of a sector (i.e. the curved portion as well as the two straight portions) of a circle of radius $14$ cm and sector angle $75°$.

Solution. A sector is bounded by three pieces: the arc and the two radii.

The arc:

$$\frac{75}{360} \times 2 \times \frac{22}{7} \times 14 = \frac{75}{360} \times 88 = \frac{55}{3} \approx 18.33 \text{ cm}$$

The two radii: $\;2 \times 14 = 28$ cm.

$$\text{perimeter} = \frac{55}{3} + 28 = \frac{55 + 84}{3} = \frac{139}{3} \approx 46.33 \text{ cm}$$

Question 5

Find the perimeters of the following shapes, taking the arcs to be quarter, half or three-quarters of a circle as appropriate (Fig. 6.14i to 6.14ix).

Solution. In each case, identify which arcs and which straight edges form the boundary.

(i) A rectangle $80$ m by $60$ m with a semicircular end on each short side

The two long sides of $80$ m stay; the two short sides of $60$ m are replaced by semicircles of diameter $60$ m. Two semicircles of the same diameter make one full circle:

$$P = 2(80) + \pi(60) = 160 + \frac{22}{7}(60) = 160 + \frac{1320}{7} = \frac{2440}{7} \approx 348.57 \text{ m}$$

(ii) A semicircular ring, outer diameter $12$ cm and inner diameter $8$ cm

The boundary is the outer semicircular arc, the inner semicircular arc, and the two short straight pieces at the bottom, each $\frac{12 – 8}{2} = 2$ cm:

$$P = \pi(6) + \pi(4) + 2(2) = \frac{22}{7}(10) + 4 = \frac{220}{7} + 4 = \frac{248}{7} \approx 35.43 \text{ cm}$$

(iii) A square of side $10$ cm with a semicircle on each side

The four straight sides are replaced by four semicircles of diameter $10$ cm. Four semicircles make two full circles:

$$P = 4 \times \pi(5) = 20\pi = \frac{440}{7} \approx 62.86 \text{ cm}$$

(iv) An equilateral triangle of side $12$ cm with a semicircle on each side

Three semicircles of diameter $12$ cm:

$$P = 3 \times \pi(6) = 18\pi = \frac{396}{7} \approx 56.57 \text{ cm}$$

(v) A $3 \times 3$ grid of squares of side $14$ cm, with rounded corners and bumps

The figure is a large square divided into nine equal squares of side $14$ cm. The boundary consists of:

• four quarter-circles of radius $14$ cm at the corners — together one full circle;
• four semicircles of diameter $14$ cm bulging out from the middle third of each side — together two full circles of radius $7$ cm.

$$P = 2\pi(14) + 4 \times \pi(7) = 28\pi + 28\pi = 56\pi = 56 \times \frac{22}{7} = 176 \text{ cm}$$

(vi) A baseline of $28$ cm in four equal parts, with a large semicircle above and four small ones alternating

• the large semicircle on the whole $28$ cm as diameter: $\pi(14) = 14\pi$;
• four semicircles of diameter $7$ cm (two below the line, two above): $4 \times \pi(3.5) = 14\pi$.

$$P = 14\pi + 14\pi = 28\pi = 28 \times \frac{22}{7} = 88 \text{ cm}$$

(Whether each small semicircle bulges up or down makes no difference to its length.)

(vii) Semicircles on the three sides of a right-angled triangle with legs $8$ cm and $6$ cm

The hypotenuse is $\sqrt{8^2 + 6^2} = 10$ cm. Semicircles stand on all three sides, and the triangle’s own sides are dashed construction lines, not part of the boundary:

$$P = \frac{\pi}{2}(8 + 6 + 10) = \frac{\pi}{2}(24) = 12\pi = \frac{264}{7} \approx 37.71 \text{ cm}$$

(viii) A semicircle of diameter $12$ cm above three semicircles of diameter $4$ cm

$$P = \pi(6) + 3 \times \pi(2) = 6\pi + 6\pi = 12\pi = \frac{264}{7} \approx 37.71 \text{ cm}$$

(ix) A semicircle of diameter $20$ cm, plus two semicircles of diameter $10$ cm

This is the familiar taijitu outline. The baseline of $20$ cm is split into two $10$ cm halves, with a large semicircle above and one small semicircle on each half — one above and one below:

$$P = \pi(10) + \pi(5) + \pi(5) = 20\pi = \frac{440}{7} \approx 62.86 \text{ cm}$$

Question 6

If the diameter of a car tyre is $56$ cm, then:

Solution.

(i) How far does the car need to travel for the tyre to complete one revolution?

One revolution rolls out exactly one circumference:

$$C = \pi d = \frac{22}{7} \times 56 = 176 \text{ cm}$$

(ii) How many revolutions does the tyre make if the car travels $10$ km?

Convert to the same unit first. $10$ km $= 10 \times 1000 \times 100 = 1{,}000{,}000$ cm.

$$\text{revolutions} = \frac{1{,}000{,}000}{176} = \frac{62500}{11} \approx 5681.8$$

So the tyre turns about $5682$ times — or, exactly, $5681$ complete revolutions with a little over four fifths of another.

Question 7

Find the total perimeter of all the petals in each of the given flowers.

Solution.

(i) Four petals in a square of side $14$ cm; the arc centres are the midpoints of the sides

Each arc is centred at the midpoint of a side, with radius half the side, $7$ cm. Such an arc runs from one corner of that side, through the centre of the square, to the other corner — a semicircle of radius $7$ cm standing on that side as its diameter.

There is one such semicircle per side, and each is split at the centre of the square into two petal edges — so every bit of every semicircle is used.

$$\text{total} = 4 \times \pi(7) = 28\pi = 28 \times \frac{22}{7} = 88 \text{ cm}$$

(ii) Six petals in a regular hexagon of side $42$ cm; the arc centres are the vertices

Each arc is centred at a vertex with radius equal to the side, $42$ cm, running from one neighbouring vertex to the other. The angle it subtends at that vertex is the hexagon’s interior angle, $120°$.

$$\text{one arc} = \frac{120}{360} \times 2\pi(42) = \frac{1}{3} \times 84\pi = 28\pi$$

Six arcs, each split at the centre into two petal edges:

$$\text{total} = 6 \times 28\pi = 168\pi = 168 \times \frac{22}{7} = 528 \text{ cm}$$

Question 8

The ratio of the perimeters of two circles is $5:4$. What is the ratio of their radii?

Solution. The circumference is directly proportional to the radius, since $C = 2\pi r$ and $2\pi$ is the same constant for both circles:

$$\frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2}$$

The $2\pi$ cancels, so the ratio of the radii equals the ratio of the circumferences:

$$\frac{r_1}{r_2} = \frac{5}{4}$$

(Note that areas would not behave this way: the areas would be in the ratio $5^2 : 4^2 = 25 : 16$, because area depends on $r^2$.)

Common mistakes

• Question 1, using the area formula. “Perimeter of a circle” is its circumference, $2\pi r$, not $\pi r^2$.
• Question 4, giving only the arc. A sector’s perimeter includes the two radii as well as the arc — that is why the question spells out “the curved portion as well as the two straight portions”.
• Question 5, including the dashed lines. In (iv) and (vii) the triangle’s sides are construction lines, not boundary. Adding them inflates the answer considerably.
• Question 5(ii), forgetting the two straight ends. A semicircular ring has two short straight pieces at the bottom, $2$ cm each here. Leaving them out loses $4$ cm.
• Question 5(iii) and (iv), using $2\pi r$ for each arc. Each is a semicircle, of length $\pi r$, not a full circle.
• Question 5(vi), worrying which way each bump faces. A semicircle has the same length whether it bulges up or down. Only the diameters matter.
• Question 6(ii), mixing units. $10$ km must become $1{,}000{,}000$ cm before dividing by a circumference measured in centimetres. Dividing $10$ by $176$ gives a meaningless answer.
• Question 7, using quarter-circles. In (i) the arcs are semicircles of radius $7$ (the caption says the centres are the side midpoints); in (ii) they are $120°$ arcs of radius $42$ (the centres are the vertices). Reading the caption is what settles the radius and angle.
• Question 8, squaring the ratio. $5^2 : 4^2$ is the ratio of the areas. Perimeter is proportional to $r$ itself, so the ratio is unchanged.

Practise next

• Exercise Set 6.2 — areas of triangles and quadrilaterals, including Heron’s formula.
• Exercise Set 6.3 — the area counterpart of this exercise: sectors, segments and the regions between shapes.
• The End-of-Chapter Exercises, Questions 7, 8 and 9 — three more wheel-and-circumference problems of exactly the kind in Question 6.