NCERT Solutions Class 9 Maths Chapter 6 Think and Reflect — Measuring Space: Perimeter and Area

NCERT Class 9 Mathematics — Ganita Manjari Part I, Measuring Space: Perimeter and Area. All 15 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

Chapter 6 opens with a genuine question — why do sprinters in outer lanes start further forward? — and its nine Think and Reflect boxes keep returning to it while the machinery is built up. Several of the later ones ask about dissection: whether one shape can be cut up and rearranged into another.

Key insight. The extra distance an outer lane adds per lap is $2\pi w$, where $w$ is the lane width — and the track’s own size does not appear in it. Two circles whose radii differ by $w$ always have circumferences differing by $2\pi w$, whether they are running tracks or planets. That one fact answers three of these prompts.

Prompt 1 comes from the box on page 118, on the chapter’s opening page about relay staggers.

Question 1

In my school, the playground is too small to have a $400$ m track, so the school constructed a $200$ m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (smaller than the stagger used in the Olympics), for the same $4 \times 100$ m relay race?

Solution. No — and in fact the opposite.

The stagger exists to cancel out the extra distance an outer-lane runner covers on the bends. For one full lap, an athlete in a lane whose radius is $w$ greater runs an extra

$$2\pi(r + w) – 2\pi r = 2\pi w$$

Notice what has vanished: $r$. The extra distance depends only on the lane width, not on how big the track is. A tight little bend and a sweeping Olympic one add exactly the same extra distance per lap for the same lane width.

So a smaller track does not mean a smaller stagger per lap. And a $4 \times 100$ m relay covers $400$ m either way:

• on a $400$ m track that is one lap, so the stagger is $2\pi w$;
• on a $200$ m track it is two laps, so the extra distance is doubled: $4\pi w$.

With the chapter’s lane width of $1.22$ m, that is about $7.67$ m on the big track and about $15.33$ m on the small one.

Prompt 2 comes from the box on page 119, just after the question “what is the perimeter of a circle?” is posed.

Question 2

What is the connection between this question and the one about the $400$ m athletics track?

Solution. The stagger question cannot be answered until the circle question is.

A running track is two straight sections joined by two semicircular bends. On the straights, every lane is the same length — the lanes run side by side. All the difference is on the bends, and each bend is half a circle.

So working out how much further an outer lane runs means comparing the lengths of two semicircular arcs of different radii. That is impossible without knowing how to find the length around a circle — which is exactly the question just asked.

Once the formula $C = 2\pi r$ is available, the comparison is immediate: two lanes whose radii differ by $w$ have circumferences differing by $2\pi w$, and the stagger follows.

Prompt 3 comes from the box on page 127, after the $400$ m track has been analysed in detail.

Question 3

What is the difference in radius between the first and second lanes? Use Fig. 6.11 to find the stagger needed by the runner in the second lane. Will an equal stagger be needed between the third and second lanes?

Solution. The difference in radius. The chapter gives the lane width as $1.22$ m. Moving out one lane moves the runner $1.22$ m further from the centre of the bend, so

$$\text{difference in radius} = 1.22 \text{ m}$$

The stagger. The two semicircular bends together make one full circle. The runner in lane 2 covers a circle of radius $1.22$ m greater, and so runs an extra

$$2\pi(r + 1.22) – 2\pi r = 2\pi(1.22) = 2 \times 3.1416 \times 1.22 \approx 7.67 \text{ m}$$

So the lane-2 runner must start about $7.67$ m ahead.

Between lanes 2 and 3? Yes, exactly the same. The radius again increases by $1.22$ m, and the calculation is identical — the starting radius cancelled out, so it makes no difference that lane 3 is further out. Every pair of adjacent lanes needs the same stagger of about $7.67$ m, which is why the staggered start marks on a real track are evenly spaced.

Prompts 4 and 5 come from the two boxes on page 131, during the derivation of the area of a parallelogram.

Question 4

What happens if the parallelogram is ‘thin’ (Fig. 6.18) and the foot of the perpendicular from $C$ to $AD$ does not lie on side $AD$? The construction then does not seem to work. How do we fix it?

Solution. The standard derivation cuts a right-angled triangle off one end of the parallelogram and slides it to the other, turning it into a rectangle. For a very slanted (“thin”) parallelogram, the perpendicular lands outside the side, and there is no triangle of the right shape to cut.

The fix: extend the base. Treat $AD$ not as a segment but as a whole line, and drop the perpendicular onto that line. The height is still well defined — it is the distance between the two parallel lines containing the opposite sides — even when the foot lies beyond $D$.

The formula is unaffected:

$$\text{area} = \text{base} \times \text{height}$$

Why it still holds. Two arguments work where the cut-and-slide fails:

• Split it differently. A diagonal cuts any parallelogram into two congruent triangles, each of area $\frac{1}{2} \times \text{base} \times \text{height}$. Together they give $\text{base} \times \text{height}$, with no picture-dependent construction at all.
• Shear it. Sliding the top side along its own line, keeping the base and the height fixed, changes neither. Slide far enough and the thin parallelogram becomes an ordinary one, for which the original construction works.

The general lesson is worth keeping: a proof that depends on the picture being typical is not finished until the awkward cases are handled.

Question 5

The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides?

Solution. No. The side lengths alone do not determine a parallelogram’s area.

Imagine a parallelogram made of four hinged rods. Push it over: the four sides keep their lengths, but the shape flattens and the area shrinks. Push it all the way and the area approaches zero; stand it up square and the area is greatest.

Concretely. Take sides $a$ and $b$ with angle $\theta$ between them. The height is $b\sin\theta$, so

$$\text{area} = ab\sin\theta$$

With $a = 5$ and $b = 4$:

• at $\theta = 90°$ (a rectangle): area $= 20$;
• at $\theta = 30°$: area $= 10$;
• at $\theta = 5°$: area $\approx 1.7$.

Same sides every time, wildly different areas.

Why the rectangle is different. A rectangle’s angles are fixed at $90°$, so the “missing” piece of information is already supplied. For a general parallelogram you need the sides and one angle — or equivalently a base and the height.

This is also why a rectangle is rigid enough to build with and a parallelogram is not: a four-bar frame collapses unless braced with a diagonal.

Prompt 6 comes from the box on page 133, after two triangles of equal area have been compared.

Question 6

Since $\triangle ABD$ and $\triangle ACD$ have equal area, you may wonder — can we divide $\triangle ABD$ using straight cuts into two or more pieces that we can then rearrange to exactly cover $\triangle ACD$?

Solution. Yes, always — and remarkably, this is true for any two polygons of equal area, not just these triangles.

For two triangles of equal area, here is a method.

1. Turn each triangle into a parallelogram: cut along the line joining the midpoints of two sides and rotate the top piece through $180°$. This gives a parallelogram with the same base and half the height.
2. Turn each parallelogram into a rectangle of a chosen width, by cutting off a triangle at one end and sliding it to the other (extending the base if the parallelogram is thin, as Question 4 describes).
3. Make both rectangles have the same width. Since their areas are equal, their heights must then match too — so the two rectangles are congruent.

Running the steps backwards on the second triangle turns that rectangle back into it. So the pieces from the first triangle can be reassembled into the second.

The general result. This is the Wallace–Bolyai–Gerwien theorem: any two polygons of equal area are equidecomposable — one can always be cut into finitely many straight-edged pieces that reassemble into the other. It was proved in the early nineteenth century, and the next prompt asks exactly this question in general.

Prompts 7 to 13 come from the two panels on page 134. The first asks about dissecting shapes of equal area; the second is about rectangles of a fixed perimeter.

Question 7

Suppose we are given two polygons $P$ and $Q$ with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other?

Solution. Yes, always. This is the general form of Question 6, and the answer is the Wallace–Bolyai–Gerwien theorem (1807–1833).

Why it works, in outline. Every polygon can be cut into triangles. Each triangle can be turned into a rectangle, as in Question 6. Each rectangle can be re-cut into a rectangle of any chosen width. So bring every piece of $P$ to rectangles of one common width, stack them into a single rectangle, and do the same for $Q$ — and since the areas are equal, those two big rectangles are congruent.

Worth knowing: the corresponding statement in three dimensions is false. Two solids of equal volume cannot always be cut into finitely many pieces that reassemble — a regular tetrahedron and a cube of the same volume cannot. That was Hilbert’s third problem, settled by Dehn in 1900.

Question 8

A square and a non-square rectangle with equal area.

Solution. Take a rectangle $a \times b$ with $a \neq b$ and a square of side $\sqrt{ab}$ — equal areas.

A three-piece dissection (the “step” method). Suppose the rectangle is $9 \times 4 = 36$ and the square is $6 \times 6$.

1. Draw a staircase across the rectangle with steps $3$ wide (the difference between $9$ and the target $6$) and $2$ high.
2. Cut along the staircase.
3. Slide the upper piece one step across and one step up.

The staircase interlocks with itself in the new position and the result is the $6 \times 6$ square.

In general, for any rectangle a dissection into at most three pieces exists, found by the rectangle-to-rectangle method of Question 6 with the target width set to $\sqrt{ab}$.

Question 9

Two triangles with different shapes but equal area.

Solution. Possible, by the route of Question 6: turn each into a parallelogram, then each parallelogram into a rectangle of the same width. Equal areas then force equal heights, so the two rectangles coincide and the pieces transfer.

A quicker special case. If the two triangles happen to share a base length, only one cut is needed. Triangles on the same base with equal areas have equal heights, so one can be sheared into the other — cut along the midline, rotate the top piece, and the resulting parallelograms are related by a single shear, which one more cut resolves.

Question 10

A triangle and a square with equal area. Formulate a conjecture of your own about this.

Solution. Possible again — and this case is the one with a name, “squaring the triangle”, which prompt 14 below asks about directly.

The route. Triangle $\to$ parallelogram $\to$ rectangle $\to$ square. The last step is the only new one: to convert an $a \times b$ rectangle into a square, the side must be $\sqrt{ab}$, the geometric mean of the two sides, and that length is constructible with ruler and compass (see Question 14).

A conjecture worth making, and it is the right one: any two polygons of the same area can be cut into finitely many pieces that rearrange into one another. That is exactly the Wallace–Bolyai–Gerwien theorem of Question 7, and the three cases in this box — rectangle to square, triangle to triangle, triangle to square — are the building blocks of its proof.

Prompts 11 to 13 are the second panel on page 134: think of various rectangles with perimeter $40$ units (the sides do not have to be integers).

Question 11

How many such rectangles are there?

Solution. A perimeter of $40$ means

$$2(\ell + w) = 40 \quad\Longrightarrow\quad \ell + w = 20$$

So the two sides must add to $20$, and any positive $w$ less than $20$ gives a rectangle, with $\ell = 20 – w$.

Since $w$ may be any real number strictly between $0$ and $20$ — the question explicitly says the sides need not be whole numbers — there are infinitely many such rectangles: $1 \times 19$, $2 \times 18$, $3.5 \times 16.5$, $\sqrt2 \times (20 – \sqrt2)$, and so on.

(If the sides were required to be whole numbers there would be just $10$ of them, from $1 \times 19$ to $10 \times 10$, counting $a \times b$ and $b \times a$ as the same rectangle.)

Question 12

Among them, is there one whose area is the largest? What are its dimensions?

Solution. With $\ell + w = 20$, write $\ell = 20 – w$ and express the area in terms of $w$ alone:

$$A = w(20 – w) = 20w – w^2$$

Complete the square:

$$A = -\left( w^2 – 20w \right) = -\left( w^2 – 20w + 100 \right) + 100 = 100 – (w – 10)^2$$

The subtracted term $(w-10)^2$ is never negative, so $A$ is at most $100$, and equals $100$ exactly when $w = 10$.

So the largest area is $100$ square units, from the $10 \times 10$ rectangle — a square.

$$\text{largest area} = 100 \text{ square units, at } 10 \times 10$$

Checking a few: $9 \times 11 = 99$, $8 \times 12 = 96$, $5 \times 15 = 75$, $1 \times 19 = 19$. All below $100$, and falling away faster the further from square.

Question 13

Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?

Solution. There is no smallest. As $w$ gets closer to $0$, the rectangle becomes a longer and longer thin sliver:

$$w = 1: \; A = 19 \qquad w = 0.1: \; A = 1.99 \qquad w = 0.01: \; A = 0.1999$$

The area can be made as close to $0$ as you like — but never reaches it, because $w = 0$ gives no rectangle at all, just a segment. There is no smallest positive area, so no minimising rectangle exists.

The surprise. Two things are worth noticing.

• Equal perimeter does not mean equal area. A $10 \times 10$ square and a $0.01 \times 19.99$ sliver both have perimeter $40$, yet their areas differ by a factor of five hundred. Perimeter and area are genuinely independent quantities.
• The maximum exists but the minimum does not. The largest area is achieved, by the square; the smallest is approached but never attained. A quantity can have a greatest value without having a least one.

The first point is the practical one: a farmer with a fixed length of fencing should make the field square, and a long thin field wastes almost all of the enclosed space.

Prompt 14 comes from the box on page 142, just before Exercise Set 6.2.

Question 14

What procedure would you use to square a given triangle? The task is to construct a square whose area is equal to the area of some given triangle. How would you proceed?

Solution. Three steps, each a standard construction.

Step 1 — triangle to rectangle. Let the triangle have base $b$ and height $h$, so its area is $\frac{1}{2}bh$. Cut along the line joining the midpoints of the two other sides and rotate the top piece through $180°$: the result is a parallelogram of base $b$ and height $\frac{h}{2}$. Shear it into a rectangle of sides $b$ and $\frac{h}{2}$ — same area, $\frac{1}{2}bh$.

Step 2 — find the geometric mean. A square of the same area must have side

$$s = \sqrt{b \times \frac{h}{2}}$$

the geometric mean of the rectangle’s two sides. This is where the construction gets clever.

Step 3 — construct that length. Place the two lengths $p = b$ and $q = \frac{h}{2}$ end to end along a straight line, giving a segment of length $p + q$. Draw a semicircle on that segment as diameter. At the point where the two lengths meet, erect a perpendicular to meet the semicircle at $X$.

The length of that perpendicular is exactly $\sqrt{pq}$.

Why. The angle in a semicircle is $90°$ (Chapter 5), so the perpendicular is the altitude to the hypotenuse of a right-angled triangle. That altitude is the geometric mean of the two segments it divides the hypotenuse into — which is the relation $x^2 = pq$.

Build a square on that perpendicular, and its area is $pq = \frac{1}{2}bh$: the triangle, squared.

A note of history. This works for any polygon, by first cutting it into triangles. But the famously impossible version — “squaring the circle”, constructing a square equal in area to a given circle — was proved impossible in 1882, because $\pi$ is not merely irrational but transcendental.

Prompt 15 comes from the box on page 144, opening the discussion of the circle.

Question 15

Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circle?

Solution. Both — and the practical reasons are mathematical ones.

Practical reasons, with the mathematics behind them.

• Rolling. A circle has the same radius in every direction, so a wheel’s axle stays at a constant height. No other shape of constant width does this with a central axle — which is why the wheel is round and not oval.
• Most area for the least boundary. Among all shapes with a given perimeter, the circle encloses the greatest area. A circular hut needs the least wall for a given floor, and a circular fence the least fencing. (Question 12 above is the rectangle-only version of the same phenomenon, where the square wins.)
• Strength. A circular arch or pipe distributes pressure evenly, because every point of the wall is the same distance from the centre.
• Fairness. Everyone seated around a circular table is equally placed — no head of the table. The same property makes a circle the natural shape for a shared cooking fire or a dance.
• Ease of construction. A rope and a peg draw a perfect circle. No measuring is needed, which matters when there are no rulers.

Other reasons. The circle has no beginning and no end, and it is unchanged by rotation through any angle at all. That unmatched symmetry has made it a symbol of completeness, of cycles, and of the heavens in cultures all over the world — the sun and the full moon are the two most prominent objects in the sky, and both are circles.

Uses. Wheels and gears; pots, plates and grinding stones; arches, domes and wells; coins; clock faces and the division of the day; roundabouts; the chakra on the national flag; and — from this chapter’s own opening — the bends of a running track.

Common mistakes

• Question 1, assuming a smaller track needs a smaller stagger. The extra distance is $2\pi w$, with the radius cancelling out — so track size is irrelevant. The number of laps is what changes.
• Question 3, computing the stagger from the radii themselves. The stagger is the difference of two circumferences, and only the difference in radius survives. There is no need to know $36.5$ m at all.
• Question 5, answering “yes” by analogy with rectangles. A rectangle’s angles are fixed, so its sides determine it; a parallelogram’s do not. Its area is $ab\sin\theta$.
• Question 7, thinking the theorem extends to three dimensions. It does not. Equal volume does not imply equidecomposability for solids.
• Question 11, counting only whole-number sides. The question explicitly allows any lengths, so the answer is infinitely many.
• Question 12, thinking a long thin rectangle has the larger area. It has the smaller. Among shapes of equal perimeter, the most “regular” encloses the most.
• Question 13, answering that the smallest is a $1 \times 19$ rectangle. Sides need not be whole numbers, so $0.1 \times 19.9$ is smaller, and $0.01 \times 19.99$ smaller again. Nothing is smallest.
• Question 14, taking the square’s side as half the base or similar. The side must be the geometric mean $\sqrt{pq}$, not an arithmetic average — and the semicircle construction is what produces it.

Practise next

• Exercise Set 6.1 — the circumference work these track questions depend on, including several composite-shape perimeters.
• Exercise Set 6.2, Questions 6 to 11 — the equal-area arguments behind Questions 6 to 10 here.
• The End-of-Chapter Exercises, Questions 11 to 15 — the trapezium, kite and scaling dissections, which is Question 7’s theorem in practice.