NCERT Solutions for Class 9 Maths Chapter 6 Measuring Space Perimeter and Area (PDF) – 2027

Because the outer lanes are longer. On the straights all lanes are equal, but on the bends each outer lane runs round a semicircle of larger radius, and a semicircle of radius r has length πr. The finish line is common, so each outer runner starts ahead — the stagger — by exactly the extra distance.

No. The stagger is worked out so that every athlete runs exactly the same distance: it equals the extra length of the outer lane on the bends. So it removes an unfairness rather than creating one.

On the formula for the length of a circle. That is the whole reason the chapter starts here.

So all the organisers need is the lane width w and the value of π. For the standard track, w = 1.22 m, so

No — the stagger does not become smaller. On each bend the next lane out runs a semicircle of radius (r + w) instead of r: an extra π(r + w) − πr = πw, which does not depend on r.

So, with lanes of the same width (1.22 m), the school track actually needs a larger (double) stagger for the same race, because the runners go round twice as many bends.

The perimeter of a circle — its circumference — is 2πr units. Finding it is not like finding the perimeter of a square, because you cannot add up straight pieces.

All circles are also scaled copies of one another. So the ratio

must be the same number for every circle. That number is what we call π. Once we know it, C = πD = 2πr. The work of the chapter is to find out what π is.

They are the same question. The bends of the track are semicircles, so the length of a lap is two straights plus one full circle — and you cannot compute that until you know how to find the length round a circle.

Both expressions contain π. Without a value for the C/D ratio the track designer cannot mark a single starting line.

Yes, it is the same for every circle — that is exactly why the constant π exists.

Reason: any circle is an enlargement of any other. Enlarging by a factor k multiplies both C and D by k, so C/D does not change.

The value is π = 3.14159265… — a number whose decimal expansion never ends and never repeats. There are two honest ways to estimate it: measure, or reason geometrically.

• By measurement: wrap a thread many times round a cylinder, measure its length, and divide by (number of turns × diameter).
• By geometry: trap the circle between an inscribed and a circumscribed polygon whose perimeters can be calculated. Hexagons give 3 < π < 2√3; polygons with more sides give closer bounds.

You should get a value between 3 and 4, and with reasonable care between 3.1 and 3.2.

Yes — by trapping the circle between polygons whose perimeters you can compute exactly.

Take a circle of radius 1, so C = 2π. A regular hexagon inscribed in it has each side equal to the radius, 1.

A regular hexagon drawn round the circle (its sides touching the circle) has each side 2/√3, by the Baudhāyana–Pythagoras theorem (worked out in Fig. 6.7 below).

Yes. The hexagon inscribed in a circle of radius 1 has perimeter exactly 6, while the circle that contains it is longer.

The inscribed hexagon gives the lower bound (previous question). The circumscribed hexagon gives the upper bound, and that is where the Baudhāyana–Pythagoras theorem is needed.

Difference in radius = 1.22 m (one lane width). Stagger ≈ 7.67 m, and yes, the same stagger is needed between lanes 2 and 3, and between every other adjacent pair.

Lanes 2 and 3 also differ in radius by 1.22 m, so the stagger between them is again ≈ 7.67 m.

r = 7 cm.

Using C = 2πr with π ≈ ²²⁄₇:

An arc that turns through θ° at the centre is the fraction θ/360 of the whole circle.

(i) r = 3.5 cm, θ = 60°

(ii) r = 6.3 m, θ = 120°

Perimeter = 46.33 cm (exactly ¹³⁹⁄₃ cm).

Every one of these is built from arcs whose lengths you already know. Recall that a semicircle of radius r has arc length πr, and a quarter circle has arc length πr/2. Use π ≈ ²²⁄₇ throughout.

(i) 176 cm = 1.76 m. (ii) About 5682 revolutions.

(i) 88 cm. (ii) 528 cm.

5 : 4 — the same ratio.

The formula is still base × height; only the proof needs repair. Replace ABCD by a less slanted parallelogram of the same area, and repeat:

Each step keeps the base, height and area but reduces the slant. After enough steps the foot of the perpendicular lies on the base, the cut-and-slide proof works, and area = base × height.

No. The side lengths do not determine the area of a parallelogram. Two parallelograms can have exactly the same four sides and completely different areas.

There is a gap: when ∠EFG is obtuse, the foot J of the perpendicular from E falls outside FG (on GF produced), so FG is not b₁ + b₂. We close it by subtracting instead of adding.

Yes. In Fig. 6.21B the copy A′B′C′ is turned round and placed along the side AC, with C′ on A, A′ on C and B′ at D.

It is made of two congruent triangles, so ar(ΔABC) = ½ × ar(ABCD) = ½ × base × height = ½ bh.

Two pieces (one straight cut) are enough. One piece cannot work, because ΔABD and ΔACD are in general not congruent.

Yes, it is possible — and here just one straight cut suffices.

• Let M be the midpoint of AB. Cut ΔABD along MD.
• Turn the piece MBD through 180° about M. B lands on A, and the two pieces form a triangle ADD′ with sides AD, DD′ = AC and AD′ = DC (midpoint theorem) — a copy of ΔACD.
• Move this triangle onto ΔACD: the two pieces cover it with no gap and no overlap.

Conjecture: yes — any two polygons with equal area can be cut into pieces that rearrange into each other.

• Square and rectangle: e.g. an 8 × 2 rectangle cut in half gives two 4 × 2 pieces, which together make a 4 × 4 square.
• Two triangles: cut each along a midline (joining the midpoints of two sides) and turn the top piece through 180° to get a parallelogram; a parallelogram can be cut and slid into a rectangle, and rectangles of equal area into one another.
• Triangle and square: triangle → parallelogram → rectangle → square, each step by cutting and rearranging.

1. Infinitely many. 2. Yes — the 10 × 10 square, area 100 sq. units. 3. No — there is no smallest.

Smallest area: none. As x gets closer to 20 (or to 0), the rectangle becomes a thinner and thinner strip and its area gets closer to 0, but a rectangle of area 0 does not exist. Whatever rectangle you pick, one with x closer to 20 has a smaller area.

Do it in two steps: triangle → rectangle → square.

Step 1. In ΔABC take the base BC = b and the height h. Construct a rectangle with sides b and h/2 (bisect the altitude to get h/2). Its area is b × h/2 = ½bh = area of the triangle.

Step 2. Square this rectangle by Baudhāyana’s construction (Fig. 6.30), using its longer side as AD and its shorter side as AB. The square HPQS obtained has area ½bh, the same as the triangle; its side is √(½bh).

Area = 40 cm² — and the position of E on BC makes no difference at all.

Area = 720 cm².

Area = 8√30 ≈ 43.82 cm². No height is given, so this is Heron’s formula.

Area = 1500√3 ≈ 2598.08 m².

Shorter diagonal = 8√2 ≈ 11.31 cm.

The ratio is 1 : 1 — the two triangles always have the same area, wherever P and Q sit on AB.

In fact each is half the parallelogram:

Proof. Let the diagonals PR and QS of the parallelogram meet at M.

Let ABCD be the 4-gon and P, Q, R, S the midpoints of AB, BC, CD, DA. Joining them cuts off four corner triangles; we show that together they make up half of ABCD.

Use the median theorem twice — once on ΔABC and once on ΔPBC — and subtract.

The ratio is 1 : 1 — each pair is exactly half the square, wherever P is placed inside it.

The plan: show ar(ΔBPQ) = ar(ΔBCD), and then note that CD is a median.

Both — practical reasons and other reasons.

• Practical: for a given length of boundary (a wall or a fence) a circle encloses the most area; it is the easiest shape to mark out, with just a peg and a rope; and it rolls smoothly, which gave us wheels, potter’s wheels, pulleys and grinding stones. Round wells, pots, granaries and huts are strong and use less material.
• Other reasons: the Sun, the full Moon and the horizon look circular, so the circle became a symbol; it looks the same from every side, so people sit or dance in a circle (as in garba) and draw rangolis and maṇḍalas with it.

Because the slices are laid out alternately — one pointing up, the next pointing down — so exactly half of them contribute their arcs to the top edge and the other half to the bottom edge.

Area = ⁷⁷⁄₃ ≈ 25.67 cm².

Area = 38.5 cm².

Area swept = ⁷⁷⁄₃ ≈ 25.67 cm².

(i) 78.5 cm² (ii) 235.5 cm².

Minor segment ≈ 20.44 cm²; major segment ≈ 686.06 cm².

A segment is a sector with the triangle cut off.

Total area cleaned = ⁴⁹²⁸⁄₃ ≈ 1642.67 cm².

Take the sector and subtract the triangle.

First find the side of the triangle in terms of r.

The diagonal of an inscribed square is a diameter.

A regular hexagon inscribed in a circle is six equilateral triangles glued at the centre.

Why exactly twice the answer to Question 8.

Fig. 6.41: the square of side a + b is made of a², b² and two rectangles ab, so (a + b)² = a² + 2ab + b².

(a + b)(a − b) = a² − b²: remove a b × b square from a corner of an a × a square; cut the L-shaped rest into two rectangles of width a − b and put them end to end.

(a + b + c)²: divide each side of a square of side a + b + c into a, b, c. The 9 pieces are a², b², c² and two each of ab, bc, ca.

Area = 50√2 ≈ 70.71 cm².

Check by half base × height. The height to the base bisects it (isosceles):

Each equal side is 13 cm.

Perimeter = 36 cm.

Area = ^75√15⁄₄ ≈ 72.62 cm².

Area = 84 cm² by both routes.

Way 1 — spot the right angle.

Way 2 — Heron’s formula.

Distance ≈ 18,857.14 cm = 188.57 m.

Area = 86.625 cm².

One turn = 176 cm = 1.76 m; about 568 turns per kilometre.

Yes. For rectangles, equal area together with equal perimeter forces the same pair of side lengths.

Cut the trapezium into a parallelogram and a triangle, exactly as Fig. 6.42 shows.

Draw one diagonal. It splits the trapezium into two triangles that share the same height.

Take two copies of the trapezium ABCD (AB ∥ DC, AB = a, DC = b, height h). Turn one copy through 180° and join it to the other along a slanting side.

Let the kite be ABCD with AB = AD and CB = CD. Its diagonal AC is the perpendicular bisector of BD; they meet at M. Let d₁ = AC and d₂ = BD.

(i) Using algebra.

(ii) Using geometry. Draw lines through A and C parallel to BD, and through B and D parallel to AC. They make a rectangle with sides d₁ and d₂ round the kite.

Do the copies fit? Yes. (i) Halve PQRS both ways. (ii) Join the midpoints of the sides of ΔPQR: by the midpoint theorem all 4 small triangles have sides a, b, c. (iii) Trisect each side of ΔPQR and draw lines through these points parallel to the sides: 9 small triangles, each with sides a, b, c.

Fig. 6.43: one half. Fig. 6.44: one fifth.

Fig. 6.43. In ΔABC (A at the top, B at the bottom left), M is the midpoint of AB and AP = PQ = QC on AC; the shaded part is BMPQ. Let T = ar(ΔABC).

Fig. 6.44. The lines cut the square into the shaded square, 4 small corner triangles and 4 four-sided pieces. Turning each small triangle through 180° about the midpoint of its side fits it onto the next four-sided piece, making a square equal to the shaded one. So the whole square = 5 shaded squares, and the shaded part is ⅕.

Both answers are the same: π/4 ≈ 0.785, that is about 78.5%.

Conjecture: however many circles are used, they always cover exactly π/4 of the rectangle — the fraction does not depend on the number of circles.

Proof.

Perimeter of each small rectangle = ^18√10⁄₅ = 3.6√10 ≈ 11.38 cm.

Let BC be trisected at D and E (BD = DE = EC); the blue triangle is ABD and the red one is AEC.

• Step 1: AD is a median of ΔABE. As on page 133, cut ΔABD along DM (M = midpoint of AB) and turn piece MBD through 180° about M: the two pieces make a copy of ΔADE.
• Step 2: AE is a median of ΔADC. Cut this copy along EN (N = midpoint of AD) and turn piece NDE through 180° about N: the pieces now make a copy of ΔAEC, which covers the red triangle.

So the blue triangle can be cut into at most 4 pieces that exactly cover the red triangle.

Let the square have side a, and let the quarter circle Q have its centre at the corner O and radius a. The semicircles S₁ and S₂ stand on the two sides through O (diameter a each) and lie inside Q.

Perimeter = 4π = ⁸⁸⁄₇ ≈ 12.57 units. Area = 2π − 4 = ¹⁶⁄₇ ≈ 2.29 sq. units.

Let the outer radius be R and the inner radius r. The green region is the annulus between them.

Let the right angle be at Q, with legs QP = b, QR = a and hypotenuse PR = c, so a² + b² = c². The semicircle on PR passes through Q (the angle in a semicircle is a right angle), so it is made of the triangle C and the two segments S₁ (cut off by chord QP) and S₂ (cut off by chord QR).

Area of the shaded common region = r²(^2π⁄₃ − ^√3⁄₂) ≈ 1.23 r².

Let the rectangle be W × H with bottom-left corner O. Let T (on the top side) be p from the left side and R (on the right side) be q above the bottom; M, the third vertex of C, is below T and level with R, so ∠TMR = 90°.

Let OA = OB = OC = r. B is directly above O, so ∠AOB = 90° and AB² = r² + r² = 2r².