NCERT Solutions for Class 9th Maths Chapter 6 Whole chapter — End-of-Chapter Exercises
Book page 149–153 Updated on2026-09-08
Q1.
In the problems below, unless stated otherwise, use the approximation 22/7 for π. Identities in algebra can sometimes be shown as area relationships. For example, the figure shown (Fig. 6.41: a square of side a + b cut into a², ab, ab and b²) corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a – b) = a² – b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
Answer
Yes. In Fig. 6.41 the outer square has side a + b, so its area is (a + b)². Cutting each side at the point that divides it into a and b chops the square into four rectangles of areas a², ab, ab and b². The pieces make up the whole, so (a + b)² = a² + 2ab + b².
Figure for (a + b)(a – b) = a² – b². Start with a square of side a and cut a square of side b out of one corner. What remains has area a² – b². Slice the L-shape into a rectangle b × (a – b) and a rectangle a × (a – b)… more usefully, cut it as below and slide one piece to the side.
The L-shaped region a² − b² is cut into two rectangles of height a − b, which lay end to end into one rectangle (a + b)(a − b).
L-shaped region = a² − b² = [a × (a − b)] + [b × (a − b)] = (a − b)(a + b) ✓
Figure for (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. Take a square of side a + b + c and cut every side into pieces a, b, c. That gives a 3 × 3 grid of nine rectangles.
Nine pieces: three squares on the diagonal and three matching pairs of rectangles.
Why it happens: an area model turns an algebraic identity into a statement you can see: the whole equals the sum of its parts. It also explains why the "cross terms" come with a 2 — each mixed rectangle appears twice, once above the diagonal and once below it, because ab and ba are different pieces of the picture but the same number. Put c = 0 in the second figure and the bottom row and right column vanish, leaving Fig. 6.41 — the first identity is a special case of the second.
Q2.
An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Answer
Area = 50√2 ≈ 70.71 cm².
base = 40 − (15 + 15) = 10 cm
By Heron's formula: s = ½ × 40 = 20 Area = √(20 × (20−15) × (20−15) × (20−10)) = √(20 × 5 × 5 × 10) = √5000 = √(2500 × 2) = 50√2 ≈ 70.71 cm²
Check by half base × height. The height to the base bisects it (isosceles):
h = √(15² − 5²) = √(225 − 25) = √200 = 10√2 Area = ½ × 10 × 10√2 = 50√2 ✓
Why it happens: for an isosceles triangle both methods are available, and they must agree — Heron's formula is not a different geometry, only a different route. Heron's is faster when the height is awkward; the height method is faster when it is not.
Q3.
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Answer
Each equal side is 13 cm.
Area = ½ × base × height 60 = ½ × 10 × h 60 = 5h ⟹ h = 12 cm
The height to the base of an isosceles triangle bisects the base, so it makes a right triangle with legs 5 and 12.
equal side = √(5² + 12²) = √(25 + 144) = √169 = 13 cm
Why it happens: this is Question 2 run backwards. Because the triangle is isosceles, the foot of the altitude is the midpoint of the base — that is the only extra fact needed, and it comes from the reflection symmetry of the figure in that altitude.
Q4.
The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Answer
Perimeter = 36 cm.
In a right-angled triangle the two legs are perpendicular, so one is the base and the other is the height.
Check it yourself: could the 12 cm side have been the hypotenuse instead? Then the legs p, q would satisfy pq = 108 and p² + q² = 144. But p² + q² ≥ 2pq = 216 > 144 — impossible. So 12 cm must be a leg, as the question says.
Why it happens: the right angle is what makes this easy: it hands you a base and a perpendicular height for free, so no separate height calculation is needed. 9–12–15 is the 3–4–5 triple multiplied by 3.
Q5.
The sides of a triangle are in the ratio 2: 3: 4, and its perimeter is 45 cm. Find its area.
Answer
Area = 75√15⁄4 ≈ 72.62 cm².
Let the sides be 2k, 3k, 4k. 2k + 3k + 4k = 45 ⟹ 9k = 45 ⟹ k = 5 sides = 10 cm, 15 cm, 20 cm
s = 45⁄2 = 22.5 s − a = 22.5 − 10 = 12.5 s − b = 22.5 − 15 = 7.5 s − c = 22.5 − 20 = 2.5
Why it happens: 84375 = 5⁵ × 3³, so √84375 = 5²·3·√(5 × 3) = 75√15. Pulling out the perfect squares is worth the effort — it turns an unrecognisable number into an exact answer you can check. Note also that 10 + 15 = 25 > 20, so these three lengths really do close up into a triangle.
Q6.
The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Answer
Area = 84 cm² by both routes.
Way 1 — spot the right angle.
7² + 24² = 49 + 576 = 625 = 25² By the converse of the Baudhāyana–Pythagoras theorem, the triangle is right-angled, with the right angle between the sides 7 and 24.
Area = ½ × 7 × 24 = 84 cm²
Way 2 — Heron's formula.
s = ½(7 + 24 + 25) = 56⁄2 = 28 s − a = 21, s − b = 4, s − c = 3
Area = √(28 × 21 × 4 × 3) = √7056 = 84 cm² ✓
Why it happens: both must give the same number, because both compute the same area. The first way is far quicker when the numbers cooperate — 7, 24, 25 is a Pythagorean triple. Heron's formula never needs that luck; it works on any three sides. This pair of methods is exactly the "test the formula against known cases" strategy the chapter uses for Examples 3, 4 and 5.
Tip: Pythagorean triples worth memorising: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29 — and every multiple of them.
Q7.
If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.
Answer
Distance ≈ 18857.14 cm = 188.57 m.
distance in one rotation = circumference = πd = 22⁄7 × 60 = 1320⁄7 ≈ 188.57 cm
distance in 100 rotations = 100 × 1320⁄7 = 132000⁄7 ≈ 18857.14 cm = 188.57 m (to 2 d.p.)
Why it happens: a rolling wheel lays its rim on the road exactly once per turn, so distance = (number of turns) × circumference. This is why a bicycle with larger wheels covers more ground per pedal stroke in a given gear.
Q8.
Find the area of a quadrant of a circle whose circumference is 66 cm.
Answer
Area = 86.625 cm².
2πr = 66 2 × 22⁄7 × r = 66 44⁄7 r = 66 ⟹ r = 66 × 7⁄44 = 21⁄2 = 10.5 cm
area of circle = πr² = 22⁄7 × (10.5)² = 22⁄7 × 110.25 = 22 × 15.75 = 346.5 cm²
area of quadrant = ¼ × 346.5 = 86.625 cm²
Tip: keep 10.5 as 21⁄2 and the arithmetic stays exact: πr² = 22⁄7 × 441⁄4 = 22 × 63⁄4 = 1386⁄4 = 346.5.
Q9.
The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.
Answer
One turn = 176 cm = 1.76 m; about 568 turns per kilometre.
distance in one turn = 2πr = 2 × 22⁄7 × 28 = 2 × 22 × 4 = 176 cm = 1.76 m
1 km = 1000 m = 100 000 cm number of turns = 100000⁄176 = 568.18… ≈ 568 complete turns
Why it happens: the wheel completes 568 whole turns and is 0.18 of the way through the 569th. Since a wheel cannot turn a fraction of a time and then stop the journey, "how many times does it turn" is best answered with the whole number 568 — with the remark that a little of the next turn is also used.
Check it yourself: 568 × 1.76 m = 999.68 m, which is 32 cm short of a kilometre — exactly the 0.18 of a turn left over.
Q10.
*Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Answer
Yes. For rectangles, equal area together with equal perimeter forces the same pair of side lengths.
Let the first rectangle have sides a, b and the second p, q. equal perimeters: a + b = p + q = S equal areas: ab = pq = A
Then a and b are the two roots of x² − Sx + A = 0 and p and q are the roots of the very same equation.
A quadratic has at most two roots, so {a, b} = {p, q}. ∴ the rectangles have the same length and the same width — they are congruent.
Why it happens: a rectangle is completely determined by its two side lengths, and the sum and product of two numbers determine those two numbers (up to order). This is the "sum and product" idea you meet again in factorising quadratics. Note how sharply this contrasts with the rest of the chapter: for triangles, equal area and equal perimeter do not force congruence, and for parallelograms even all four sides do not fix the area.
Try This: find two non-congruent triangles with the same perimeter and the same area — for example the triangles with sides (17, 25, 28) and (20, 21, 29). Both have perimeter 70 and area 210. So the answer would have been "no" for triangles.
Q11.
You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., ½(a + b)h. (Fig. 6.42: a trapezium with parallel sides a and b, height h.)
Answer
Cut the trapezium into a parallelogram and a triangle, exactly as Fig. 6.42 shows.
Let the parallel sides be a (shorter, on top) and b (longer, at the bottom), with height h. From the left-hand end of the top side, draw a line parallel to the slanting left side down to the bottom side.
This cuts the trapezium into • a parallelogram of base a and height h → area = ah • a triangle of base (b − a) and height h → area = ½(b − a)h
The single cut parallel to a slanting side splits the trapezium into a parallelogram of base a and a triangle of base b − a.
Why it happens: ½(a + b) is the average of the two parallel sides. So the formula says: a trapezium has the same area as a rectangle whose width is the average of its two parallel sides. That average width is the length of the segment joining the midpoints of the two slanting sides — the mid-line — which is why the formula feels right the moment you see it.
Q12.
By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).
Answer
Draw one diagonal. It splits the trapezium into two triangles that share the same height.
Let the trapezium be ABCD with AB ∥ DC, AB = a, DC = b, and distance h between them. Draw the diagonal AC.
ΔABC: base AB = a; the apex C lies on the line DC, which is at distance h from AB ⟹ height h ar(ΔABC) = ½ a h
ΔACD: base DC = b; the apex A lies on the line AB, again at distance h ⟹ height h ar(ΔACD) = ½ b h
area of trapezium = ½ah + ½bh = ½(a + b)h ∎
One diagonal, two triangles — one standing on each parallel side, both with the same height h.
Why it happens: the two parallel sides lie on two parallel lines a distance h apart. A triangle with its base on one of those lines and its apex on the other always has height h — no matter where along the line the apex sits. That single observation makes both triangles easy, and it is the same "same base, between the same parallels" idea used throughout Exercise Set 6.2.
Q13.
Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?
Answer
Take the trapezium ABCD with AB ∥ DC, AB = a, DC = b, height h. Make an identical copy, turn it through 180° and fit it against the original along one of the slanting sides.
A half-turn of the copy makes the two slanting sides coincide; the top and bottom of the new figure are each a + b long.
A half-turn maps every line to a parallel line, so the copy's side DC lands parallel to and in line with AB. The new figure has bottom edge = b + a top edge = a + b and the same height h
Opposite sides equal and parallel ⟹ it is a parallelogram.
area of parallelogram = base × height = (a + b) h But it is made of two identical trapezia: 2 × area of trapezium = (a + b)h ∴ area of trapezium = ½(a + b)h ∎
Why it happens: this is the same doubling trick used earlier for the triangle — two congruent copies of a triangle make a parallelogram, hence the ½. It works here for exactly the same reason: a half-turn about the midpoint of the slanting side swaps the endpoints of that side and turns the two parallel sides into one straight edge of length a + b. Of the three proofs (Q11, Q12, Q13), this is the one that shows most clearly where the ½ comes from.
Q14.
Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.
Answer
Let the kite be ABCD with AB = AD and CB = CD, so AC is its axis of symmetry. Then AC is the perpendicular bisector of BD. Write d₁ = AC and d₂ = BD, and let AC meet BD at M.
(i) Using algebra.
AC ⊥ BD, and M is the midpoint of BD, so BM = MD = d₂⁄2 Let AM = p and MC = q, so p + q = d₁.
ar(ΔABD) = ½ × BD × AM = ½ d₂ p ar(ΔCBD) = ½ × BD × CM = ½ d₂ q
ar(kite) = ½ d₂ p + ½ d₂ q = ½ d₂ (p + q) = ½ d₁ d₂
(ii) Using geometry. Through each vertex draw a line parallel to the diagonal that does not pass through it. The four lines make a rectangle whose sides are d₁ and d₂ (because AC ⊥ BD).
Each of the four small right triangles inside the kite is congruent to the one just outside it, so the kite is exactly half the d₁ × d₂ rectangle.
The two diagonals cut the kite into 4 right triangles. Each of those triangles has, on the other side of a kite edge, a congruent copy inside the rectangle (same two legs, right angle between them). So the 4 triangles inside the kite and the 4 outside it pair up.
ar(kite) = ½ ar(rectangle) = ½ d₁ d₂ ∎
Why it happens: only one property is used — the diagonals are perpendicular. The formula ½d₁d₂ therefore holds for every 4-gon with perpendicular diagonals: kites, rhombuses and squares alike. That is why Question 5 of Exercise Set 6.2, about a rhombus, used the same formula.
Q15.
Three problems about fitting congruent shapes together: (i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see! (ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c. Show that ∆PQR has 4 times the area of ∆ABC. Does this mean that 4 copies of ∆ABC will fit into ∆PQR? Check and see! (iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a, 3b, 3c. Show that ∆PQR has 9 times the area of ∆ABC. Does this mean that 9 copies of ∆ABC will fit into ∆PQR? Check and see!
Answer
In all three parts the areas are in the ratio k², and in all three parts the copies really do fit — but the second and third need a construction, not just an arithmetic check.
(i) Rectangles.
ar(ABCD) = ab ar(PQRS) = 2a × 2b = 4ab = 4 ab ✓
And yes, 4 copies fit: cut PQRS in half both ways, into a 2 × 2 array of a × b rectangles.
(ii) Triangles, scale factor 2. ΔPQR has all three sides double those of ΔABC, so the two triangles are similar with ratio 2. Every length in ΔPQR — including the height on any side — is doubled.
ar(ΔABC) = ½ b h ar(ΔPQR) = ½ (2b)(2h) = 4 × ½ b h = 4 ar(ΔABC) ✓
Four copies do fit. Join the midpoints of the three sides of ΔPQR. This cuts it into four triangles: the middle one (the medial triangle) and three corner ones. Each has sides a, b, c by the midpoint theorem, so all four are congruent to ΔABC.
Joining the midpoints cuts the double-size triangle into four congruent copies of the original — the middle one is upside down.
(iii) Triangles, scale factor 3.
ar(ΔPQR) = ½ (3b)(3h) = 9 × ½ b h = 9 ar(ΔABC) ✓
Nine copies fit. Divide each side of ΔPQR into three equal parts and draw all the lines through the division points parallel to the sides. You get 9 small triangles — 6 the same way up as ΔPQR and 3 upside down — every one with sides a, b, c.
Why it happens: "area ratio = k²" and "k² copies fit" are two different statements, and the second does not follow from the first. (Two shapes can have areas in the ratio 4 : 1 and yet the small one may not tile the big one — a circle of radius 2r and a circle of radius r, for instance.) For triangles, the fit works because the medial subdivision exists: the midpoint theorem produces parallel lines, and parallel lines produce congruent copies. Notice that the upside-down triangles are essential — with only right-way-up copies you could never fill the gaps.
Try This: what about 2 copies? A triangle scaled by √2 has twice the area, but 2 copies of the original will not in general fit inside it. Scaling factors that give a clean tiling are the whole numbers.
Q16.
*Fig. 6.43: What fraction of the triangle is shaded? (In ΔABC, M is the midpoint of AB, and AC is divided into three equal parts at P and Q; the shaded region is the 4-gon B, M, P, Q.) Fig. 6.44: What fraction of the square is shaded? (Each side of the square is bisected, and four lines are drawn, each joining a vertex to the midpoint of a non-adjacent side, in rotational order; the shaded region is the small quadrilateral they enclose.)
Answer
Fig. 6.43: one half. Fig. 6.44: one fifth.
Fig. 6.43. In ΔABC, let M be the midpoint of AB, and let P and Q trisect AC with AP = PQ = QC. The shaded region is the 4-gon BMPQ. Split it with the diagonal BP.
Write T = ar(ΔABC).
ΔABP: base AP = 1⁄3 AC, same apex B as ΔABC on base AC ∴ ar(ΔABP) = 1⁄3 T
ΔAMP: base AM = ½ AB, and it shares the apex P with ΔABP on base AB ∴ ar(ΔAMP) = ½ ar(ΔABP) = 1⁄6 T
Fig. 6.44. The four cevians from the vertices to the midpoints of the non-adjacent sides enclose a small square in the middle. Take the big square to have side 2 and area 4.
Place the square with corners (0,0), (2,0), (2,2), (0,2). The four lines are (0,2) → (1,0) (2,2) → (0,1) (2,0) → (1,2) (0,0) → (2,1)
Solving each adjacent pair gives the inner vertices (0.4, 0.8), (0.8, 1.6), (1.6, 1.2), (1.2, 0.4)
By the shoelace rule the enclosed area is 4⁄5
fraction = 4/5⁄4 = 1⁄5
Without coordinates: the four cevians cut the square into the central square, four triangles and four more triangles. Rotating the whole picture through 90° about the centre maps the configuration onto itself, so the four large right triangles are congruent, and each of the four leftover triangles is congruent to the others. Adding up gives the same one fifth.
Why it happens: both answers come from the single fact that a triangle's area scales with its base when the apex is fixed. Fractions of the sides therefore become fractions of the area, and no lengths or angles ever need measuring. Fig. 6.44 is a famous configuration — the inner square is always exactly one fifth of the outer one, whatever the size.
Try This: in Fig. 6.44, replace "midpoint" by "the point one third along". The inner region is then a smaller square. Repeat the coordinate calculation and see what fraction you get.
Q17.
Fig. 6.45: What fraction of the rectangle is covered by the circles? (Three equal circles in a row, each touching its neighbours and the two long sides of the rectangle.) Fig. 6.46: What fraction of the rectangle is covered by the circles? (The same arrangement with four circles.)
Answer
Both answers are the same: π/4 ≈ 0.785, that is about 78.5%.
Let each circle have diameter d, so radius d⁄2.
Fig. 6.45 — three circles. The circles touch the top and bottom sides, so the rectangle has height d. They touch each other in a row, so the length is 3d.
area of rectangle = 3d × d = 3d² area of circles = 3 × π(d⁄2)² = 3 × πd²⁄4 = 3πd²⁄4
Why it happens: the number of circles cancels. Each circle sits inside its own d × d square, and covers π⁄4 of it; putting n such squares side by side changes nothing about that ratio. In other words the whole figure is just n copies of one picture — "circle in a square" — repeated.
Tip: with π ≈ 22⁄7 the fraction is 22⁄28 = 11⁄14 ≈ 0.7857. About 21.5% of the rectangle is always wasted — which is why packing round tins into a box always leaves gaps.
Q18.
Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!
Answer
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.
Number of circles n
Rectangle
Total circle area
Fraction
3
3d × d = 3d²
3 × πd²/4
π/4
4
4d × d = 4d²
4 × πd²/4
π/4
10
10d²
10 × πd²/4 = 2.5πd²
π/4
20
20d²
5πd²
π/4
50
50d²
12.5πd²
π/4
Proof.
Let there be n equal circles of diameter d in a row, each touching its neighbours and both long sides of the rectangle.
height of rectangle = d length of rectangle = nd area of rectangle = nd × d = nd²
area of one circle = π(d⁄2)² = πd²⁄4 area of n circles = n × πd²⁄4 = nπd²⁄4
fraction = nπd²/4⁄nd² = π⁄4 (both n and d² cancel) ∎
Why it happens: the answer is independent of n and of d because the figure is built by repeating one unit — a circle inscribed in a square — and a ratio computed on one unit is the ratio for any number of copies. This is a useful habit of mind: when a quantity does not change as a parameter varies, look for the repeated unit that explains it.
Did you know? If you are allowed to stagger the rows instead of stacking them squarely, circles can be packed more densely — the hexagonal packing reaches π/(2√3) ≈ 0.9069, and Thue proved in 1890 that no arrangement does better.
Q19.
*The figure (Fig. 6.47) shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle. (Four rectangles lie side by side in the upper row and five lie side by side in the lower row.)
Answer
Perimeter of each small rectangle = 18√10⁄5 = 3.6√10 ≈ 11.38 cm.
The two rows must span the same width, so the four upper rectangles are lying on their long sides and the five lower ones are standing on their short sides.
Let the small rectangle be l long and w wide, and let the big rectangle be W wide and H high.
Upper row: 4 rectangles across ⟹ each is W⁄4 wide, and its height is the row height. Lower row: 5 rectangles across ⟹ each is W⁄5 wide.
All nine are identical, and W⁄4 ≠ W⁄5, so the two rows use the two different sides: l = W⁄4 and w = W⁄5 upper row height = w = W⁄5, lower row height = l = W⁄4
H = W⁄5 + W⁄4 = 9W⁄20
Area = W × H = W × 9W⁄20 = 9W²⁄20 = 72 W² = 160 ⟹ W = √160 = 4√10 cm
Check it yourself: area of one small rectangle = l × w = √10 × 4√10⁄5 = 4 × 10⁄5 = 8 cm². Nine of them give 72 cm² ✓. And the sides are in the ratio l : w = √10 : 4√10⁄5 = 5 : 4, as the picture suggests.
Why it happens: the whole puzzle is fixed by the single condition 4l = 5w, which comes from the two rows having to be equally wide. That pins the shape of the small rectangle (sides in the ratio 5 : 4) without knowing its size; the total area 72 cm² then pins the size. The answer is irrational because 72 is not a multiple of the "natural" area 180 that a 5 : 4 rectangle of whole-number sides would give.
Q20.
*Show that the areas of the shaded blue triangle and the shaded red triangle are equal (Fig. 6.48: lines are drawn from one vertex of a triangle to the two points of trisection of the opposite side; the blue triangle is the leftmost of the three parts and the red one is the rightmost). Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.
Answer
Let the triangle be ABC with the base BC trisected at D and E, so BD = DE = EC, and let the cevians AD and AE be drawn. The blue triangle is ABD and the red one is AEC.
BD = DE = EC = 1⁄3 BC (given: points of trisection)
Both ΔABD and ΔAEC have their base on the line BC, and both have the same apex A. ∴ both have the same height h, the distance from A to BC.
ar(ΔABD) = ½ · BD · h ar(ΔAEC) = ½ · EC · h BD = EC ⟹ ar(ΔABD) = ar(ΔAEC) ∎
(Each is one third of ΔABC, so the three parts are equal.)
Cutting one into the other. The two triangles have equal bases on one line and a common apex, but they are usually not congruent — the blue one may be tall and thin, the red one squat. Here is a three-step dissection.
Step 1. Cut ΔABD along its midline — the segment joining the midpoints of AB and AD. Rotate the small top triangle through 180° about the midpoint of AD. The two pieces now form a parallelogram of base BD and height h/2.
Step 2. Do the same to ΔAEC: it becomes a parallelogram of base EC = BD and height h/2.
Step 3. The two parallelograms have the same base and the same height and lie between the same pair of parallel lines, so one is a shear of the other. A single straight cut across the first parallelogram, followed by sliding the cut piece along, produces the second. (If the two lean very differently, repeat the slide — the "staircase" cut — a few times.)
So 3 pieces are enough in the usual case.
Why it happens: everything rests on the same principle as before — triangles on equal bases between the same parallels have equal areas, because area depends on base and height only. The dissection then works because equal area is sufficient for a plane dissection (the Wallace–Bolyai–Gerwien theorem); the midline-and-half-turn move is the standard first step, since it converts any triangle into a parallelogram of half the height.
Q21.
*The figure (Fig. 6.49) shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.
Answer
Let the square have side a, with the quarter circle centred at the corner O and of radius a. The two semicircles are drawn inwards on the two sides through O, each of diameter a.
Call the two semicircular discs S₁ and S₂.
area of the quarter disc Q = ¼ πa² area of each semicircular disc = ½ π(a⁄2)² = πa²⁄8 area(S₁) + area(S₂) = πa²⁄8 + πa²⁄8 = πa²⁄4 = area(Q) … (1)
Both semicircular discs lie entirely inside the quarter disc. (Every point of S₁ is at most a⁄2 + a⁄2 = a from O, and the same for S₂.) Region A is where the two semicircular discs overlap; region B is the part of the quarter disc that neither of them covers.
For any two regions: (S₁ ∪ S₂) + (S₁ ∩ S₂) = S₁ + S₂ Here S₁ ∩ S₂ = A, so area(S₁ ∪ S₂) + A = πa²⁄4 … (2)
Also, since S₁ ∪ S₂ lies inside Q and B is what is left over, area(S₁ ∪ S₂) + B = area(Q) = πa²⁄4 … (3)
Comparing (2) and (3): A = B ∎
A is the overlap of the two semicircular discs; B is the part of the quarter disc left uncovered. Since the two semicircles together have exactly the quarter disc's area, A and B must match.
Why it happens: the coincidence that makes it work is (1): two semicircles on half-length diameters have together exactly the same area as the quarter circle on the full-length radius, because area scales with the square of length — halving the diameter quarters the area, and two quarters plus the extra factor from "semicircle vs quarter circle" balance out. Once the totals match, whatever one region double-counts (A) the other must leave out (B). No integration and no π-value is needed; π cancels completely.
Q22.
*In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.
Put the square on coordinates with corners (0,0), (2,0), (2,2), (0,2). The four semicircles have centres (1,0), (2,1), (1,2), (0,1) and radius 1, drawn inwards.
Why it happens: the four semicircles all pass through the centre of the square, because the distance from the midpoint of a side to the centre is 1 — exactly the radius. That is what makes the petals meet at a single point and makes every arc a clean quarter circle. The flower takes up 2π − 4 ≈ 2.28 out of the square's 4 square units, that is about 57%.
Check it yourself: the four semicircular discs have total area 4 × ½π(1)² = 2π. The square has area 4. The parts of the square covered exactly twice are the four petals, and 2π − 4 is precisely the "excess" — a neat second route to the same answer.
Q23.
*In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is ¼πl².
Answer
Let the outer radius be R and the inner radius r. The green region is the annulus between them.
BC touches the inner circle at A, so OA ⊥ BC and OA = r. OA is therefore the perpendicular from the centre O to the chord BC of the big circle, and the perpendicular from the centre bisects the chord: AB = AC = l⁄2
In right triangle OAB, by the Baudhāyana–Pythagoras theorem: OB² = OA² + AB² R² = r² + (l⁄2)² R² − r² = l²⁄4
area of the annulus = πR² − πr² = π(R² − r²) = π · l²⁄4 = ¼ πl² ∎
The tangent chord makes a right triangle with legs r and l/2 and hypotenuse R, so R² − r² is fixed by l alone.
Why it happens: the individual radii never appear in the answer — only their difference of squares does, and that is exactly what the right triangle supplies. So all pairs of concentric circles with the same tangent-chord length have annuli of the same area: a thin annulus of huge radius and a fat one of small radius can have equal area, provided the chord that grazes the inner circle has the same length. It also means the answer is the area of a circle whose diameter is l.
Q24.
*In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C). (A and B are the two crescent-shaped regions between the semicircles on the legs and the semicircle on the hypotenuse; C is the triangle itself.)
Answer
This is the classical result known as the lunes of Hippocrates. Let the right angle be at Q, with legs QP = b, QR = a, and hypotenuse PR = c, so a² + b² = c².
Three semicircles are drawn: on QP and on QR outwards, and on PR on the same side as Q. Note that the third one passes through Q, because ∠PQR = 90° and the angle in a semicircle is a right angle.
Semicircular disc on PR (the one containing Q) is made of the triangle C, plus the segment S₁ cut off by the chord QP, plus the segment S₂ cut off by the chord QR.
½π(c⁄2)² = C + S₁ + S₂ … (1)
Lune A = (semicircular disc on QP) − S₁ = ½π(b⁄2)² − S₁ Lune B = (semicircular disc on QR) − S₂ = ½π(a⁄2)² − S₂
Why it happens: two facts do all the work. First, the areas of the three semicircles are proportional to a², b², c² (with the same constant π/8), so the Pythagoras relation carries straight over: semicircle on the hypotenuse = sum of the semicircles on the legs. Second, the semicircle on the hypotenuse passes exactly through the right-angle vertex, so it can be split cleanly into the triangle plus the two segments that the lunes have to give up. Subtracting the same two segments from both sides leaves the identity.
Did you know? Hippocrates of Chios (c. 440 BCE) found this while trying to square the circle. Its striking feature is that the answer C is a straight-sided area even though A and B are bounded entirely by arcs — every π cancels. He hoped this would let him square the circle itself; it does not, as Lindemann finally proved in 1882.
Q25.
*Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius r. (The shaded region is the lens-shaped part common to both circles, with vertices C and D.)
Answer
Area of the shaded common region = r²(2π⁄3 − √3⁄2) ≈ 1.228 r².
Let the centres be A and B, so AB = r, and let the circles meet at C and D.
AC = BC = r (radii) and AB = r ⟹ ΔABC is equilateral ∴ ∠CAB = 60°, and by the same argument ∠DAB = 60° ∴ ∠CAD = 120° (and likewise ∠CBD = 120°)
The lens is made of two identical circular segments — one cut from each circle by the chord CD, each with central angle 120°.
sector of A with angle 120° = πr² × 120⁄360 = πr²⁄3
triangle ACD: AC = AD = r with 120° between them CD = 2 · r sin 60° = √3 r, and the height from A is r cos 60° = r⁄2 ar(ΔACD) = ½ × √3 r × r⁄2 = √3⁄4 r²
Tip: if instead you read "the region enclosed by the two circles" as the whole shape they bound on the outside (the figure of Example 1, whose perimeter was found to be 8πr⁄3), then its area is
Why it happens: everything follows from AB = r, which forces the two triangles ABC and ABD to be equilateral and hence pins every angle at 60° or 120°. That is why the answer contains √3 — it is the height of an equilateral triangle showing up. The same configuration gave the arcs of 120° in Example 1 of the chapter, which is why that perimeter came out as 8πr⁄3.
Q26.
*In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is 2(A + C)(B + C)/C.
Answer
Set up coordinates. Let the rectangle have width W and height H, with the bottom-left corner at O = (0, 0). Let the marked point on the top edge be (p, H), the marked point on the right edge be (W, q), and let M = (p, q) be the interior point where the three triangles meet.
Triangle A has vertices O, (p, H), M. Its side from (p, H) to M is vertical, of length H − q, and O is p away from it. A = ½ p (H − q)
Triangle C has vertices (p, H), (W, q), M — right-angled at M. C = ½ (H − q)(W − p)
Triangle B has vertices O, M, (W, q). Its side MW is horizontal, of length W − p, at height q; O is q below it. B = ½ (W − p) q
A + C = ½ p(H − q) + ½ (H − q)(W − p) = ½ (H − q)[p + W − p] = ½ W (H − q)
B + C = ½ (W − p) q + ½ (H − q)(W − p) = ½ (W − p)[q + H − q] = ½ H (W − p)
(A + C)(B + C) = ¼ W H (H − q)(W − p) = ¼ W H × 2C [since (H − q)(W − p) = 2C] = ½ W H · C
∴ WH = 2(A + C)(B + C)⁄C ∎
Why it happens: the two grouped sums are the two "half-rectangles" the picture hides. A + C is exactly half of the rectangle of width W and height H − q (the strip above the level of the marked right-edge point), and B + C is half of the rectangle of width W − p and height H. Multiplying them brings in the product of the two strips, and the factor (H − q)(W − p) that is left over is nothing but 2C — so dividing by C removes it and leaves the whole rectangle. The moral is the one from Question 10 of Exercise Set 6.2: group the triangles before computing.
Check it yourself: take W = 4, H = 3, p = 3, q = 2. Then A = ½·3·1 = 1.5, C = ½·1·1 = 0.5, B = ½·1·2 = 1. The formula gives 2(1.5+0.5)(1+0.5)/0.5 = 2 × 2 × 1.5 / 0.5 = 12 = 4 × 3 ✓
Q27.
*In the figure (Fig. 6.55) we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal. (O is the centre of a semicircle on diameter AC; B is the point of that semicircle directly above O; D is the midpoint of AB, and a semicircle is drawn on AB as diameter, on the far side from O. One shaded region is the crescent between that semicircle and the arc AB of the big circle; the other is the triangle AOB.)
Answer
Another lune of Hippocrates. Let OA = OB = OC = r.
B is directly above O, so OB ⊥ AC and ∠AOB = 90°. By the Baudhāyana–Pythagoras theorem, AB = √(r² + r²) = r√2
The quarter disc AOB (centre O, radius r, angle 90°): = ¼ πr²
The semicircular disc on AB (radius AB⁄2 = r√2⁄2): = ½ π(r√2⁄2)² = ½ π · 2r²⁄4 = πr²⁄4
These two are equal! … (1)
Now split the quarter disc by the chord AB:
quarter disc AOB = triangle AOB + segment S where S is the piece between the chord AB and the arc AB. πr²⁄4 = ΔAOB + S … (2)
The lune (the crescent) is the semicircular disc on AB with the same segment S removed: lune = πr²⁄4 − S … (3)
Comparing (2) and (3): lune = ΔAOB
and ΔAOB = ½ × OA × OB = ½ r · r = r²⁄2
So both shaded regions have area r²⁄2. ∎
The semicircle on AB and the quarter disc AOB have exactly the same area, so removing the same segment from each leaves the lune equal to the triangle.
Why it happens: line (1) is the whole trick, and it is a scaling fact. AB is √2 times OA, so the semicircle on AB has 2 times the area of a semicircle on OA — but a semicircle is twice a quarter circle, and those two factors of 2 undo each other. The moment two curved regions have equal area, subtracting a shared piece leaves equal remainders, and one of the remainders happens to be straight-sided. As in Question 24, every π cancels.