Key Concepts & Formulae at a Glance
- Perimeter of a square (side a) = 4a; equilateral triangle (side a) = 3a; rectangle (sides a, b) = 2(a+b).
- π = C/D ratio, constant for every circle, ≈ 22/7 or 3.14; π is irrational (Lambert, 1761).
- Circumference \(C=2\pi r\). Arc length subtending angle \(\theta^\circ\) at the centre: \(l = 2\pi r \times \dfrac{\theta^\circ}{360^\circ}\).
- Area of rectangle = ab; parallelogram = base × height = bh; triangle = \(\tfrac12 bh\).
- Heron's formula: area of a triangle with sides a, b, c is \(\sqrt{s(s-a)(s-b)(s-c)}\), where \(s=\tfrac12(a+b+c)\).
- Brahmagupta's formula: area of a cyclic 4-gon with sides a, b, c, d is \(\sqrt{(s-a)(s-b)(s-c)(s-d)}\), where \(s=\tfrac12(a+b+c+d)\). It generalises Heron's formula (set d = 0).
- Area of circle = \(\pi r^2\); sector (angle θ°) = \(\pi r^2 \times \dfrac{\theta^\circ}{360^\circ}\); segment = sector − triangle.
- A median divides a triangle into two triangles of equal area. A diagonal divides a parallelogram into two triangles of equal area.
- Throughout this chapter's solutions, unless a question states otherwise, π is taken as 22/7.
Think and Reflect (Chapter Opener)
TRIn 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 (i.e., smaller than the stagger used in the Olympics), for the same 4 × 100 m relay race?
A 200 m track has a smaller radius on its curves than a 400 m track — so it needs a proportionally smaller stagger.
Yes. The stagger between lanes depends directly on the radius of the curved part of the track (specifically, on the difference in radius between adjacent lanes), and a 200 m track has smaller curved sections (smaller radius) than a 400 m track, since both tracks are built to fit the same standard lane width but the 200 m track is scaled down overall.
Since the stagger is essentially the extra circumference an outer lane's semicircle has compared to an inner lane's semicircle — and circumference is proportional to radius — a track with a smaller radius will need a smaller stagger between its lanes to keep the race fair for the same distance.
Think and Reflect
TRWhat is the connection between this question (the perimeter of a circle with radius r) and the one about the 400 m athletics track?
The 400 m athletics track is built from two straight sections joined by two semicircular curves. To design the track (and to work out how much longer the outer lanes are, i.e., the stagger), the organisers must first know how to calculate the perimeter of a circle in terms of its radius.
In other words, finding the perimeter of a circle is the essential piece of mathematics needed to answer the athletics-track question: once we know the circumference formula, we can compute the semicircular portions of each lane and hence the exact stagger required between adjacent lanes.
Think and Reflect
TRWhat 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?
Sketch — the two lanes as concentric semicircular arcs around centre O (not to scale).
The innermost semicircle has radius 36.5 m, and the athlete in lane 1 is assumed to run 0.3 m from the inner border, giving an effective radius of 36.8 m (as computed in the text). Each lane is 1.22 m wide, so the runner in lane 2 runs on a semicircle with effective radius \(36.8+1.22=38.02\) m.
Difference in radius between lane 1 and lane 2 = 1.22 m (exactly the width of one lane).
Stagger for lane 2 = extra distance run on the two semicircular curves compared to lane 1. Since the two semicircles together make a full circle, the extra distance is the difference in circumference between the two full circles:
Stagger \(= 2\pi(38.02) - 2\pi(36.8) = 2\pi(38.02-36.8) = 2\pi(1.22)\)
\(= 2 \times 3.1416 \times 1.22 \approx 7.67\) m
Will lane 3 need the same stagger as lane 2? Yes — since each lane is 1.22 m wider than the previous one, the difference in radius between any two consecutive lanes is always the same (1.22 m). Because the stagger formula \(2\pi \times (\text{difference in radius})\) depends only on this constant difference (not on the absolute radius itself), the stagger between every pair of consecutive lanes is exactly the same: \(2\pi(1.22) \approx 7.67\) m.
Stagger needed for lane 2 ≈ 7.67 m
Yes — an equal stagger (≈7.67 m) is needed between every pair of consecutive lanes, since the radius always increases by the same 1.22 m lane width.
Exercise Set 6.1
1The perimeter of a circle is 44 cm. What is its radius?
Rough sketch — circle of unknown radius r, circumference 44 cm.
\(C = 2\pi r \Rightarrow 44 = 2 \times \dfrac{22}{7} \times r\)
\(44 = \dfrac{44}{7} r \Rightarrow r = 44 \times \dfrac{7}{44} = 7\)
2Calculate, correct to 3 significant figures, the circumference of a circle with:
(i) radius 7 cm
(ii) radius 10 cm
(iii) radius 12 cm.
Rough sketch — circles of radius 7, 10, and 12 cm.
(i) \(C=2\times\dfrac{22}{7}\times7 = 44\) cm (exactly, so 44.0 cm to 3 s.f.)
(ii) \(C=2\times\dfrac{22}{7}\times10 = \dfrac{440}{7} = 62.857\ldots \approx 62.9\) cm
(iii) \(C=2\times\dfrac{22}{7}\times12 = \dfrac{528}{7} = 75.428\ldots \approx 75.4\) cm
3Calculate the length of the arc of a circle if:
(i) the radius is 3.5 cm and the angle at the centre is 60°, and
(ii) the radius is 6.3 m and the angle at the centre is 120°.
Rough sketch — arc subtending angle θ at the centre.
(i) \(l = 2\pi r \times \dfrac{\theta^\circ}{360^\circ} = 2\times\dfrac{22}{7}\times3.5\times\dfrac{60}{360}\)
\(2\times\dfrac{22}{7}\times3.5 = 22\) (the full circumference); arc \(=22\times\dfrac{1}{6}=\dfrac{22}{6}=\dfrac{11}{3}\approx3.67\) cm
(ii) \(2\times\dfrac{22}{7}\times6.3 = 39.6\) m (full circumference); arc \(=39.6\times\dfrac{120}{360}=39.6\times\dfrac{1}{3}=13.2\) m
4Find 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°.
Rough sketch — sector of radius 14 cm, angle 75°.
Arc length \(= 2\pi r\times\dfrac{75}{360}=2\times\dfrac{22}{7}\times14\times\dfrac{75}{360}\)
\(2\times\dfrac{22}{7}\times14=88\); arc \(=88\times\dfrac{75}{360}=\dfrac{6600}{360}=\dfrac{55}{3}\approx18.33\) cm
Perimeter of sector = arc length + 2 radii \(= \dfrac{55}{3}+2(14)=\dfrac{55}{3}+28=\dfrac{55+84}{3}=\dfrac{139}{3}\approx46.33\) cm
5Find the perimeters of the shapes in Fig. 6.14 (i)–(ix), taking the arcs to be quarter, half or three-quarters of a circle as appropriate.
Reading each figure as follows (based on the markings shown): straight edges are noted where present, and every curved edge is a semicircular arc unless stated otherwise.
Schematic sketches of shapes (i)–(ix) — not to scale, showing the arrangement of straight edges and semicircular arcs used in each calculation.
(i) Stadium shape, overall length 80 m, height 60 m (diameter of the semicircular ends, radius 30 m).
Straight sections: \(80-60=20\) m each, two of them = 40 m.
The two semicircular ends together form one full circle of radius 30 m:
\(2\pi(30)=2\times\dfrac{22}{7}\times30=\dfrac{1320}{7}\approx188.57\) m
Total \(\approx 40+188.57=228.57\) m.
(ii) A large semicircle (diameter 12 cm, radius 6) with a smaller semicircular notch (diameter 8 cm, radius 4) cut from its base, leaving \((12-8)/2=2\) cm of flat base exposed on each side.
Perimeter = large arc + small arc + 2 flat bits
\(= \pi(6)+\pi(4)+2(2) = \dfrac{22}{7}(6+4)+4=\dfrac{220}{7}+4\approx31.43+4=35.43\) cm
(iii) A 4-lobed "flower" outline: 4 semicircular bulges, each of diameter 10 cm (radius 5), arranged around a 10 cm square.
Perimeter = 4 semicircle arcs
\(=4\times\pi(5)=4\times\dfrac{22}{7}\times5=\dfrac{440}{7}\approx62.86\) cm
(iv) An equilateral triangle of side 12 cm sitting on a semicircle of the same diameter (12 cm, radius 6) — the base of the triangle is replaced by the semicircular arc.
Perimeter = 2 triangle sides + semicircle arc
\(=2(12)+\pi(6)=24+\dfrac{22}{7}\times6=24+\dfrac{132}{7}\approx24+18.86=42.86\) cm
(v) Same 4-lobed pattern as (iii), but with side 14 cm (radius 7).
Perimeter \(=4\times\pi(7)=4\times22=88\) cm
(vi) A wavy border made of 4 alternating semicircular bumps along a baseline of total length 28 cm, so each semicircle has diameter \(28/4=7\) cm (radius 3.5).
Perimeter (the wavy curve) \(=4\times\pi(3.5)=4\times11=44\) cm
(vii) A right triangle with legs 6 cm and 8 cm (hypotenuse 10 cm by Pythagoras), with a semicircle drawn on the hypotenuse (radius 5) and a smaller semicircle on the 6 cm leg (radius 3), the 8 cm leg left exposed.
Perimeter \(= 8 + \pi(5)+\pi(3) = 8+\dfrac{22}{7}(8)=8+\dfrac{176}{7}\approx8+25.14=33.14\) cm
(viii) A large semicircle (diameter 12 cm, radius 6) sitting above 3 small semicircular scallops along its base (each diameter 4 cm, radius 2).
Perimeter \(=\pi(6)+3\pi(2)=\pi(6+6)=12\times\dfrac{22}{7}=\dfrac{264}{7}\approx37.71\) cm
(ix) An "S-curve" made of 2 semicircles, each of diameter 10 cm (radius 5), one bulging up and one down along a 20 cm baseline.
Perimeter \(=2\times\pi(5)=10\times\dfrac{22}{7}=\dfrac{220}{7}\approx31.43\) cm
(ii) ≈35.43 cm
(iii) ≈62.86 cm
(iv) ≈42.86 cm
(v) 88 cm
(vi) 44 cm
(vii) ≈33.14 cm
(viii) ≈37.71 cm
(ix) ≈31.43 cm
6If the diameter of a car tyre is 56 cm, then:
(i) How far does the car need to travel for the tyre to complete one revolution?
(ii) How many revolutions does the tyre make if the car travels 10 km?
Rough sketch — car tyre, diameter 56 cm.
(i) One revolution covers a distance equal to the circumference: \(C=\pi d=\dfrac{22}{7}\times56=176\) cm.
(ii) 10 km \(=10\times100000=1{,}000{,}000\) cm. Number of revolutions \(=\dfrac{1000000}{176}\approx5681.82\)
7Find the total perimeter of all the petals in each of the given flowers (Fig. 6.15A, 6.15B).
Fig. 6.15A — 4-petal flower, arc centres at square midpoints.
Fig. 6.15B — 6-petal flower, arc centres at hexagon vertices.
(i) Fig. 6.15A: square of side 14 cm, with the centres of the arcs at the midpoints of the sides. This creates a 4-petal flower where each side contributes one full semicircular arc (diameter = side = 14 cm, radius 7) to the total boundary of the flower.
Total perimeter \(=4\times\pi(7)=4\times\dfrac{22}{7}\times7=4\times22=88\) cm.
(ii) Fig. 6.15B: regular hexagon of side 42 cm, with the centres of the arcs at the vertices of the hexagon. Each vertex draws an arc of radius = side = 42 cm, sweeping an angle of 60° (matching the hexagon's construction from 6 equilateral triangles), giving 6 arcs total.
Each arc length \(= 2\pi(42)\times\dfrac{60}{360}=\dfrac{2\pi(42)}{6}=\pi(14)=\dfrac{22}{7}\times14=44\) cm.
Total for 6 arcs \(=6\times44=264\) cm.
8The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Two circles with radii in ratio 5:4 (not to scale).
Since \(C=2\pi r\), the circumference is directly proportional to the radius. So if \(C_1:C_2=5:4\), then \(r_1:r_2\) must also be \(5:4\) (the constant \(2\pi\) cancels out in the ratio).
Think and Reflect
TRWhat 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 this 'gap'?
Rough sketch — thin parallelogram ABCD where the perpendicular from C falls outside AD.
The textbook itself gives the fix (Fig. 6.19): instead of dropping a perpendicular that lands outside segment AD, we extend DA beyond A and mark a point A' such that AA' equals a small extension D'D taken from the other end, forming a new parallelogram A'BCD' with the same base and height as ABCD.
Since triangles CDD' and BAA' are congruent (matching sides and angles from the parallel sides and equal extensions), removing triangle BAA' and adding congruent triangle CDD' leaves the total area unchanged. So A'BCD' has exactly the same area as ABCD, but now the perpendicular from C properly lands within the new base A'D'.
If a single such shift still isn't enough (for an extremely "thin," elongated parallelogram), the same shifting step can simply be repeated as many times as necessary until the foot of the perpendicular lands on the (repeatedly extended) base.
TRThe 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? Why or why not? (Hint: What happens to the area of a parallelogram if we decrease or increase the angle between the adjacent sides while keeping the lengths fixed?)
Same four side lengths, two different angles between them — very different areas.
No — unlike a rectangle, knowing only the side lengths of a parallelogram is not enough to determine its area.
This is because the area formula is \(bh\), and while the base \(b\) is one of the given side lengths, the height \(h\) also depends on the angle between the two adjacent sides — which is not fixed by the side lengths alone. A parallelogram is "hinge-able": you can keep both pairs of sides the same length while changing the angle between them (imagine a rectangular picture frame with hinged corners being pushed sideways into a slanted shape).
As you decrease the angle between the adjacent sides (making the parallelogram "leaner" or more slanted, approaching a flattened shape), the height \(h\) decreases towards 0, so the area shrinks towards 0 — even though the side lengths never change. Conversely, the area is largest (equal to the rectangle's area, \(ab\)) exactly when the angle between adjacent sides is 90°.
Think and Reflect
TRIs there a gap in our argument? What would we do if angle EFG is obtuse and the triangle were shaped like triangle EFG in Fig. 6.20B?
Fig. 6.20B — Obtuse triangle EFG; the perpendicular from E meets the extended base at H, outside segment FG.
When angle EFG is obtuse, the foot of the perpendicular from E to line FG falls outside segment FG (beyond F), rather than between F and G. The rectangle-enclosure method needs a small adjustment in this case.
Extend line FG beyond F, and drop the perpendicular from E to this extended line, meeting it at a point H (outside segment FG). The rectangle is now built using this extended base.
The area of triangle EFG can then be found as: (area of triangle EHG, the larger triangle using the full extended base HG) minus (area of triangle EHF, the extra triangle beyond F). Since both of these use the "base × height ÷ 2" formula with the same height (EH), and \(HG = HF+FG\), the subtraction works out to exactly \(\tfrac12 \times FG \times EH\), i.e., the same formula \(\tfrac12 bh\) still applies, using FG as the base and EH as the height.
TRSince ΔABD and ΔACD have equal area, you may wonder — Can we divide ΔABD using straight cuts into two or more pieces that we can then rearrange to exactly cover ΔACD? What do you think? Is it possible?
Fig. 6.22 — Median AD divides ΔABC into ΔABD and ΔACD, with BD = DC.
Yes, it is possible. Since ΔABD and ΔACD have the same base length (BD = DC) and the same height (the perpendicular distance from A to line BC), a "shearing" dissection can convert one into the other.
One way: cut ΔABD along the line joining A to the midpoint of BD. This splits ΔABD into two smaller triangles. Rotating one of these two pieces by 180° about that midpoint rearranges the two pieces into a parallelogram with base \(\tfrac12 BD\) and height equal to the original height. Doing the same construction for ΔACD (cutting along the line from A to the midpoint of DC) produces a congruent parallelogram, since \(BD=DC\) and the height is the same for both triangles. Since the two resulting parallelograms are congruent, the pieces of one triangle can indeed be rearranged to exactly cover the other.
TRSuppose 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 polygon? Try this for: (1) a square and non-square rectangle with equal area, (2) two triangles with different shapes but equal area, (3) a triangle and a square with equal area. Formulate your own conjecture. — Also: Think of various rectangles with perimeter 40 units. (1) How many such rectangles are there? (2) Is there one whose area is largest? What are its dimensions? (3) Is there one whose area is smallest? What are its dimensions? Do either answer surprise you?
Part 1 — Dissection conjecture: Yes, this is always possible for any two polygons of equal area — this is a famous result known as the Bolyai–Gerwien theorem. Testing the given cases:
(1) A square and a non-square rectangle of equal area: cut the rectangle with straight vertical/horizontal cuts into strips and rearrange (a classic "staircase" dissection) to form the square.
(2) Two differently-shaped triangles of equal area: both can be dissected (e.g. via an intermediate rectangle or parallelogram of the same area) and reassembled into each other.
Two triangles with the same base b and height h (hence equal area) but different shapes — a simple example for test case (2).
(3) A triangle and a square of equal area: convert the triangle into a rectangle of the same area (using the parallelogram/triangle-area relationship), then convert that rectangle into a square using the staircase method from (1).
Conjecture: Any two polygons of equal area can always be cut into finitely many pieces by straight cuts and rearranged to form the other polygon.
Part 2 — Rectangles with perimeter 40 units: If the sides are \(x\) and \(20-x\) (since \(2(x+(20-x))=40\)), then as \(x\) ranges over all values strictly between 0 and 20, there are infinitely many such rectangles (since the sides don't have to be integers).
Area \(A(x) = x(20-x) = 20x-x^2\). This is a downward parabola in \(x\), maximised at \(x=10\) (the vertex), giving the largest area when the rectangle is actually a square of side 10 units, with area 100 sq. units.
Three rectangles, all with perimeter 40 units — as the shape gets closer to a square, the area increases.
As \(x\to 0\) or \(x\to 20\) (an increasingly long, thin rectangle), the area \(A(x)\to 0\), so there is no smallest area — the area can be made as small as we like (though never actually reaching 0, since a rectangle needs positive side lengths), but there is no single rectangle with the minimum area.
Rectangles with perimeter 40: infinitely many exist.
Largest area = 100 sq. units, achieved by the square of side 10 (a square, not a "typical" rectangle, gives the maximum — often the surprising part).
No smallest area exists, since the area shrinks towards 0 as the rectangle gets thinner and longer.
Think and Reflect
TRWhat procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. How would you proceed?
The two-step conversion: triangle → rectangle (same area) → square (Baudhāyana's construction).
This can be done in two steps, both of which the chapter has already built the tools for:
Turn the triangle into a rectangle of the same area. If the triangle has base \(b\) and height \(h\), its area is \(\tfrac12 bh\). This is exactly the area of a rectangle with sides \(b\) and \(\tfrac{h}{2}\) (or equivalently \(\tfrac{b}{2}\) and \(h\)) — so construct a rectangle with these dimensions (halving one side is a simple ruler-and-compass bisection).
Square that rectangle. Apply Baudhāyana's construction (Section 6.9) directly to this rectangle to construct a square with exactly the same area.
Exercise Set 6.2
1Find the area of triangle ADE in Fig. 6.31 (rectangle ABCD-style figure, width 10 cm, height 8 cm, with E on side BC).
Fig. 6.31 — Triangle ADE always has base AD = 10 cm and height 8 cm, whichever point E is on BC.
Triangle ADE has base AD = 10 cm. Since E lies on side BC, which is parallel to AD at a perpendicular distance of 8 cm (the full height of the rectangle), the height of the triangle (measured from E down to line AD) is 8 cm — regardless of exactly where E sits along BC.
Area \(= \tfrac12 \times \text{base} \times \text{height} = \tfrac12 \times 10 \times 8 = 40\) sq cm
2The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Rough sketch — trapezium with parallel sides 40 cm, 20 cm and equal legs 26 cm.
Since the trapezium is isosceles, dropping perpendiculars from the ends of the shorter parallel side (20 cm) to the longer one (40 cm) creates two congruent right triangles on either side, each with base \(\tfrac{40-20}{2}=10\) cm and hypotenuse 26 cm (the equal non-parallel side).
Height \(h = \sqrt{26^2-10^2}=\sqrt{676-100}=\sqrt{576}=24\) cm
Area \(= \tfrac12(a+b)h = \tfrac12(40+20)(24)=\tfrac12(60)(24)=720\) sq cm
3Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Rough sketch — triangle with sides 8 cm, 11 cm, and the third side 13 cm.
Third side \(= 32-8-11=13\) cm.
\(s=\dfrac{32}{2}=16\)
Area \(=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{16(16-8)(16-11)(16-13)}=\sqrt{16\times8\times5\times3}=\sqrt{1920}\)
\(\sqrt{1920}=\sqrt{64\times30}=8\sqrt{30}\approx43.82\) sq cm
4The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.
Rough sketch — triangle with sides in ratio 3:5:7.
Let sides = 3k, 5k, 7k. Sum \(=15k=300 \Rightarrow k=20\). Sides = 60 m, 100 m, 140 m.
\(s=\dfrac{300}{2}=150\)
Area \(=\sqrt{150(150-60)(150-100)(150-140)}=\sqrt{150\times90\times50\times10}=\sqrt{6{,}750{,}000}\)
\(=\sqrt{675\times10^4}=100\sqrt{675}=100\sqrt{225\times3}=100\times15\sqrt3=1500\sqrt3\approx2598.08\) sq m
5One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Rough sketch — rhombus with one diagonal (2d) twice the other (d).
Let the shorter diagonal = d, so the longer diagonal = 2d. Area of rhombus \(=\tfrac12 d_1 d_2\):
\(\tfrac12 (d)(2d)=128 \Rightarrow d^2=128 \Rightarrow d=\sqrt{128}=8\sqrt2\approx11.31\) cm
6ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (ΔPCD):area (ΔQCD)?
Rough sketch — parallelogram ABCD with P, Q on side AB.
Both triangles PCD and QCD share the same base CD. Since P and Q both lie on side AB, which is parallel to CD (opposite side of the parallelogram), both P and Q are at the same perpendicular distance from line CD — namely, the height of the parallelogram.
Since both triangles have the same base and the same height, they have equal area.
7O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Rough sketch — parallelogram PQRS with O on diagonal PR.
Given: PQRS is a parallelogram, and O is any point on diagonal PR.
To Prove: Area(ΔPSO) = Area(ΔPQO)
Proof:
The diagonal PR divides parallelogram PQRS into two triangles, PQR and PSR, of equal area (a diagonal always bisects a parallelogram's area).
Let \(h_1\) = perpendicular distance from Q to line PR, and \(h_2\) = perpendicular distance from S to line PR.
Since area(PQR) \(=\tfrac12 \times PR \times h_1\) and area(PSR)\(=\tfrac12\times PR\times h_2\) are equal, we get \(h_1=h_2\).
Since O lies on PR, triangles PQO and PSO share the same base PO. Their heights are exactly \(h_1\) and \(h_2\) respectively (since Q and S maintain the same perpendicular distance to line PR regardless of where O sits on it).
Area(PQO) \(=\tfrac12\times PO\times h_1\)
Area(PSO)\(=\tfrac12\times PO\times h_2\)
Since \(h_1=h_2\), these areas are equal.
Hence proved: Area(ΔPSO) = Area(ΔPQO).
8If the mid-points of the sides of a 4-gon are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.
Rough sketch — 4-gon ABCD with midpoints P, Q, R, S forming the inner (Varignon) parallelogram.
Given: ABCD is a 4-gon; P, Q, R, S are the midpoints of AB, BC, CD, DA respectively.
To Prove: Area(parallelogram PQRS) = ½ × Area(ABCD)
Proof:
Draw diagonal BD. This splits ABCD into triangles ABD and CBD, with
area(ABD) + area(CBD) = area(ABCD)
In triangle ABD, P and S are midpoints of AB and AD, so triangle APS is similar to triangle ABD with ratio \(\tfrac12\) (matching sides AP:AB = AS:AD = 1:2).
So area(APS) \(=\left(\tfrac12\right)^2\)area(ABD)\(=\tfrac14\)area(ABD)
Similarly, in triangle CBD, Q and R are midpoints of CB and CD, so
area(CQR) \(=\tfrac14\)area(CBD)
Adding Steps 2 and 3:
area(APS) + area(CQR) \(=\tfrac14[\text{area(ABD)}+\text{area(CBD)}]=\tfrac14\)area(ABCD)
By the same argument using diagonal AC instead (splitting ABCD into ABC and ACD):
area(BPQ) + area(DRS) \(=\tfrac14\)area(ABCD)
The 4 corner triangles (APS, BPQ, CQR, DRS) together make up \(\tfrac14+\tfrac14=\tfrac12\) of area(ABCD). What remains — the parallelogram PQRS — must be the other half:
area(PQRS) \(=\) area(ABCD) \(- \tfrac12\)area(ABCD) \(= \tfrac12\)area(ABCD)
Hence proved: Area of parallelogram PQRS = ½ × Area of the 4-gon ABCD.
9In ΔABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).
Fig. 6.32 — Median AD, with P any point along it.
Given: In ΔABC, D is the midpoint of BC, AD is the median, and P is any point on AD.
To Prove: Area(ΔABP) = Area(ΔACP)
Proof:
Since AD is a median (D is the midpoint of BC), area(ABD) = area(ACD) — this is the "median property" theorem proved earlier in the chapter (both triangles have equal bases BD = DC and the same height from A).
Consider triangles PBD and PCD. They share vertex P, and their bases BD and DC are equal and lie along the same line BC — so they have the same height (the perpendicular distance from P to line BC) and the same base length.
Hence area(PBD) = area(PCD).
Subtracting:
area(ABP) = area(ABD) − area(PBD)
area(ACP) = area(ACD) − area(PCD)
Since area(ABD) = area(ACD) (shown above) and area(PBD) = area(PCD) (shown above), it follows that area(ABP) = area(ACP).
Hence proved: Area(ΔABP) = Area(ΔACP).
10Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB and ΔPCD) and the green region (ΔPBC and ΔPDA)?
Fig. 6.33 — Square ABCD with interior point P joined to all four vertices.
Given: ABCD is a square, and P is a point inside it; PA, PB, PC, PD are joined.
To Find: Ratio of area(ΔPAB)+area(ΔPCD) to area(ΔPBC)+area(ΔPDA)
Solution:
Let the square have side length a. Drop perpendiculars from P to sides AB and CD (a pair of opposite, parallel sides); call these distances \(h_1\) and \(h_2\).
Since AB and CD are a apart, \(h_1+h_2=a\).
Area(PAB) + Area(PCD) \(=\tfrac12 a h_1 + \tfrac12 a h_2 = \tfrac12 a(h_1+h_2)=\tfrac12 a \times a = \tfrac12 a^2\)
— exactly half the square's area, regardless of where P is.
Since the total area is \(a^2\), the remaining pair, Area(PBC)+Area(PDA), must also equal \(\tfrac12 a^2\).
11In ΔABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined (Fig. 6.34). Prove that Area (ΔBPQ) = ½ Area (ΔABC).
Fig. 6.34 — D is midpoint of AB; CQ is drawn parallel to PD.
Given: In ΔABC, D is the midpoint of AB; P is any point on BC; Q is on AB such that CQ ∥ PD.
To Prove: Area(ΔBPQ) = ½ Area(ΔABC)
Construction: Join CD.
Proof:
Since D is the midpoint of AB, CD is a median of ΔABC, so by the median property:
area(BDC) \(=\tfrac12\)area(ABC) ...(i)
Since CQ ∥ PD (given), triangles DPC and DPQ lie on the same base DP and between the same pair of parallels (DP and CQ), so they have equal area:
area(DPC) = area(DPQ) ...(ii)
Since D, B, and Q all lie on line AB, with D between B and Q, the segment PD splits triangle BPQ into two smaller triangles: BPD and DPQ. So:
area(BPQ) = area(BPD) + area(DPQ)
= area(BPD) + area(DPC) [using (ii)]
= area(BDC) [since cevian DP splits triangle BDC into BPD and DPC, as P lies on BC]
= \(\tfrac12\)area(ABC) [using (i)]
Hence proved: Area(ΔBPQ) = ½ Area(ΔABC).
Think and Reflect
TRWhy 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 circular shape?
Practical reasons: A circular shape encloses the maximum possible area for a given perimeter (of all shapes with the same boundary length, the circle has the largest area) — making it ideal for storage structures like granaries, pots, and towers, where you want maximum capacity for the material used to build the wall. Circular wheels, pots, and cylindrical towers are also easy to construct with a simple rope and peg, and circular objects can roll smoothly (wheels) or be turned uniformly (pottery wheels).
Non-practical (aesthetic/symbolic) reasons: Circles have also held deep symbolic and aesthetic significance across cultures — representing wholeness, the cycle of time (sun, moon, seasons), infinity, and unity, and appearing widely in art, religious architecture (domes, mandalas, rose windows), and design purely for their pleasing symmetry.
Uses: wheels and gears, pots and storage silos, coins, clock faces, astronomical instruments, domes and arches in architecture, and decorative motifs in art and textiles.
Exercise Set 6.3
1Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Rough sketch — sector of radius 7 cm, angle 60°.
Area \(=\pi r^2 \times \dfrac{\theta^\circ}{360^\circ}=\dfrac{22}{7}\times49\times\dfrac{60}{360}\)
\(\dfrac{22}{7}\times49=154\); Area \(=154\times\dfrac16=\dfrac{154}{6}=\dfrac{77}{3}\approx25.67\) sq cm
2Find the area of a quadrant of a circle whose circumference is 44 cm.
Rough sketch — quadrant (90° sector) of the circle.
\(C=2\pi r=44 \Rightarrow r=7\) cm (same computation as Ex 6.1, Q1).
Quadrant area \(=\tfrac14\pi r^2=\tfrac14\times\dfrac{22}{7}\times49=\tfrac14\times154=38.5\) sq cm
3The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Rough sketch — the minute hand sweeping through 60° in 10 minutes.
In 60 minutes the minute hand sweeps a full 360°, so in 10 minutes it sweeps \(\dfrac{10}{60}\times360^\circ=60^\circ\).
The minute hand's length (7 cm) is the radius. Area swept \(=\pi r^2\times\dfrac{60}{360}=\dfrac{22}{7}\times49\times\dfrac16=\dfrac{154}{6}=\dfrac{77}{3}\approx25.67\) sq cm
4A 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 π ≈ 3.14.)
Rough sketch — the 90° minor sector (major sector is the remaining 270°).
(i) Minor sector (90°): Area \(=\pi r^2\times\dfrac{90}{360}=3.14\times100\times0.25=78.5\) sq cm
(ii) Major sector (270°): Area \(=3.14\times100\times0.75=235.5\) sq cm
Check: \(78.5+235.5=314=\pi r^2\) ✓ (matches the full circle's area).
5A 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 π ≈ 3.14, √3 ≈ 1.73.)
Rough sketch — minor segment (teal) cut off by the chord, inside the 60° sector.
Minor sector (60°): Area \(=3.14\times225\times\dfrac16=117.75\) sq cm
Since the two radii are equal (15 cm each) and the included angle is 60°, the triangle formed is equilateral with side 15 cm:
Triangle area \(=\dfrac{\sqrt3}{4}(15)^2=\dfrac{1.73}{4}\times225=97.3125\) sq cm
Minor segment = minor sector − triangle \(=117.75-97.3125=20.4375\approx20.44\) sq cm
Major segment = full circle area − minor segment \(=3.14\times225-20.4375=706.5-20.4375=686.0625\approx686.06\) sq cm
Major segment ≈ 686.06 sq cm
6A 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.
Rough sketch — one wiper's sweep, radius 28 cm, angle 120°.
Area swept by one wiper (a sector, radius = blade length = 28 cm, angle 120°):
\(=\dfrac{22}{7}\times28^2\times\dfrac{120}{360}=\dfrac{22}{7}\times784\times\dfrac13\)
\(\dfrac{22}{7}\times784=22\times112=2464\); one wiper \(=\dfrac{2464}{3}\approx821.33\) sq cm
Total for 2 wipers (no overlap) \(=2\times\dfrac{2464}{3}=\dfrac{4928}{3}\approx1642.67\) sq cm
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 equal to r²(π/6 − √3/4).
Rough sketch — minor segment cut off by a chord subtending 60° at the centre.
Minor sector area (60°) \(=\pi r^2\times\dfrac{60}{360}=\dfrac{\pi r^2}{6}\)
Since the two radii (length r) enclose a 60° angle, the triangle is equilateral with side r; its area \(=\dfrac{\sqrt3}{4}r^2\)
Minor segment = minor sector − triangle \(=\dfrac{\pi r^2}{6}-\dfrac{\sqrt3}{4}r^2 = r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)\)
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 equal to 3√3/(4π) ≈ 0.413.
Rough sketch — equilateral triangle inscribed in a circle of radius r.
For an equilateral triangle inscribed in a circle of radius r, the circumradius formula gives \(r=\dfrac{a}{\sqrt3}\) (where a is the side length), so \(a=r\sqrt3\).
Area of triangle \(=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}(r\sqrt3)^2=\dfrac{\sqrt3}{4}(3r^2)=\dfrac{3\sqrt3}{4}r^2\)
Area of circle \(=\pi r^2\)
Ratio \(=\dfrac{\frac{3\sqrt3}{4}r^2}{\pi r^2}=\dfrac{3\sqrt3}{4\pi}\approx0.413\)
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 equal to 2/π ≈ 0.637.
Rough sketch — square inscribed in a circle of radius r.
For a square inscribed in a circle of radius r, the diagonal of the square equals the diameter of the circle: diagonal \(=2r\).
If the side of the square is s, then \(s\sqrt2=2r \Rightarrow s=\dfrac{2r}{\sqrt2}=r\sqrt2\)
Area of square \(=s^2=(r\sqrt2)^2=2r^2\)
Ratio \(=\dfrac{2r^2}{\pi r^2}=\dfrac{2}{\pi}\approx0.637\)
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 equal to 3√3/(2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
Rough sketch — regular hexagon inscribed in a circle of radius r.
A regular hexagon inscribed in a circle of radius r can be divided into 6 equilateral triangles, each with side length equal to r (the circumradius of a regular hexagon equals its own side length).
Area of hexagon \(=6\times\dfrac{\sqrt3}{4}r^2=\dfrac{6\sqrt3}{4}r^2=\dfrac{3\sqrt3}{2}r^2\)
Ratio \(=\dfrac{\frac{3\sqrt3}{2}r^2}{\pi r^2}=\dfrac{3\sqrt3}{2\pi}\approx0.827\)
Why exactly twice Q8's answer? The equilateral triangle in Q8 (inscribed in the same circle) can be formed by joining every other vertex of this same regular hexagon — a well-known fact is that this "alternate-vertex" triangle has exactly half the area of the hexagon it's inscribed in. Since Q8's ratio (triangle:circle) is half of Q10's ratio (hexagon:circle), Q10's answer is exactly double Q8's.
It is exactly twice the Q8 answer, because the equilateral triangle of Q8 is formed by alternating vertices of this same hexagon, and such a triangle always has exactly half the hexagon's area.
End-of-Chapter Exercises
1Draw figures corresponding to the identities (a+b)(a−b) = a²−b² and (a+b+c)² = a²+b²+c²+2ab+2bc+2ca.
(a+b)(a−b) = a²−b²: Start with a big square of side a. Cut a small square of side b from one corner. The remaining L-shaped region has area \(a^2-b^2\). This same L-shape can be cut into 2 rectangular strips and rearranged into a single rectangle of dimensions \((a+b)\times(a-b)\), showing the two expressions represent the same area.
Big square (side a) minus small corner square (side b) = the L-shaped region, area a²−b².
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca: Draw a big square of side \((a+b+c)\), and divide each side into 3 segments of length a, b, c. This creates a 3×3 grid of 9 regions: three squares (\(a^2, b^2, c^2\)) along the diagonal, and six rectangles (\(ab\) appearing twice, \(bc\) twice, \(ca\) twice) filling in the rest — matching every term in the expansion.
2An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Rough sketch — isosceles triangle, equal sides 15 cm, base 10 cm.
Base \(=40-15-15=10\) cm.
Height \(=\sqrt{15^2-5^2}=\sqrt{225-25}=\sqrt{200}=10\sqrt2\) cm
Area \(=\tfrac12\times10\times10\sqrt2=50\sqrt2\approx70.71\) sq cm
3An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Rough sketch — isosceles triangle, base 10 cm, area 60 cm².
Area \(=\tfrac12\times\text{base}\times\text{height}=60 \Rightarrow \tfrac12\times10\times h=60 \Rightarrow h=12\) cm
Equal side \(=\sqrt{(\text{base}/2)^2+h^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\) cm
4The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Rough sketch — right triangle, one leg 12 cm, area 54 cm².
Area \(=\tfrac12\times12\times(\text{other leg})=54 \Rightarrow \text{other leg}=\dfrac{54\times2}{12}=9\) cm
Hypotenuse \(=\sqrt{12^2+9^2}=\sqrt{144+81}=\sqrt{225}=15\) cm
Perimeter \(=12+9+15=36\) cm
5The sides of a triangle are in the ratio 2:3:4, and its perimeter is 45 cm. Find its area.
Rough sketch — triangle with sides in ratio 2:3:4.
Sides \(=2k,3k,4k\). Sum \(=9k=45\Rightarrow k=5\). Sides = 10, 15, 20 cm.
\(s=\dfrac{45}{2}=22.5\)
Area \(=\sqrt{22.5(22.5-10)(22.5-15)(22.5-20)}=\sqrt{22.5\times12.5\times7.5\times2.5}\)
\(=\sqrt{5273.4375}=\dfrac{75\sqrt{15}}{4}\approx72.62\) sq cm
6The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Rough sketch — the 7-24-25 right triangle.
Method 1 (spotting the right angle): Check \(7^2+24^2=49+576=625=25^2\). So this is a right triangle with legs 7, 24 and hypotenuse 25.
Area \(=\tfrac12\times7\times24=84\) sq cm
Method 2 (Heron's formula): \(s=\dfrac{7+24+25}{2}=28\)
Area \(=\sqrt{28(28-7)(28-24)(28-25)}=\sqrt{28\times21\times4\times3}=\sqrt{7056}=84\) sq cm
7If 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.
Rough sketch — bicycle wheel, diameter 60 cm.
Distance per rotation = circumference \(=\pi d=\dfrac{22}{7}\times60=\dfrac{1320}{7}\) cm
Distance for 100 rotations \(=100\times\dfrac{1320}{7}=\dfrac{132000}{7}\approx18857.14\) cm \(\approx188.57\) m
8Find the area of a quadrant of a circle whose circumference is 66 cm.
Rough sketch — quadrant of the circle.
\(C=2\pi r=66 \Rightarrow r=\dfrac{66\times7}{44}=\dfrac{462}{44}=10.5\) cm
Quadrant area \(=\tfrac14\pi r^2=\tfrac14\times\dfrac{22}{7}\times(10.5)^2=\tfrac14\times\dfrac{22}{7}\times110.25=\tfrac14\times346.5=86.625\) sq cm
9The 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.
Rough sketch — car wheel, outer radius 28 cm.
Distance per turn = circumference \(=2\pi r=2\times\dfrac{22}{7}\times28=176\) cm
1 km \(=100{,}000\) cm. Number of turns \(=\dfrac{100000}{176}\approx568.18\)
10*Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Two rectangles with the same area and perimeter.
Yes — surprisingly, for rectangles specifically, this always forces them to be congruent.
If a rectangle has sides \(x, y\), then its perimeter fixes the sum \(x+y=S\) and its area fixes the product \(xy=P\). By Vieta's formulas, \(x\) and \(y\) are exactly the two roots of the quadratic equation \(t^2-St+P=0\).
A quadratic equation has at most 2 roots — so if a second rectangle has sides \(x', y'\) with the same sum \(S\) and same product \(P\), then \(x', y'\) must satisfy the exact same equation \(t^2-St+P=0\), meaning \(\{x',y'\}\) is the very same pair of numbers as \(\{x,y\}\) (just possibly swapped, i.e. length and width exchanged).
11Using the fact that area of a parallelogram is base × height, show using Fig. 6.42 that the area of a trapezium is half the sum of the parallel sides × height, i.e., ½(a+b)h.
Fig. 6.42 — Trapezium with parallel sides a, b and height h.
Given: A trapezium with parallel sides a and b, and height h (Fig. 6.42).
To Prove: Area of trapezium = ½(a+b)h
Construction: Take two identical copies of the trapezium. Rotate one copy by 180° and join it to the other along one of the slanted (non-parallel) sides.
Proof:
This construction creates a parallelogram whose base is \(a+b\) (the two parallel sides of the trapezium, now placed end to end) and whose height is still h.
Area of this parallelogram \(=\text{base}\times\text{height}=(a+b)h\)
Since this parallelogram is made of exactly 2 copies of the original trapezium, the area of one trapezium is half of this:
Area of trapezium \(=\tfrac12(a+b)h\)
Hence proved: Area of trapezium = ½(a+b)h.
12By dividing a trapezium into two triangles show that its area is half the sum of the parallel sides multiplied by the height.
Trapezium split by a diagonal into two triangles.
Given: A trapezium with parallel sides a and b, and height h.
To Prove: Area of trapezium = ½(a+b)h
Construction: Draw a diagonal of the trapezium, splitting it into two triangles.
Proof:
One triangle has base = the parallel side of length a and height h; the other has base = the parallel side of length b and height h (both triangles share the same height h, the perpendicular distance between the two parallel sides).
Area of first triangle \(=\tfrac12 a h\)
Area of second triangle \(=\tfrac12 b h\)
Total area \(=\tfrac12 ah+\tfrac12 bh=\tfrac12(a+b)h\)
Hence proved: Area of trapezium = ½(a+b)h.
13Show 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?
Two trapeziums (one rotated 180°) joined to form a parallelogram.
Construction: Take two identical copies of a trapezium with parallel sides a and b and height h. Rotate one copy by 180° (turning it upside-down) and place it next to the other, joining them along one of the slanted sides.
Proof:
The two trapezium shapes interlock — the parallel side of length a on one copy lines up with the parallel side of length b on the rotated copy (and vice versa on the other end), so together they form one continuous base of length \(a+b\).
The result is a parallelogram with base \((a+b)\) and height h.
Its area \(=(a+b)h\).
Since this parallelogram is exactly 2 trapeziums joined together, one trapezium's area is half of this:
Area of trapezium \(=\tfrac12(a+b)h\)
14Show that the area of a kite is half the product of its diagonals. Show this:
(i) using algebra, and
(ii) using geometry.
Fig. — Kite ABCD with perpendicular diagonals AC and BD meeting at O.
Given: A kite ABCD, with diagonal AC being the axis of symmetry (so AC ⊥ BD, and AC bisects BD at their intersection point O, i.e. \(OB=OD\)). Let \(AC=p\), \(BD=q\).
To Prove: Area of kite = ½ × (product of diagonals) = ½pq
Proof (i) — using algebra:
Since the diagonals are perpendicular, split the kite into 4 right triangles by both diagonals: AOB, BOC, COD, DOA, where \(AO+OC=p\) and \(OB=OD=\dfrac{q}{2}\).
Total area \(=\tfrac12(AO)(OB)+\tfrac12(OC)(OB)+\tfrac12(OC)(OD)+\tfrac12(AO)(OD)\)
\(=\tfrac12(OB)(AO+OC)+\tfrac12(OD)(OC+AO)=\tfrac12(OB+OD)(AO+OC)=\tfrac12(BD)(AC)=\tfrac12 pq\)
Proof (ii) — using geometry:
Diagonal AC splits the kite into triangles ABC and ADC, both sharing base AC. Their heights (the perpendicular distances from B and D to line AC) are OB and OD respectively.
Area(ABC) + Area(ADC) \(=\tfrac12(AC)(OB)+\tfrac12(AC)(OD)=\tfrac12(AC)(OB+OD)=\tfrac12(AC)(BD)=\tfrac12 pq\)
Hence proved (both ways): Area of kite = ½pq.
15(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?
Rectangle ABCD (a×b) alongside rectangle PQRS (2a×2b), split into 4.
Given: Rectangle ABCD has sides a, b; rectangle PQRS has sides 2a, 2b.
To Show: Area(PQRS) = 4 × Area(ABCD); and whether 4 copies of ABCD tile PQRS.
Area(ABCD) \(=ab\)
Area(PQRS) \(=2a\times2b=4ab=4\times\)Area(ABCD)
Ratio = 4:1
Does it fit? Yes — cut PQRS exactly in half along both its length and width (through the midpoints of its sides). This divides PQRS into exactly 4 rectangles, each of size \(a\times b\) — identical to ABCD. So 4 copies of ABCD do fit perfectly into PQRS.
15(ii)ΔABC has sides a, b, c, and ΔPQR has sides 2a, 2b, 2c. Show that ΔPQR has 4 times the area of ΔABC. Does this mean that 4 copies of ΔABC will fit into ΔPQR?
ΔPQR (sides doubled) split by its midsegments into 4 triangles congruent to ΔABC.
Given: ΔABC has sides a, b, c; ΔPQR has sides 2a, 2b, 2c.
To Show: Area(PQR) = 4 × Area(ABC); and whether 4 copies of ABC tile PQR.
By Heron's formula, if ΔABC has semi-perimeter \(s=\tfrac12(a+b+c)\), then ΔPQR (with all sides doubled) has semi-perimeter \(2s\).
Area(PQR) \(=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{2s\times2(s-a)\times2(s-b)\times2(s-c)}\)
\(=\sqrt{16\times s(s-a)(s-b)(s-c)}=4\sqrt{s(s-a)(s-b)(s-c)}=4\times\)Area(ABC)
Does it fit? Yes — join the midpoints of the sides of ΔPQR. By the midsegment theorem, this divides ΔPQR into exactly 4 smaller triangles, each congruent to ΔABC (this is the classic "medial triangle" dissection). So 4 copies of ΔABC do fit perfectly into ΔPQR.
15(iii)ΔABC has sides a, b, c, and ΔPQR has sides 3a, 3b, 3c. Show that ΔPQR has 9 times the area of ΔABC. Does this mean that 9 copies of ΔABC will fit into ΔPQR?
ΔPQR (sides tripled) split by a triangular grid into 9 triangles congruent to ΔABC.
Given: ΔABC has sides a, b, c; ΔPQR has sides 3a, 3b, 3c.
To Show: Area(PQR) = 9 × Area(ABC); and whether 9 copies of ABC tile PQR.
By the same Heron's-formula scaling argument as in 15(ii), with semi-perimeter scaled by 3:
Area(PQR) \(=9\times\)Area(ABC) (each factor of 3 inside the square root contributes, and \(\sqrt{3\times3\times3\times3}=9\)).
Does it fit? Yes — divide each side of ΔPQR into 3 equal parts and draw lines through these points parallel to the sides. This creates a triangular grid that divides ΔPQR into exactly 9 smaller triangles, each congruent to ΔABC.
16*What fraction of the triangle is shaded (Fig. 6.43)? What fraction of the square is shaded (Fig. 6.44)?
Fig. 6.43 — cevians to the ⅓ points create a small inner triangle.
Fig. 6.44 — joining the 1:2 division points creates a rotated inner square.
Reading the tick marks as dividing each side in the ratio 1:2 (a single tick vs. a double tick), matching the classic versions of these two puzzles:
Fig. 6.43 (triangle): When each vertex of a triangle is joined to the point 1/3 of the way along the opposite side (in the same rotational order), the three cevians enclose a small central triangle whose area is exactly 1/7 of the original triangle — this is a famous classical result, provable using mass-point geometry or coordinate geometry (assign coordinates, find the 3 cevian intersection points, and compute the inner triangle's area via the shoelace formula).
Fig. 6.44 (square): When each side of a square is divided in the ratio 1:2 and these division points are joined in order, the resulting inner (rotated) square has area exactly 5/9 of the original square — provable by coordinate geometry: for a unit square with vertices (0,0),(1,0),(1,1),(0,1), joining the points at 1/3 along each side (in rotational order) gives an inner square of side \(\dfrac{\sqrt5}{3}\), hence area \(\left(\dfrac{\sqrt5}{3}\right)^2=\dfrac{5}{9}\).
17What fraction of the rectangle is covered by the circles (Fig. 6.45: 3 circles; Fig. 6.46: 4 circles)?
Fig. 6.45 — 3 circles in a rectangle.
Fig. 6.46 — 4 circles in a rectangle.
In both figures, identical circles are packed in a single row, each touching the top and bottom of the rectangle (so the rectangle's height = the circle's diameter, \(2r\)) and each touching its neighbours (so each circle "owns" a width of \(2r\) along the rectangle).
For n circles: rectangle dimensions \(= (2r) \times (n\times2r)\), so rectangle area \(=4nr^2\). Total circle area \(=n\times\pi r^2\).
Fraction covered \(=\dfrac{n\pi r^2}{4nr^2}=\dfrac{\pi}{4}\) — the n's cancel out completely!
Fig. 6.45 (3 circles): fraction \(=\dfrac{\pi}{4}\approx0.785\) (78.5%). Fig. 6.46 (4 circles): fraction \(=\dfrac{\pi}{4}\approx0.785\) (78.5%) — exactly the same fraction.
18Use 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!
A row of n identical circles — each occupies the same fixed π/4 fraction of its own 2r×2r cell.
Conjecture: No matter how many identical circles are packed in a single row inside a rectangle (each touching top and bottom, and touching its neighbours), the fraction of the rectangle's area covered by the circles is always exactly \(\dfrac{\pi}{4}\), regardless of the number of circles.
Testing: For n = 10, 20, or 50 circles, using the same formula as Q17: rectangle area \(=4nr^2\), total circle area \(=n\pi r^2\), fraction \(=\dfrac{n\pi r^2}{4nr^2}=\dfrac{\pi}{4}\) in every case — confirming the conjecture holds regardless of n.
Proof: Each circle, together with the small square section of the rectangle it occupies (a \(2r\times2r\) square), always has the same fixed ratio \(\dfrac{\pi r^2}{(2r)^2}=\dfrac{\pi}{4}\) of circle-area to square-area. Since the whole rectangle is just n copies of this same \(2r\times2r\) square placed side by side, and each copy has exactly the same π/4 coverage ratio, the overall rectangle must also have exactly π/4 coverage.
19*The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.
The solved arrangement: 3 rectangles (4×2 cm, horizontal) on top, 6 rectangles (2×4 cm, vertical) on the bottom — both rows span the same 12 cm width.
Reading Fig. 6.47 as: 3 small rectangles placed side-by-side in one row (in their "long" orientation), and 6 small rectangles placed side-by-side in the other row (rotated 90°, in their "short" orientation) — with both rows having the same total width.
Let the small rectangle have long side L and short side S. The top row (3 rectangles, long side horizontal) has width \(3L\); the bottom row (6 rectangles, short side horizontal) has width \(6S\). Since both rows form the same big rectangle, their widths must match:
\(3L=6S \Rightarrow L=2S\)
Total area of all 9 small rectangles \(=9\times(L\times S)=72 \Rightarrow LS=8\)
Substituting \(L=2S\): \(2S\times S=8 \Rightarrow S^2=4 \Rightarrow S=2\) cm, so \(L=4\) cm.
Check: top row width \(=3(4)=12\);
bottom row width\(=6(2)=12\) ✓ matches.
Big rectangle height \(=S+L=2+4=6\);
big rectangle area \(=12\times6=72\) ✓ matches the given area exactly.
Perimeter of one small rectangle \(=2(L+S)=2(4+2)=12\) cm
20*Lines from a vertex to the points of trisection of the opposite side (Fig. 6.48). Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle and rearranging the pieces to cover the red triangle.
Fig. 6.48 — Triangle ABC with base trisected at D, E; blue triangle ABD and red triangle AEC each have base ⅓ BC.
Given: Triangle ABC, with base BC trisected at points D and E (so \(BD=DE=EC=\tfrac13 BC\)); lines AD (blue) and AE (red) are drawn.
To Prove: Area(ABD) = Area(AEC); and find a way to rearrange the pieces of one to cover the other.
Proof:
Both triangles ABD and AEC share the same height (the perpendicular distance from A to line BC).
Their bases (BD and EC) are equal in length, each being \(\tfrac13\) of BC.
Since Area \(=\tfrac12\times\text{base}\times\text{height}\), and both the base and height match, the two triangles have equal area (each equal to \(\tfrac13\) of the whole triangle ABC).
Hence proved: Area(ABD) = Area(AEC).
Rearranging the pieces: Cut triangle ABD along the segment joining A to the midpoint of BD. This splits it into two smaller triangles.
Rotating one of these two pieces by 180° about that midpoint rearranges them into a parallelogram with base \(\tfrac12 BD\) and the same height as before.
Repeating the identical construction on triangle AEC (cutting along the line from A to the midpoint of EC) produces a congruent parallelogram, since \(BD=EC\) and the height is the same.
Because the two parallelograms are congruent, the pieces from triangle ABD can be rearranged to exactly cover triangle AEC.
21*The figure shows a quarter circle in a square (centre at one vertex, passing through two adjacent vertices) and two semicircles on two adjacent sides as diameters, creating shaded regions A and B. Show that A and B have equal area.
Fig. 6.49 — Quarter circle (blue) centred at O, radius a, and two semicircles (teal) on the sides meeting at O; A and B are the shaded regions exclusive to the semicircles and quarter circle respectively.
Given: A square of side a, with a quarter circle centred at one corner O (radius a, passing through the two adjacent vertices) and two semicircles drawn on the two sides meeting at O (each with diameter a, radius \(\dfrac{a}{2}\)).
To Prove: Area(A) = Area(B), where A and B are the shaded regions shown.
Proof:
Area of quarter circle \(=\tfrac14\pi a^2=\dfrac{\pi a^2}{4}\)
Area of each semicircle \(=\tfrac12\pi\left(\dfrac{a}{2}\right)^2=\dfrac{\pi a^2}{8}\)
So the sum of the two semicircles \(=2\times\dfrac{\pi a^2}{8}=\dfrac{\pi a^2}{4}\) — exactly equal to the quarter circle's area found above!
Let C be the region covered by both the quarter circle and the two semicircles (the overlap).
Region B is the part of the quarter circle not covered by either semicircle, so:
Area(quarter circle) = Area(C) + Area(B)
Region A is the part of the two semicircles not covered by the quarter circle, so:
Area(semicircle₁) + Area(semicircle₂) = Area(C) + Area(A)
Since Area(quarter circle) = Area(semicircle₁)+Area(semicircle₂) (both equal \(\dfrac{\pi a^2}{4}\), shown above):
Area(C) + Area(B) = Area(C) + Area(A)
So Area(A) = Area(B)
Hence proved: Area(A) = Area(B).
22*Four semicircles have been drawn within a square of side 2 units, centred at the midpoints of the sides, creating a 4-petalled flower. Find the perimeter and the area of this flower.
Fig. 6.50 — 4-petal flower formed by semicircles centred at the midpoints of a side-2 square.
Perimeter: Each petal's boundary is made of 2 quarter-circle arcs (90° each) from two adjacent semicircles (radius 1), so each petal's boundary length \(=\tfrac{\pi}{2}+\tfrac{\pi}{2}=\pi\). With 4 petals: total perimeter \(=4\pi\approx12.57\) units.
Area: Consider one petal, formed by the overlap of two semicircles of radius 1 whose centres are a distance \(\sqrt2\) apart (the two midpoints of adjacent sides of the unit square formed by quartering the big square). Using the standard circle-overlap ("lens") formula with \(r=1\), \(d=\sqrt2\):
Lens area \(=2r^2\cos^{-1}\!\left(\dfrac{d}{2r}\right)-\dfrac{d}{2}\sqrt{4r^2-d^2} = 2\cos^{-1}\!\left(\dfrac{1}{\sqrt2}\right)-\dfrac{\sqrt2}{2}\sqrt{4-2}\)
\(=2\times\dfrac{\pi}{4}-\dfrac{\sqrt2}{2}\times\sqrt2=\dfrac{\pi}{2}-1\)
Each petal has area \(\dfrac{\pi}{2}-1\). With 4 petals: total flower area \(=4\left(\dfrac{\pi}{2}-1\right)=2\pi-4\approx2.28\) sq units.
23*Two concentric circles have common centre O. A chord BC of the larger circle touches (is tangent to) the smaller circle at A. The length of BC is l. Show that the area of the region enclosed between the two circles is ¼πl².
Fig. 6.51 — Concentric circles with common centre O; chord BC of the larger circle is tangent to the smaller circle at A.
Given: Two concentric circles with common centre O (radii R and r, R > r). Chord BC of the larger circle is tangent to the smaller circle at A, and BC = l.
To Prove: Area of the region enclosed between the two circles \(=\tfrac14\pi l^2\)
Proof:
Since BC is tangent to the smaller circle at A, the radius OA is perpendicular to BC (a tangent is always perpendicular to the radius at the point of contact), and OA = r.
Since the perpendicular from the centre of a circle to a chord always bisects that chord, A is the midpoint of BC:
\(AB=AC=\dfrac{l}{2}\)
In right triangle OAB:
\(OB^2=OA^2+AB^2 \Rightarrow R^2=r^2+\left(\dfrac{l}{2}\right)^2 \Rightarrow R^2-r^2=\dfrac{l^2}{4}\)
Area of the region between the circles (the annulus):
\(=\pi R^2-\pi r^2=\pi(R^2-r^2)=\pi\times\dfrac{l^2}{4}=\dfrac14\pi l^2\)
Hence proved: Area of the annular region = ¼πl².
24*Semicircles have been drawn on all the sides of a right-angled triangle. Show that Area(A) + Area(B) = Area(C).
Given: A right-angled triangle with legs p, q and hypotenuse r (so \(p^2+q^2=r^2\) by the Baudhāyana–Pythagoras theorem). Semicircles are drawn outward on each leg, and a semicircle is drawn on the hypotenuse (passing through the right-angle vertex, by Thales' theorem). C is the triangle itself, and A, B are the two crescent-shaped "lunes" formed between each leg's outward semicircle and the hypotenuse semicircle's arc.
Fig. 6.52 — Semicircles on all three sides of a right triangle. A and B label the two crescent-shaped lunes; C labels the triangle itself.
To Prove: Area(A) + Area(B) = Area(C)
Proof:
Since the area of a semicircle is proportional to the square of its diameter, multiply the Pythagorean identity \(p^2+q^2=r^2\) by \(\dfrac{\pi}{8}\):
\(\dfrac{\pi p^2}{8}+\dfrac{\pi q^2}{8}=\dfrac{\pi r^2}{8}\)
i.e. Area(semicircle on p) + Area(semicircle on q) = Area(semicircle on r)
The semicircle on the hypotenuse (r) splits into the triangle C plus two circular segments (the parts beyond each leg).
Each leg's outward semicircle, in turn, consists of exactly its corresponding lune plus that same circular segment.
Substituting these decompositions into the semicircle-area identity above, and cancelling the two (equal, shared) segment areas from both sides, leaves exactly:
Lune(A) + Lune(B) = Area(C)
Hence proved: Area(A) + Area(B) = Area(C) — this is the famous "Lune of Hippocrates" theorem (c. 440 BCE), one of the earliest results in the history of geometry to show that a curved (lune-shaped) region can have an area exactly equal to a straight-edged shape.
25*Two circles pass through each other's centres (Fig. 6.53). Find the area of the region enclosed by the two circles, in terms of the common radius r.
Fig. 6.53 — Two congruent circles centred at A and B, each passing through the other's centre and intersecting at C and D; the shaded lens is the overlap region.
Given: Two congruent circles of radius r, each passing through the other's centre (Fig. 6.53), intersecting at points C and D.
To Find: The area enclosed by the two circles (the shaded lens/overlap region), in terms of r.
Solution:
This uses the same configuration as Example 1 in Section 6.5, where it was already shown that each dotted arc (bounding the central lens shape) is \(\tfrac13\) of its circle's circumference, corresponding to a central angle of \(120^\circ\) (since \(\angle CAD = \angle CBD = 120^\circ\), as derived from the two equilateral triangles ABC and ABD formed because all sides equal r).
The enclosed lens-shaped region is made of 2 circular segments (one from each circle), each cut off by the common chord CD, with central angle 120°.
Sector area (120° of radius r):
\(=\pi r^2\times\dfrac{120}{360}=\dfrac{\pi r^2}{3}\)
Triangle area (isosceles, two sides r, included angle 120°, e.g. triangle ACD with AC=AD=r):
\(=\tfrac12(r)(r)\sin(120^\circ)=\tfrac12 r^2\times\dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{4}r^2\)
One segment area = sector − triangle:
\(=\dfrac{\pi r^2}{3}-\dfrac{\sqrt3}{4}r^2\)
Total lens area (2 segments):
\(=2\left(\dfrac{\pi r^2}{3}-\dfrac{\sqrt3}{4}r^2\right)=\dfrac{2\pi r^2}{3}-\dfrac{\sqrt3}{2}r^2=r^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\right)\)
26*Three triangles A, B, C are formed within a rectangle by cevians from a common point. Show that the area of the rectangle is 2(A+C)(B+C)/C.
Illustrative sketch — rectangle with a point on one side joined to two opposite corners, forming three triangles A, B, C.
This result depends on the exact construction shown in the figure (which cevians are drawn, and from which points). The general strategy for this style of problem is as follows — set up coordinates for the rectangle and the point(s) defining the cevians, express each triangle's area (A, B, C) in terms of the rectangle's dimensions and the cevian point's position, and then verify the target identity algebraically.
As a guide to the method: label the rectangle's dimensions and the position of the dividing point using variables, write each of A, B, C as \(\tfrac12\times\text{base}\times\text{height}\) using those same variables, and substitute into the right-hand side \(\dfrac{2(A+C)(B+C)}{C}\) — the algebra should simplify to the rectangle's width × height once the correct construction (matching your figure) is used.
27*The figure shows two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.
Schematic (illustrative) — semicircle on AC (centre O) and a quarter circle centred at A passing through B and E. Points F and D mark plausible interior construction points (intersections of the diagonals AB, EC and EO shown dashed); their exact position in the original figure could not be confirmed, so treat this diagram as a guide to the overall structure rather than an exact reproduction.
This is another equal-area puzzle in the same family as Q21 and Q24 (Lune of Hippocrates), which this chapter uses repeatedly: two curved shapes built from circular arcs of related radii often turn out to have matching areas because of an underlying Pythagorean-style identity between the arc radii (semicircle/quarter-circle areas scale with the square of their radius, exactly like the Baudhāyana–Pythagoras theorem).
General method: identify the quarter circle, the semicircle, and the triangle in your figure;
express each one's area in terms of the shared lengths (e.g. if the triangle is right-angled and inscribed so that a semicircle's diameter is its hypotenuse, by Thales' theorem the semicircle passes through the right-angle vertex);
then, exactly as in Q21 and Q24, decompose the two shaded regions so that they differ from two DIFFERENT combinations of the same known-equal areas — leaving the two shaded regions equal once the common (overlapping) pieces are cancelled from both sides.
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