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.
\[ \text{Area of rectangle} = ab \qquad \text{Area of parallelogram} = bh \qquad \text{Area of triangle} = \tfrac12 bh \]
\[ \text{Heron's: } \sqrt{s(s-a)(s-b)(s-c)} \qquad \text{Circle: } \pi r^2 \qquad \text{Sector: } \pi r^2 \times \tfrac{\theta^\circ}{360^\circ} \]
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.
Yes, a smaller (200 m) track needs a proportionally smaller stagger, because the stagger scales directly with the radius of the track's curved sections.
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?
400 m track = 2 straight sections + 2 semicircular curves
Semicircular curves are found using the circle perimeter formula[C = 2πr]
Same formula also gives the extra distance (stagger) for outer lanes[difference of circumferences]
Both questions rely on the same idea — the perimeter (circumference) formula of a circle — since the curved parts of the track are semicircles.
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).
Lane 1 effective radius = 36.8 m[36.5 m + 0.3 m from inner border]
Lane width = 1.22 m[given]
Lane 2 effective radius \(=36.8+1.22=38.02\) m[add one lane width]
Difference in radius = 38.02 − 36.8 = 1.22 m[= one lane width]
Two semicircular curves per lane = 1 full circle
Stagger \(=2\pi(38.02)-2\pi(36.8)\)[difference of circumferences]
\(=2\pi(38.02-36.8)=2\pi(1.22)\)
\(=2\times3.1416\times1.22\approx7.67\) m[π ≈ 3.1416]
Lane 3 vs lane 2: same stagger, ≈7.67 m[radius always increases by the same 1.22 m]
Difference in radius = 1.22 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.
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.
\(C=2\pi r\)[circumference formula]
\(44=2\times\dfrac{22}{7}\times r\)[substituting C = 44]
\(44=\dfrac{44}{7}r\)
\(r=44\times\dfrac{7}{44}=7\)[dividing both sides by 44/7]
Radius = 7 cm.
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[exact — 44.0 cm to 3 s.f.]
(ii) \(C=2\times\dfrac{22}{7}\times10=\dfrac{440}{7}=62.857\ldots\)[C = 2πr]
≈ 62.9 cm[rounded to 3 s.f.]
(iii) \(C=2\times\dfrac{22}{7}\times12=\dfrac{528}{7}=75.428\ldots\)[C = 2πr]
≈ 75.4 cm[rounded to 3 s.f.]
(i) 44.0 cm (ii) 62.9 cm (iii) 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) \(l=2\pi r\times\dfrac{\theta}{360^\circ}\)[arc length formula]
Full circumference \(=2\times\dfrac{22}{7}\times3.5=22\) cm[C = 2πr]
Arc \(=22\times\dfrac{60}{360}=22\times\dfrac16=\dfrac{11}{3}\approx3.67\) cm[θ = 60°]
(ii) Full circumference \(=2\times\dfrac{22}{7}\times6.3=39.6\) m[C = 2πr]
Arc \(=39.6\times\dfrac{120}{360}=39.6\times\dfrac13=13.2\) m[θ = 120°]
(i) ≈ 3.67 cm (ii) 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 \(=2\pi r\times\dfrac{75}{360}\)[arc length formula]
Full circumference \(=2\times\dfrac{22}{7}\times14=88\) cm[C = 2πr]
Arc \(=88\times\dfrac{75}{360}=\dfrac{55}{3}\approx18.33\) cm[θ = 75°]
Perimeter of sector = arc + 2 radii[curved + 2 straight edges]
\(=\dfrac{55}{3}+2(14)=\dfrac{55}{3}+28\)
\(=\dfrac{55+84}{3}=\dfrac{139}{3}\approx46.33\) cm
Perimeter of sector ≈ 46.33 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: length 80 m, semicircular ends of radius 30 m[shape read from figure]
Straight sections \(=80-60=20\) m each, ×2 = 40 m[80 − diameter of ends]
2 semicircular ends = 1 full circle, radius 30 m[joining the two halves]
\(2\pi(30)=2\times\dfrac{22}{7}\times30=\dfrac{1320}{7}\approx188.57\) m[C = 2πr]
Total \(\approx40+188.57=228.57\) m[straight + curved]
(ii) Large semicircle (d = 12 cm) with small notch (d = 8 cm) cut out[shape read from figure]
Flat base exposed each side \(=(12-8)/2=2\) cm[gap left by the notch]
Perimeter = large arc + small arc + 2 flat bits[tracing the outline]
\(=\pi(6)+\pi(4)+2(2)=\dfrac{22}{7}(10)+4\)[radii 6 and 4]
\(=\dfrac{220}{7}+4\approx31.43+4=35.43\) cm
(iii) 4-lobed flower, side 10 cm (radius 5)[shape read from figure]
Perimeter = 4 semicircle arcs[one per side]
\(=4\times\pi(5)=4\times\dfrac{22}{7}\times5=\dfrac{440}{7}\approx62.86\) cm[πr per arc]
(iv) Equilateral triangle (side 12 cm) on a semicircle (d = 12 cm)[base replaced by the arc]
Perimeter = 2 triangle sides + semicircle arc[3rd side is the arc]
\(=2(12)+\pi(6)=24+\dfrac{132}{7}\approx24+18.86=42.86\) cm[radius 6]
(v) Same 4-lobed flower, side 14 cm (radius 7)[shape read from figure]
\(=4\times\pi(7)=4\times22=88\) cm[same method as (iii)]
(vi) Wavy border, 4 bumps along a 28 cm baseline[shape read from figure]
Each semicircle diameter \(=28/4=7\) cm (radius 3.5)[splitting the baseline]
\(=4\times\pi(3.5)=4\times11=44\) cm[πr per bump]
(vii) Right triangle, legs 6 cm & 8 cm[shape read from figure]
Hypotenuse \(=\sqrt{6^2+8^2}=10\) cm[Pythagoras]
Semicircles on hypotenuse (r=5) and 6 cm leg (r=3); 8 cm leg exposed[shape read from figure]
\(=8+\pi(5)+\pi(3)=8+\dfrac{176}{7}\approx8+25.14=33.14\) cm[straight leg + 2 arcs]
(viii) Large semicircle (d=12) with 3 small scallops (d=4 each) along its base[shape read from figure]
\(=\pi(6)+3\pi(2)=\pi(12)=12\times\dfrac{22}{7}=\dfrac{264}{7}\approx37.71\) cm[large arc + 3 small arcs]
(ix) S-curve: 2 semicircles (d=10 each) along a 20 cm baseline[shape read from figure]
\(=2\times\pi(5)=10\times\dfrac{22}{7}=\dfrac{220}{7}\approx31.43\) cm[2 arcs, radius 5]
(i) ≈228.57 m
(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
(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) 1 revolution = circumference of tyre[distance per revolution]
\(C=\pi d=\dfrac{22}{7}\times56=176\) cm[C = πd]
(ii) 10 km \(=10\times100000=1{,}000{,}000\) cm[1 km = 100000 cm]
Revolutions \(=\dfrac{1000000}{176}\approx5681.82\)[total distance ÷ distance per revolution]
(i) 176 cm per revolution (ii) ≈ 5681.82 revolutions (so the tyre completes 5681 full revolutions, with part of one more).
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: each semicircle has one side of the square as its diameter, so all four bulge inward and meet exactly at the centre.
(i) Square side 14 cm, arcs centred at midpoints of the sides[Fig. 6.15A]
Each side gives 1 semicircular arc (diameter = side = 14, radius 7)
Total \(=4\times\pi(7)=4\times\dfrac{22}{7}\times7=88\) cm[4 semicircular arcs]
Fig. 6.15B — 6-petal flower: each arc is centred at a vertex, radius = side, connecting that vertex's two neighbours through the centre.
(ii) Hexagon side 42 cm, arcs centred at the vertices[Fig. 6.15B]
Vertex-to-centre distance = side = 42 cm[property of a regular hexagon]
So a vertex-centred arc of radius 42 cm passes exactly through the centre
Each arc sweeps 120° (not 60°) — the centre bisects it into two 60° halves
Arc length \(=2\pi(42)\times\dfrac{120}{360}=\dfrac{2\times\frac{22}{7}\times42}{3}=88\) cm[arc length formula]
Total \(=6\times88=528\) cm[6 such arcs, one per vertex]
Cross-check: each petal = 2 × 60° half-arc = 2×44 = 88 cm; 6×88=528 ✓
(i) 88 cm (ii) 528 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).
\(C=2\pi r\)[circumference formula]
C is directly proportional to r[2π is a constant]
\(C_1:C_2=5:4 \Rightarrow r_1:r_2=5:4\)[2π cancels out in the ratio]
Ratio of radii = 5:4.
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.
Extend DA beyond A, mark A' with AA' = a matching extension D'D[Fig. 6.19 construction]
New parallelogram A'BCD' has the same base and height as ABCD
△CDD' ≅ △BAA'[parallel sides, equal extensions]
Removing △BAA' and adding congruent △CDD' leaves area unchanged[congruent triangles, equal area]
Perpendicular from C now lands within the new base A'D'[base extended far enough]
Repeat the shift as many times as needed for very "thin" parallelograms[general fix]
We repeatedly extend the base by matching amounts on either end (creating congruent corner triangles that cancel out in area) until the perpendicular foot lands on the new base — this preserves the parallelogram's area while fixing the construction.
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 — side lengths alone are not enough[unlike a rectangle]
Area = \(bh\), and base b is a given side length[area formula]
But height h depends on the angle between adjacent sides[not fixed by side lengths]
A parallelogram is "hinge-able": sides stay fixed, angle can change[like a pushed picture frame]
Decreasing the angle → h → 0 → area → 0[same side lengths throughout]
Area is largest (= ab) when the angle is 90°[parallelogram becomes a rectangle]
No — the same two side lengths can form parallelograms of many different areas, since the height (and hence the area) also depends on the angle between the sides, which the side lengths alone do not fix.
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.
Foot of perpendicular from E falls outside segment FG (beyond F)[since ∠EFG is obtuse]
Extend line FG beyond F; drop perpendicular from E, meeting it at H[fix: extend the base]
Area(△EFG) = Area(△EHG) − Area(△EHF)[larger triangle minus the extra piece]
Both use the same height EH, and HG = HF + FG
Subtraction simplifies to \(\tfrac12\times FG\times EH\)
Same formula \(\tfrac12 bh\) still holds, with base FG, height EH
For an obtuse triangle, extend the base until the perpendicular from the opposite vertex meets it; the required triangle's area is then (area of the larger right-angled triangle) minus (area of the small extra right-angled triangle), which still simplifies to ½ × base × 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[BD = DC and same height, from A to BC]
Cut ΔABD along the line from A to the midpoint of BD[splits it into 2 smaller triangles]
Rotate one piece 180° about that midpoint[rearranges into a parallelogram]
Parallelogram: base \(\tfrac12 BD\), same height as ΔABD[result of the rotation]
Same construction on ΔACD gives a congruent parallelogram[since BD = DC, same height]
Congruent parallelograms ⇒ pieces of ΔABD can cover ΔACD
Yes — because the two triangles share the same base length and height, cutting each along the median to its base and rotating one piece 180° turns both into congruent parallelograms, showing the pieces of ΔABD can be rearranged to exactly cover ΔACD.
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 — Yes, always possible for equal-area polygons[Bolyai–Gerwien theorem]
(1) Square & rectangle: cut rectangle into strips, rearrange[classic "staircase" dissection]
(2) Two differently-shaped triangles, equal area[dissect via a common intermediate rectangle]
Two triangles with the same base b and height h (hence equal area) but different shapes — a simple example for test case (2).
(3) Triangle & square, equal area[triangle → rectangle → square, via (1)]
Conjecture: any 2 equal-area polygons can be dissected into each other[finitely many straight cuts]
Part 2 — Let sides be \(x\) and \(20-x\)[since 2(x+(20−x))=40]
x ranges over (0, 20) → infinitely many rectangles[sides need not be integers]
Area \(A(x)=x(20-x)=20x-x^2\)[area = product of sides]
Downward parabola, maximised at x = 10 (the vertex)[standard result for A(x)]
Largest area = 100 sq. units, at the square (10 × 10)[x = 10 ⇒ both sides equal]
Three rectangles, all with perimeter 40 units — as the shape gets closer to a square, the area increases.
As x → 0 or x → 20, A(x) → 0[long, thin rectangle]
So there is no smallest area[area can be made as small as we like, but never reaches 0]
Dissection conjecture: yes, always possible (Bolyai–Gerwien theorem).
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.
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).
Step 1: Turn the triangle into a rectangle of the same area[two-step plan]
Triangle area \(=\tfrac12 bh\)[base b, height h]
Same as a rectangle with sides b and \(\tfrac h2\)[½bh = b × (h/2)]
Construct this rectangle[bisecting a side, ruler & compass]
Step 2: Square that rectangle[Baudhāyana's construction, Section 6.9]
First construct a rectangle with the same area as the triangle (using half the base and the full height, or vice versa), then apply Baudhāyana's rectangle-squaring construction to that rectangle.
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.
Base AD = 10 cm[given]
Height = 8 cm, wherever E lies on BC[BC ∥ AD, distance apart = 8 cm]
Area \(=\tfrac12\times10\times8=40\) sq cm[½ × base × height]
Area of triangle ADE = 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.
Drop perpendiculars from the ends of the 20 cm side[isosceles trapezium]
2 congruent right triangles, base \(\tfrac{40-20}{2}=10\) cm each[splitting the extra 20 cm equally]
Hypotenuse = 26 cm[given, non-parallel side]
\(h=\sqrt{26^2-10^2}=\sqrt{576}=24\) cm[Pythagoras]
Area \(=\tfrac12(40+20)(24)=\tfrac12(60)(24)=720\) sq cm[½(a+b)h]
Area of the trapezium = 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[perimeter − known sides]
\(s=\dfrac{32}{2}=16\)[semi-perimeter]
Area \(=\sqrt{16(16-8)(16-11)(16-13)}\)[Heron's formula]
\(=\sqrt{16\times8\times5\times3}=\sqrt{1920}\)[multiplying]
\(=\sqrt{64\times30}=8\sqrt{30}\approx43.82\) sq cm
Area = 8√30 ≈ 43.82 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[given ratio]
\(15k=300 \Rightarrow k=20\)[sum of sides = perimeter]
Sides = 60 m, 100 m, 140 m[substituting k = 20]
\(s=\dfrac{300}{2}=150\)[semi-perimeter]
Area \(=\sqrt{150(90)(50)(10)}=\sqrt{6{,}750{,}000}\)[Heron's formula]
\(=100\sqrt{675}=1500\sqrt3\approx2598.08\) sq m
Area = 1500√3 ≈ 2598.08 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 shorter diagonal = d, longer = 2d[given ratio]
Area \(=\tfrac12 d_1d_2=\tfrac12(d)(2d)\)[rhombus area formula]
\(d^2=128\)[equating to given area]
\(d=\sqrt{128}=8\sqrt2\approx11.31\) cm[taking square root]
Shorter diagonal = 8√2 ≈ 11.31 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.
△PCD and △QCD share the same base CD[common side]
P, Q both lie on AB, and AB ∥ CD[opposite sides of the parallelogram]
So P and Q are equidistant from line CD[= height of the parallelogram]
Same base, same height ⇒ equal area[area = ½ × base × height]
Ratio area(ΔPCD) : area(ΔQCD) = 1 : 1 (the two areas are always equal, no matter where P and Q are on AB).
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 a parallelogram, O on diagonal PR
To Prove: Area(△PSO) = Area(△PQO)
Area(△PQR) = Area(△PSR)[diagonal PR bisects the parallelogram's area]
Let \(h_1,h_2\) = distances of Q, S from line PR
\(\tfrac12 PR\cdot h_1=\tfrac12 PR\cdot h_2 \Rightarrow h_1=h_2\)[equal triangle areas, same base PR]
△PQO, △PSO share base PO[O lies on PR]
Their heights are still \(h_1,h_2\)[Q, S distance to PR unchanged]
Area(PQO)\(=\tfrac12 PO\cdot h_1\), Area(PSO)\(=\tfrac12 PO\cdot h_2\)[½ × base × height]
Since \(h_1=h_2\), the areas are equal
Area(ΔPSO) = Area(ΔPQO), since both triangles share the base PO and the vertices Q, S are equidistant from line PR (because diagonal PR bisects the parallelogram's area).
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: P, Q, R, S = midpoints of AB, BC, CD, DA
To Prove: Area(PQRS) = ½ Area(ABCD)
Draw diagonal BD[splits ABCD into △ABD, △CBD]
area(ABD) + area(CBD) = area(ABCD)[the two halves]
△APS ~ △ABD, ratio ½[P, S midpoints of AB, AD]
area(APS) \(=\left(\tfrac12\right)^2\)area(ABD)\(=\tfrac14\)area(ABD)[similar triangles, ratio² area]
Similarly area(CQR) \(=\tfrac14\)area(CBD)[Q, R midpoints of CB, CD]
area(APS) + area(CQR) \(=\tfrac14\)area(ABCD)
Similarly, using diagonal AC: area(BPQ) + area(DRS) \(=\tfrac14\)area(ABCD)[same argument]
All 4 corner triangles together \(=\tfrac14+\tfrac14=\tfrac12\) area(ABCD)[sum of both pairs]
area(PQRS) = area(ABCD) − ½area(ABCD) = ½area(ABCD)[what remains, 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: D midpoint of BC, AD median, P on AD
To Prove: Area(△ABP) = Area(△ACP)
area(ABD) = area(ACD)[median property — equal bases BD = DC, same height from A]
area(PBD) = area(PCD)[△PBD, △PCD: equal bases BD = DC, same height from P]
area(ABP) = area(ABD) − area(PBD)
area(ACP) = area(ACD) − area(PCD)
area(ABP) = area(ACP)[equal minus equal, hence proved]
Area(ΔABP) = Area(ΔACP), proved by subtracting the equal-area triangles PBD and PCD from the equal-area triangles ABD and ACD.
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: Square ABCD, P inside; PA,PB,PC,PD joined
To Find: ratio of red (PAB+PCD) to green (PBC+PDA)
Let side = a; drop perpendiculars from P to AB, CD: \(h_1,h_2\)[opposite parallel sides]
\(h_1+h_2=a\)[AB, CD are a apart]
Area(PAB)+Area(PCD)\(=\tfrac12 ah_1+\tfrac12 ah_2\)[½ × base × height, base = a]
\(=\tfrac12 a(h_1+h_2)=\tfrac12 a^2\)[exactly half the square, for any P]
Remaining pair Area(PBC)+Area(PDA) \(=a^2-\tfrac12a^2=\tfrac12a^2\)[total area a²]
Ratio of red region to green region = 1 : 1 (each pair always totals exactly half the square's area, no matter where P is placed).
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: D midpoint of AB, P on BC, Q on AB with CQ ∥ PD
To Prove: Area(△BPQ) = ½ Area(△ABC)
Construction: Join CD[median of △ABC]
area(BDC) \(=\tfrac12\)area(ABC) …(i)[median property, D midpoint of AB]
area(DPC) = area(DPQ) …(ii)[same base DP, between DP ∥ CQ]
area(BPQ) = area(BPD) + area(DPQ)[PD splits △BPQ]
= area(BPD) + area(DPC)[using (ii)]
= area(BDC)[DP splits △BDC into BPD, DPC — P on BC]
\(=\tfrac12\)area(ABC)[using (i), hence proved]
Area(ΔBPQ) = ½ Area(ΔABC), as required.
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?
Circle encloses maximum area for a given perimeter[practical: max capacity for material used]
Easy to construct with a rope and peg[practical: construction]
Rolls smoothly / turns uniformly[practical: wheels, pottery wheels]
Represents wholeness, cycles of time, infinity, unity[symbolic/aesthetic]
Appears in domes, mandalas, rose windows, decorative art[symbolic/aesthetic]
Uses: wheels, gears, pots, coins, clock faces, domes, arches, textiles[summary of applications]
Both — circles were practical (maximum area for a given perimeter, ease of construction, rolling wheels) and symbolically/aesthetically significant (wholeness, cycles, religious and decorative art) across human history.
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 \(=\pi r^2\times\dfrac{\theta}{360^\circ}\)[sector area formula]
\(\dfrac{22}{7}\times49=154\)[full circle area, πr²]
Area \(=154\times\dfrac16=\dfrac{77}{3}\approx25.67\) sq cm[θ = 60°]
Area = \(\dfrac{77}{3}\) ≈ 25.67 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 as Ex 6.1, Q1]
Quadrant area \(=\tfrac14\pi r^2=\tfrac14\times\dfrac{22}{7}\times49\)[quadrant = ¼ of circle]
\(=\tfrac14\times154=38.5\) sq cm[πr² = 154]
Area of quadrant = 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.
60 min → 360°, so 10 min → \(\dfrac{10}{60}\times360^\circ=60^\circ\)[proportion]
Minute hand length = radius = 7 cm[given]
Area \(=\pi r^2\times\dfrac{60}{360}=\dfrac{22}{7}\times49\times\dfrac16\)[sector area formula]
\(=\dfrac{154}{6}=\dfrac{77}{3}\approx25.67\) sq cm
Area swept = \(\dfrac{77}{3}\) ≈ 25.67 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°): \(=\pi r^2\times\dfrac{90}{360}\)[sector area formula]
\(=3.14\times100\times0.25=78.5\) sq cm[π ≈ 3.14, r = 10]
(ii) Major sector (270°): \(=3.14\times100\times0.75=235.5\) sq cm[remaining ¾ of circle]
Check: \(78.5+235.5=314=\pi r^2\) ✓[matches full circle area]
(i) 78.5 sq cm (ii) 235.5 sq cm
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°): \(=3.14\times225\times\dfrac16=117.75\) sq cm[πr² × θ/360]
Radii equal (15 cm) + included angle 60° ⇒ equilateral triangle, side 15[isosceles with 60° apex ⇒ equilateral]
Triangle area \(=\dfrac{\sqrt3}{4}(15)^2=\dfrac{1.73}{4}\times225=97.3125\) sq cm[equilateral triangle formula]
Minor segment = sector − triangle \(=117.75-97.3125\approx20.44\) sq cm[segment = sector minus triangle]
Major segment = full circle − minor segment[remaining area]
\(=706.5-20.4375\approx686.06\) sq cm[πr² = 3.14 × 225 = 706.5]
Minor segment ≈ 20.44 sq cm
Major segment ≈ 686.06 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°.
One wiper: sector, r = 28 cm, angle 120°[blade length = radius]
\(=\dfrac{22}{7}\times28^2\times\dfrac{120}{360}\)[sector area formula]
\(\dfrac{22}{7}\times784=2464\)[πr²]
One wiper \(=\dfrac{2464}{3}\approx821.33\) sq cm[× 120/360]
Total (2 wipers, no overlap) \(=2\times\dfrac{2464}{3}=\dfrac{4928}{3}\approx1642.67\) sq cm
Total area cleaned = \(\dfrac{4928}{3}\) ≈ 1642.67 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 (60°) \(=\pi r^2\times\dfrac{60}{360}=\dfrac{\pi r^2}{6}\)[sector area formula]
Two radii (r) + 60° angle ⇒ equilateral triangle, side r[isosceles with 60° apex]
Triangle area \(=\dfrac{\sqrt3}{4}r^2\)[equilateral triangle formula]
Minor segment \(=\dfrac{\pi r^2}{6}-\dfrac{\sqrt3}{4}r^2=r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)\)[segment = sector − triangle, hence shown]
Minor segment area = \(r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)\), as required.
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.
Circumradius formula: \(r=\dfrac{a}{\sqrt3} \Rightarrow a=r\sqrt3\)[a = side of the triangle]
Triangle area \(=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}(3r^2)=\dfrac{3\sqrt3}{4}r^2\)[equilateral triangle formula]
Circle area \(=\pi r^2\)[standard formula]
Ratio \(=\dfrac{3\sqrt3/4\,r^2}{\pi r^2}=\dfrac{3\sqrt3}{4\pi}\approx0.413\)[r² cancels, hence shown]
Ratio = \(\dfrac{3\sqrt3}{4\pi}\) ≈ 0.413, as required.
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.
Diagonal of square = diameter of circle \(=2r\)[inscribed square]
\(s\sqrt2=2r \Rightarrow s=r\sqrt2\)[diagonal of a square = s√2]
Area of square \(=s^2=2r^2\)[squaring s]
Ratio \(=\dfrac{2r^2}{\pi r^2}=\dfrac{2}{\pi}\approx0.637\)[r² cancels, hence shown]
Ratio = 2/π ≈ 0.637, as required.
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.
Hexagon = 6 equilateral triangles, side = r[circumradius = side, for a regular hexagon]
Hexagon area \(=6\times\dfrac{\sqrt3}{4}r^2=\dfrac{3\sqrt3}{2}r^2\)[6 × equilateral triangle formula]
Ratio \(=\dfrac{3\sqrt3/2\,r^2}{\pi r^2}=\dfrac{3\sqrt3}{2\pi}\approx0.827\)[r² cancels, hence shown]
Q8's triangle = alternate vertices of this hexagon[same circle]
That triangle = exactly half the hexagon's area[known "alternate-vertex" fact]
So Q10's ratio is exactly double Q8's[half hexagon area ⇒ half the ratio]
Ratio = \(\dfrac{3\sqrt3}{2\pi}\) ≈ 0.827
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.
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.
Start with a square of side a; cut a square of side b from one corner
Remaining L-shape has area \(a^2-b^2\)[big square − small square]
Cut the L-shape into 2 strips; rearrange into a rectangle \((a+b)\times(a-b)\)[same area, different shape]
Big square (side a) minus small corner square (side b) = the L-shaped region, area a²−b².
Draw a square of side (a+b+c); split each side into a, b, c
3×3 grid: 3 squares (a², b², c²) on the diagonal[reading off the grid]
+ 6 rectangles: ab, ab, bc, bc, ca, ca[remaining cells, each pair appears twice]
Total = a²+b²+c²+2ab+2bc+2ca[matches the expansion]
Both identities correspond to decomposing a big square (of side a+b, or a+b+c) into smaller squares and rectangles whose areas add up to the algebraic 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[perimeter − 2 equal sides]
Height \(=\sqrt{15^2-5^2}=\sqrt{200}=10\sqrt2\) cm[Pythagoras, half-base = 5]
Area \(=\tfrac12\times10\times10\sqrt2=50\sqrt2\approx70.71\) sq cm[½ × base × height]
Area = 50√2 ≈ 70.71 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.
\(\tfrac12\times10\times h=60\)[½ × base × height = area]
\(h=12\) cm
Equal side \(=\sqrt{5^2+12^2}=\sqrt{169}=13\) cm[Pythagoras, half-base = 5]
Equal sides = 13 cm each.
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.
\(\tfrac12\times12\times(\text{other leg})=54\)[½ × base × height]
Other leg \(=\dfrac{54\times2}{12}=9\) cm[solving]
Hypotenuse \(=\sqrt{12^2+9^2}=\sqrt{225}=15\) cm[Pythagoras]
Perimeter \(=12+9+15=36\) cm[sum of 3 sides]
Perimeter = 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.
Let sides = 2k, 3k, 4k[given ratio]
\(9k=45 \Rightarrow k=5\)[sum of sides = perimeter]
Sides = 10, 15, 20 cm[substituting k = 5]
\(s=\dfrac{45}{2}=22.5\)[semi-perimeter]
Area \(=\sqrt{22.5(12.5)(7.5)(2.5)}\)[Heron's formula]
\(=\sqrt{5273.4375}=\dfrac{75\sqrt{15}}{4}\approx72.62\) sq cm
Area = (75√15)/4 ≈ 72.62 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 — check \(7^2+24^2=49+576=625=25^2\)[Pythagorean triple test]
Right triangle, legs 7 and 24[hypotenuse = 25]
Area \(=\tfrac12\times7\times24=84\) sq cm[½ × leg × leg]
Method 2 — \(s=\dfrac{7+24+25}{2}=28\)[Heron's formula]
Area \(=\sqrt{28(21)(4)(3)}=\sqrt{7056}=84\) sq cm[matches Method 1]
Area = 84 sq cm, matching both by the right-angle shortcut and by Heron's formula.
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 \(=\pi d=\dfrac{22}{7}\times60=\dfrac{1320}{7}\) cm[circumference formula]
100 rotations \(=100\times\dfrac{1320}{7}\approx18857.14\) cm
\(\approx188.57\) m
Distance travelled ≈ 188.57 m.
8Find the area of a quadrant of a circle whose circumference is 66 cm.
Rough sketch — quadrant of the circle.
\(C=2\pi r=66\)[circumference formula]
\(r=\dfrac{66\times7}{44}=10.5\) cm
Quadrant area \(=\tfrac14\pi r^2=\tfrac14\times\dfrac{22}{7}\times110.25\)[¼ of circle area]
\(=\tfrac14\times346.5=86.625\) sq cm
Area = 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 \(=2\pi r=2\times\dfrac{22}{7}\times28=176\) cm[circumference formula]
1 km \(=100{,}000\) cm[unit conversion]
Turns \(=\dfrac{100000}{176}\approx568.18\)[total distance ÷ distance per turn]
176 cm per turn; ≈568.18 turns for a 1 km journey.
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 — this always forces the rectangles to be congruent[surprising, but true for rectangles]
Sides x, y: perimeter fixes \(x+y=S\), area fixes \(xy=P\)[given]
x, y are roots of \(t^2-St+P=0\)[Vieta's formulas]
A quadratic has at most 2 roots
A second rectangle with same S, P solves the same equation[same sum & product]
So {x',y'} = {x,y} — the rectangles are congruent[same pair of side lengths]
Yes — two rectangles with equal area and equal perimeter must have the same pair of side lengths, and are therefore always congruent.
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: Trapezium, parallel sides a, b, height h[Fig. 6.42]
To Prove: Area = ½(a+b)h
Construction: Rotate a 2nd copy 180°, join along a slanted side[doubling trick]
Result: a parallelogram, base (a+b), height h[the two parallel sides now placed end to end]
Area of parallelogram \(=(a+b)h\)[base × height]
Parallelogram = 2 trapeziums, so 1 trapezium \(=\tfrac12(a+b)h\)[halving, hence proved]
Area of trapezium = ½(a+b)h, derived by doubling the trapezium into a parallelogram of base (a+b) and height 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: Trapezium, parallel sides a, b, height h
To Prove: Area = ½(a+b)h
Construction: Draw a diagonal[splits into 2 triangles]
Triangle 1: base a, height h — Triangle 2: base b, height h[both share the same perpendicular height h]
Area 1 \(=\tfrac12 ah\), Area 2 \(=\tfrac12 bh\)[½ × base × height]
Total \(=\tfrac12 ah+\tfrac12 bh=\tfrac12(a+b)h\)[adding, hence proved]
Area of trapezium = ½(a+b)h, matching the earlier formula.
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: Rotate a 2nd copy 180°, join along a slanted side[2 trapeziums, sides a, b, height h]
Side a on one copy lines up with side b on the other[interlocking shapes]
Together: one continuous base, length \(a+b\)[the two ends combine]
Result: a parallelogram, base (a+b), height h[shape formed]
Area \(=(a+b)h\)[base × height]
1 trapezium \(=\tfrac12(a+b)h\)[parallelogram = 2 trapeziums]
Two trapeziums (one rotated 180°) combine into a parallelogram of base (a+b) and height h; halving its area gives the trapezium formula ½(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: Kite ABCD, AC ⊥ BD, OB = OD, AC=p, BD=q[AC = axis of symmetry]
To Prove: Area = ½pq
(i) Diagonals split kite into 4 right triangles: AOB, BOC, COD, DOA[algebra approach]
Total \(=\tfrac12(AO)(OB)+\tfrac12(OC)(OB)+\tfrac12(OC)(OD)+\tfrac12(AO)(OD)\)[4 right-triangle areas]
\(=\tfrac12(OB+OD)(AO+OC)=\tfrac12(BD)(AC)=\tfrac12pq\)
(ii) Diagonal AC splits kite into △ABC, △ADC (same base AC)[geometry approach]
Heights = OB, OD (perpendicular distances to AC)[B, D to line AC]
Area(ABC)+Area(ADC) \(=\tfrac12(AC)(OB+OD)=\tfrac12(AC)(BD)=\tfrac12pq\)[½ × base × height, both, hence proved]
Area of kite = ½ × (product of diagonals) = ½ pq, confirmed both geometrically (splitting by one diagonal) and algebraically (splitting by both diagonals).
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: ABCD sides a,b; PQRS sides 2a,2b
Area(ABCD) \(=ab\)[l × b]
Area(PQRS) \(=2a\times2b=4ab\)[= 4 × Area(ABCD)]
Cut PQRS in half both ways (through midpoints)
Gives exactly 4 rectangles, each a × b[identical to ABCD]
Yes — 4 copies of ABCD tile PQRS perfectly
Yes, PQRS has exactly 4 times the area, and can be perfectly tiled by 4 copies of ABCD (cut PQRS in half both ways).
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: △ABC sides a,b,c; △PQR sides 2a,2b,2c
△ABC: \(s=\tfrac12(a+b+c)\); △PQR: semi-perimeter = 2s[all sides doubled]
Area(PQR) \(=\sqrt{2s\cdot2(s-a)\cdot2(s-b)\cdot2(s-c)}\)[Heron's formula]
\(=\sqrt{16\,s(s-a)(s-b)(s-c)}=4\times\)Area(ABC)[√16 = 4]
Join the midpoints of △PQR's sides
Midsegment theorem ⇒ 4 triangles, each congruent to ABC[medial-triangle dissection]
Yes — 4 copies of ABC tile PQR perfectly
Yes, ΔPQR has exactly 4 times the area, and can be perfectly tiled by 4 copies of ΔABC (using the medial-triangle dissection).
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: △ABC sides a,b,c; △PQR sides 3a,3b,3c
Same Heron's-formula scaling as 15(ii), semi-perimeter × 3[method]
\(\sqrt{3\times3\times3\times3}=9\)[4 factors of 3 inside the root]
Area(PQR) \(=9\times\)Area(ABC)[result]
Divide each side of PQR into 3 equal parts, draw lines parallel to the sides
Triangular grid ⇒ 9 triangles, each congruent to ABC
Yes, ΔPQR has exactly 9 times the area, and can be perfectly tiled by 9 copies of ΔABC (using a triangular grid with each side divided into 3 equal parts).
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.
Tick marks: each side divided in ratio 1:2[reading the figure]
Vertices joined to the ⅓ points on the opposite side[cevians, same rotational order]
Inner triangle = exactly \(\tfrac17\) of the original[classical result — mass-point / coordinate geometry]
Fig. 6.44 — joining the 1:2 division points creates a rotated inner square.
Unit square, join points at ⅓ along each side[coordinate geometry setup]
Inner square has side \(\dfrac{\sqrt5}{3}\)[distance between consecutive division points]
Area \(=\left(\dfrac{\sqrt5}{3}\right)^2=\dfrac59\)
Fraction shaded: \(\dfrac{1}{7}\) of the triangle (Fig. 6.43); \(\dfrac{5}{9}\) of the square (Fig. 6.44).
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.
n identical circles in a row, each touching top/bottom & neighbours[reading the figures]
Rectangle dimensions \(=(2r)\times(n\times2r)\)[height = diameter, width = n diameters]
Rectangle area \(=4nr^2\); circle area \(=n\pi r^2\)
Fraction \(=\dfrac{n\pi r^2}{4nr^2}=\dfrac{\pi}{4}\)[n cancels]
Fig 6.45 (n=3) and Fig 6.46 (n=4): both \(\dfrac{\pi}{4}\approx78.5\%\)
Both figures: fraction covered = π/4 ≈ 0.785 (78.5%), regardless of the number of circles.
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: fraction covered = π/4, for any n[same-row circle packing]
n = 10, 20, 50: rectangle \(=4nr^2\), circles \(=n\pi r^2\)[testing, same formula as Q17]
Fraction \(=\dfrac{n\pi r^2}{4nr^2}=\dfrac{\pi}{4}\) in every case[n cancels regardless of value]
Proof: each circle sits in its own \(2r\times2r\) cell[one cell per circle]
Each cell has ratio \(\dfrac{\pi r^2}{(2r)^2}=\dfrac{\pi}{4}\)[fixed, independent of r]
Whole rectangle = n identical cells ⇒ same π/4 overall
Conjecture: fraction covered = π/4 always (≈78.5%), independent of the number of circles — proved because each circle occupies the same fixed fraction (π/4) of its own 2r×2r square "cell," and the whole rectangle is just a row of identical cells.
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.
3 rectangles (long side horizontal) on top; 6 rectangles (short side horizontal) below[reading Fig. 6.47]
Let long side = L, short side = S
Top row width = 3L; bottom row width = 6S[3 and 6 rectangles respectively]
\(3L=6S \Rightarrow L=2S\)[both rows = same big rectangle width]
\(9\times LS=72 \Rightarrow LS=8\)[9 small rectangles, total area 72]
\(2S\times S=8 \Rightarrow S^2=4 \Rightarrow S=2\) cm[substituting L = 2S]
\(L=2S=4\) cm
Check: 3(4) = 12 = 6(2); area = 12 × 6 = 72 ✓
Perimeter \(=2(L+S)=2(4+2)=12\) cm[perimeter formula]
Each small rectangle measures 4 cm × 2 cm, giving a perimeter of 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: BC trisected at D, E; AD (blue), AE (red) drawn
To Prove: Area(ABD) = Area(AEC)
△ABD, △AEC share the same height[perpendicular distance from A to BC]
Bases BD = EC = ⅓ BC[trisection]
Same base, same height ⇒ equal area[each = ⅓ of △ABC, hence proved]
Cut △ABD along A–(midpoint of BD)[rearranging the pieces]
Rotate one piece 180° about that midpoint[forms a parallelogram, base ½BD]
Same construction on △AEC gives a congruent parallelogram[since BD = EC, same height]
Congruent parallelograms ⇒ ABD's pieces cover AEC
Area(ABD) = Area(AEC), since both share the same base length (⅓ of BC) and the same height. Cutting each along the median to its base and rotating one piece 180° turns both into congruent parallelograms, showing the blue triangle's pieces can be rearranged to exactly cover the red triangle.
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: Square side a; quarter circle at O (r=a); 2 semicircles (r=a/2) on adjacent sides
To Prove: Area(A) = Area(B)
Quarter circle area \(=\tfrac14\pi a^2\)[¼πr², r = a]
Each semicircle area \(=\tfrac12\pi\left(\tfrac a2\right)^2=\tfrac{\pi a^2}{8}\)[½πr², r = a/2]
2 semicircles together \(=\tfrac{\pi a^2}{4}\) — matches quarter circle![key equality]
Let C = the shared overlap region
Area(quarter circle) = Area(C) + Area(B)[B = quarter circle minus overlap]
Area(semi₁)+Area(semi₂) = Area(C) + Area(A)[A = semicircles minus overlap]
LHS of both equal \(\tfrac{\pi a^2}{4}\) ⇒ Area(C)+Area(B) = Area(C)+Area(A)[from the key equality above]
Area(A) = Area(B)[cancel Area(C), hence proved]
Area(A) = Area(B), because the quarter circle's area exactly equals the combined area of the two semicircles — subtracting the shared overlap region from both sides of this equality leaves the two "exclusive" regions A and B equal.
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: each semicircle has one side of the square as its diameter (radius = 1 unit = half the side), so all four bulge inward and meet exactly at the centre.
Perimeter: each petal = 2 quarter-arcs (90° each), radius 1[2 adjacent semicircles]
Each petal boundary \(=\tfrac{\pi}{2}+\tfrac{\pi}{2}=\pi\)[quarter-arc = πr/2, r=1]
4 petals: total perimeter \(=4\pi\approx12.57\) units
Area: one petal = overlap of 2 semicircles, r=1, centres \(\sqrt2\) apart[circle-overlap / "lens" setup]
Lens area \(=2\cos^{-1}\!\left(\tfrac{1}{\sqrt2}\right)-\tfrac{\sqrt2}{2}\sqrt{4-2}\)[standard lens formula, r=1, d=√2]
\(=2\times\dfrac{\pi}{4}-1=\dfrac{\pi}{2}-1\)[cos⁻¹(1/√2) = π/4]
4 petals: total area \(=4\left(\dfrac{\pi}{2}-1\right)=2\pi-4\approx2.28\) sq units
Perimeter of flower = 4π ≈ 12.57 units. Area of flower = 2π − 4 ≈ 2.28 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: concentric circles, radii R>r; chord BC (=l) tangent to smaller at A
To Prove: annulus area \(=\tfrac14\pi l^2\)
OA ⊥ BC, OA = r[tangent ⊥ radius at point of contact]
A is the midpoint of BC: \(AB=AC=\tfrac l2\)[⊥ from centre bisects a chord]
\(OB^2=OA^2+AB^2\)[Pythagoras, right △OAB]
\(R^2=r^2+\left(\tfrac l2\right)^2 \Rightarrow R^2-r^2=\tfrac{l^2}{4}\)[substituting OB=R]
Annulus \(=\pi R^2-\pi r^2=\pi(R^2-r^2)\)[area between the circles]
\(=\pi\times\tfrac{l^2}{4}=\tfrac14\pi l^2\)[substituting, hence proved]
Area of the annular region = ¼πl², as required.
24*Semicircles have been drawn on all the sides of a right-angled triangle. Show that Area(A) + Area(B) = Area(C).
Given: right triangle legs p, q, hypotenuse r[p²+q²=r², Pythagoras]
Semicircles outward on legs; semicircle on hypotenuse through right angle[Thales' theorem]
C = the triangle; A, B = the two crescent "lunes"
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)
Multiply \(p^2+q^2=r^2\) by \(\dfrac{\pi}{8}\)[semicircle area ∝ diameter²]
Semicircle(p) + Semicircle(q) = Semicircle(r)[same identity, in terms of areas]
Semicircle(r) = triangle C + 2 circular segments[splitting the hypotenuse semicircle]
Each leg's semicircle = its lune + the same segment[matching decomposition]
Cancel the equal, shared segments from both sides[substituting into the identity]
Lune(A) + Lune(B) = Area(C)[hence proved — "Lune of Hippocrates"]
Area(A) + Area(B) = Area(C), the classical Lune of Hippocrates result — following directly from the Pythagorean relation between the three semicircle areas.
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: 2 congruent circles, radius r, each through the other's centre[Fig. 6.53, meet at C, D]
To Find: area enclosed by both (the lens), in terms of r
△ABC, △ABD equilateral (all sides = r)[same setup as Example 1, Section 6.5]
∠CAD = ∠CBD = 120°[2 × 60° equilateral angles]
Lens = 2 circular segments, each cut by chord CD, angle 120°[one segment per circle]
Sector (120°, radius r) \(=\pi r^2\times\tfrac{120}{360}=\tfrac{\pi r^2}{3}\)[sector area formula]
Triangle (isosceles, sides r, r, angle 120°) \(=\tfrac12r^2\sin120^\circ=\tfrac{\sqrt3}{4}r^2\)[½ab sinC]
One segment \(=\tfrac{\pi r^2}{3}-\tfrac{\sqrt3}{4}r^2\)[segment = sector − triangle]
Lens (2 segments) \(=r^2\left(\tfrac{2\pi}{3}-\tfrac{\sqrt3}{2}\right)\)
Area enclosed by the two circles = \(r^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\right)\).
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.
Result depends on the exact cevian construction shown[note]
Set up coordinates for the rectangle and the cevian point(s)[general method]
Write A, B, C each as \(\tfrac12\times\text{base}\times\text{height}\)[using the same variables]
Substitute into \(\dfrac{2(A+C)(B+C)}{C}\)[target identity]
Simplifies to width × height[matches the rectangle's area]
Approach: express A, B, C algebraically from the figure's exact construction, then verify 2(A+C)(B+C)/C simplifies to width × height. (This problem is highly figure-specific — map your textbook's exact points onto variables and follow the same substitution method shown throughout this chapter's other area proofs.)
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.
Same family as Q21 and Q24 (Lune of Hippocrates)[matching areas via related-radii identity]
Identify the quarter circle, semicircle, and triangle[from the figure]
Express each area using the shared lengths[e.g. semicircle diameter = triangle's hypotenuse, Thales' theorem]
Decompose both shaded regions using the same known-equal areas[as in Q21, Q24]
Cancel the shared overlap from both sides[leaves the 2 shaded regions equal]
Using the same "matching areas, cancel the shared overlap" strategy as Q21 and Q24: once the quarter circle, semicircle and triangle areas are expressed in terms of the same base lengths, the identity between them (via Thales' theorem and the Pythagorean relation) forces the two shaded regions to be equal.
Extra Practice Questions
Seven extra questions in the style of the textbook's own exercises, for independent practice once you've gone through the solved questions above. Each one is shown open with its full working, since a diagram is part of the answer here — cover the solution with your hand and attempt it on paper first.
1The length of a rectangular park is 25 m more than its breadth. If the perimeter of the park is 230 m, find its length and breadth.
Rough sketch — rectangular park, breadth x m, length (x + 25) m.
Let breadth = x m, length = (x+25) m[given relation]
\(2\big[(x+25)+x\big]=230\)[perimeter = 2(l+b)]
\(2(2x+25)=230\)
\(2x+25=115\)[dividing by 2]
\(2x=90 \Rightarrow x=45\)
Length \(=45+25=70\) m[substituting back]
Length = 70 m, Breadth = 45 m.
2The diameter of a bicycle wheel is 63 cm. Find the distance covered by the wheel in 500 complete revolutions.
Rough sketch — bicycle wheel, diameter 63 cm.
1 revolution = circumference \(=\pi d=\dfrac{22}{7}\times63=198\) cm[C = πd]
500 revolutions \(=500\times198=99{,}000\) cm
\(=990\) m
Distance covered = 990 m.
3A circular field has a circumference of 176 m. Find (i) its radius, (ii) its area, and (iii) the cost of ploughing it at ₹4 per m².
Rough sketch — circular field, circumference 176 m.
(i) \(176=2\times\dfrac{22}{7}\times r\)[C = 2πr]
\(r=\dfrac{176\times7}{44}=28\) m
(ii) Area \(=\pi r^2=\dfrac{22}{7}\times28\times28=2464\) m²[πr²]
(iii) Cost \(=2464\times4=9856\)[area × rate]
(i) Radius = 28 m (ii) Area = 2464 m² (iii) Cost = ₹9,856.
4The sides of a triangular field are 41 m, 40 m and 9 m. Find its area using Heron's formula. Also find the cost of ploughing the field at ₹15 per m².
Rough sketch — triangular field, sides 41 m, 40 m, 9 m (not to scale).
\(s=\dfrac{41+40+9}{2}=45\) m[semi-perimeter]
\(s-a=4,\ s-b=5,\ s-c=36\)[45 − each side]
Area \(=\sqrt{45\times4\times5\times36}\)[Heron's formula]
\(=\sqrt{32400}=180\) m²
Cost \(=180\times15=2700\)[area × rate]
Area = 180 m²; Cost of ploughing = ₹2,700.
5Find the area of a triangle whose base is 24 cm and height is 14.5 cm. If a parallelogram stands on the same base and has the same area as this triangle, find its height.
Rough sketch — a triangle and a parallelogram on the same base of 24 cm.
Triangle area \(=\dfrac12\times24\times14.5=174\) cm²[½ × base × height]
\(174=24\times h\)[parallelogram: same base & area]
\(h=\dfrac{174}{24}=7.25\) cm
Triangle's area = 174 cm²; height of the parallelogram = 7.25 cm.
6A cyclic quadrilateral has sides 25 m, 39 m, 52 m and 60 m. Find its area using Brahmagupta's formula.
Rough sketch — cyclic quadrilateral, sides 25 m, 39 m, 52 m, 60 m.
\(s=\dfrac{25+39+52+60}{2}=88\) m[semi-perimeter]
\(s-a=63,\ s-b=49,\ s-c=36,\ s-d=28\)[88 − each side]
Area \(=\sqrt{63\times49\times36\times28}\)[Brahmagupta's formula]
\(=\sqrt{3{,}111{,}696}=1764\) m²
Area = 1764 m².
7Find the length of the arc and the area of the sector of a circle of radius 21 cm, subtending an angle of 60° at the centre.
Rough sketch — sector of radius 21 cm, angle 60° at the centre.
Arc \(l=2\pi r\times\dfrac{60}{360}=2\times\dfrac{22}{7}\times21\times\dfrac16\)[arc length formula]
\(=132\times\dfrac16=22\) cm
Sector area \(=\pi r^2\times\dfrac{60}{360}=\dfrac{22}{7}\times441\times\dfrac16\)[sector area formula]
\(=1386\times\dfrac16=231\) cm²
Arc length = 22 cm; Area of sector = 231 cm².
8A chord of a circle of radius 14 cm subtends a right angle (90°) at the centre. Find the area of the corresponding minor segment.
Rough sketch — minor segment cut off by a chord subtending 90° at the centre (radius 14 cm).
Sector (90°) \(=\pi r^2\times\dfrac{90}{360}=\dfrac{22}{7}\times14\times14\times\dfrac14\)[sector area formula]
\(=154\) cm²
Triangle: right-angled at centre, legs = radius = 14 cm[chord subtends 90°]
Area \(=\dfrac12\times14\times14=98\) cm²[½ × leg × leg]
Minor segment \(=154-98=56\) cm²[sector − triangle]
Area of the minor segment = 56 cm².
Frequently Asked Questions
Circumference = 2πr (or πd), and area = πr², where r is the radius and π ≈ 22/7 or 3.14.
Heron's formula finds the area of a triangle from its three side lengths a, b, c alone: Area = the square root of s(s-a)(s-b)(s-c), where s is the semi-perimeter, half of (a+b+c). It's especially useful when you know the sides but not the height.
Brahmagupta's formula finds the area of a cyclic quadrilateral (a 4-sided figure whose vertices all lie on one circle) from its four sides a, b, c, d: Area = the square root of (s-a)(s-b)(s-c)(s-d), where s is the semi-perimeter. It generalises Heron's formula — setting the fourth side d = 0 (collapsing the 4-gon into a triangle) turns Brahmagupta's formula exactly into Heron's.
Because such shapes can be "hinged" — the same side lengths can form many different shapes (with different angles between the sides), and area depends on both the sides and the angles. A triangle is rigid (its three sides fix its shape completely, which is exactly why Heron's formula works from sides alone), but a quadrilateral or parallelogram is not.
A sector's area is πr² times (θ/360°), where θ is the angle at the centre. A segment (the region between a chord and its arc) is found by subtracting the area of the triangle formed by the two radii and the chord from the area of the sector.