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?
Rough sketch (lane width exaggerated) — the start of a 4 × 100 m relay on a bend, as in Fig. 6.1. All runners run one bend and then the same straight to the common finish line. The bend of an outer lane is longer, so its start is moved ahead along the bend: the stagger between adjacent lanes is \(\pi w\).
What is the stagger? In a race, every runner must run the same distance and all of them finish at the same line. A runner in an outer lane has a longer path, so her starting point is placed ahead of the starting point of the inner lane by exactly this extra distance. This distance by which the start of a lane is moved ahead is the stagger. In other words, the stagger is the difference between the track lengths of the lanes.
As in Fig. 6.1, the runners start on a bend. The first leg of the 4 × 100 m relay takes each runner round one semicircular bend and then along the straight, which is the same length in every lane. So only the bend matters. Let the radius of the bend in lane 1 (innermost) be \(r\) and let each lane be \(w\) wide. Then the bends of lanes 1, 2, 3 have radii \(r,\ r+w,\ r+2w\).
Length of the track on the bend (a semicircle): lane 1: \(\pi r\); lane 2: \(\pi(r+w)\); lane 3: \(\pi(r+2w)\).
Stagger of lane 2 = difference in track lengths \(\pi(r+w)-\pi r=\pi w\).
Lane 3 must start ahead of lane 1 by \(\pi(r+2w)-\pi r=2\pi w\), which is another \(\pi w\) beyond lane 2. So the stagger between lane 3 and lane 2 is also \(\pi w\).
The radius \(r\) cancels in every case. So the stagger depends only on the width of a lane, not on the radius of the bend. It is the same on a small bend (200 m track) as on a large bend (400 m track), and it is the same between every pair of adjacent lanes.
No. For a race that starts on a bend, the stagger between adjacent lanes is \(\pi\times\) (lane width), which does not depend on the radius of the track. A 200 m track therefore needs the same stagger as a 400 m track if its lanes have the same width (for 1.22 m lanes it is \(\pi\times1.22\approx3.83\) m for a start on one bend). For a race that goes round both bends, i.e. a full lap such as the 400 m race of Fig. 6.11, the extra distance is that of a full circle, so the stagger is \(2\pi\times\) (lane width) \(\approx7.67\) m — see the second Think and Reflect after Fig. 6.11.
Think and Reflect
TR(On the perimeter of a circle with radius r) What is the connection between this question 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 in track lengths of the two lanes]
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 the 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?
Fig. 6.11 — a 400 m athletics track: two straight sections of 84.39 m and two semicircular ends with centres A and B. The inner semicircle has radius 36.5 m, each lane is 1.22 m wide and the runner in lane 1 runs 0.3 m from the inner border (radius 36.8 m). The lane width is drawn larger than its true size; the three starts are drawn staggered by 2π × lane width, with a common finish line.
Lane 1 effective radius \(=36.5+0.3=36.8\) m[inner radius 36.5 m; runner is 0.3 m from the 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]
The two straight sections are \(2\times84.39=168.78\) m in every lane[same in all lanes]
The two semicircular ends together make one full circle[one full circuit]
Track length of lane 1 \(=168.78+2\pi(36.8)=168.78+231.22=400\) m[straights + circle]
Track length of lane 2 \(=168.78+2\pi(38.02)=168.78+238.89=407.67\) m[straights + circle]
Stagger = difference in track lengths \(=407.67-400\approx7.67\) m[lane 2 must start this much ahead]
In one step: \(2\pi(38.02)-2\pi(36.8)=2\pi(1.22)=2\times3.1416\times1.22\approx7.67\) m[straights cancel; only the radius difference matters]
Lane 3 vs lane 2: the radius again increases by 1.22 m, so the stagger is again \(2\pi(1.22)\approx7.67\) m[same lane width]
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
(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
(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 following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):
Reading the figures: every curved edge is a semicircle unless it is marked as a quarter circle below; the dashed lines are only construction lines and are not part of the perimeter.
Fig. 6.14 (i)–(ix) from the textbook. Dashed lines are construction lines, not part of the perimeter.
(i) A rectangle 80 m long with a semicircle on each of its two shorter sides (diameter 60 m, radius 30 m)[shape read from figure]
Straight parts \(=2\times80=160\) m[top and bottom sides]
The two semicircles together make one circle of radius 30 m: \(2\pi r=2\times\dfrac{22}{7}\times30=\dfrac{1320}{7}\approx188.57\) m[C = 2πr]
Perimeter \(=160+\dfrac{1320}{7}=\dfrac{2440}{7}\approx348.57\) m[straight + curved]
(ii) Outer semicircle (diameter 12 cm, radius 6 cm) and inner semicircle (diameter 8 cm, radius 4 cm), with 2 flat edges on the base[shape read from figure]
Each flat edge \(=\dfrac{12-8}{2}=2\) cm[base left on either side of the inner semicircle]
Perimeter = outer arc + inner arc + 2 flat edges[tracing the outline]
\(=\pi(6)+\pi(4)+2(2)=\dfrac{22}{7}\times10+4=\dfrac{220}{7}+4=\dfrac{248}{7}\approx35.43\) cm[πr for each semicircle]
(iii) A square of side 10 cm with a semicircle on each side (diameter 10 cm, radius 5 cm)[shape read from figure]
Perimeter = 4 semicircular arcs[the sides of the square are inside the shape]
\(=4\times\pi(5)=4\times\dfrac{22}{7}\times5=\dfrac{440}{7}\approx62.86\) cm[πr per arc]
(iv) An equilateral triangle of side 12 cm with a semicircle on each side (diameter 12 cm, radius 6 cm)[shape read from figure]
Perimeter = 3 semicircular arcs[the dashed sides of the triangle are not part of the boundary]
\(=3\times\pi(6)=18\times\dfrac{22}{7}=\dfrac{396}{7}\approx56.57\) cm[πr per arc]
(v) 3 × 3 grid of squares, each of side 14 cm; 4 semicircles (diameter 14 cm, radius 7 cm) on the middle edges and 4 quarter circles (radius 14 cm) at the corners[shape read from figure]
4 semicircles \(=4\times\pi(7)=4\times22=88\) cm[πr per arc]
4 quarter circles \(=4\times\dfrac14\times2\pi(14)=4\times22=88\) cm[each arc is ¼ of 2πr]
Perimeter \(=88+88=176\) cm
(vi) A large semicircle of diameter 28 cm, and 4 small semicircles each of diameter \(\dfrac{28}{4}=7\) cm (radius 3.5 cm) along the baseline[shape read from figure]
Large arc \(=\pi(14)=\dfrac{22}{7}\times14=44\) cm[πr, r = 14]
4 small arcs \(=4\times\pi(3.5)=4\times11=44\) cm[πr, r = 3.5]
Perimeter \(=44+44=88\) cm
(vii) A right triangle with legs 6 cm and 8 cm, and a semicircle on each of its three sides[shape read from figure]
Hypotenuse \(=\sqrt{6^2+8^2}=\sqrt{100}=10\) cm[Pythagoras]
Radii of the semicircles: 5 cm (hypotenuse), 3 cm (6 cm side), 4 cm (8 cm side)[radius = half of the side]
Perimeter \(=\pi(5)+\pi(3)+\pi(4)=12\pi=12\times\dfrac{22}{7}=\dfrac{264}{7}\approx37.71\) cm[the sides of the triangle are inside the shape]
(viii) A large semicircle of diameter 12 cm and 3 small semicircles of diameter 4 cm each[shape read from figure]
\(=\pi(6)+3\times\pi(2)=12\pi=12\times\dfrac{22}{7}=\dfrac{264}{7}\approx37.71\) cm[large arc + 3 small arcs]
(ix) A large semicircle of diameter 20 cm, a semicircle of diameter 10 cm above the baseline and a semicircle of diameter 10 cm below the baseline[shape read from figure]
\(=\pi(10)+\pi(5)+\pi(5)=20\pi=20\times\dfrac{22}{7}=\dfrac{440}{7}\approx62.86\) cm[three arcs: radii 10, 5, 5]
(i) ≈348.57 m
(ii) ≈35.43 cm
(iii) ≈62.86 cm
(iv) ≈56.57 cm
(v) 176 cm
(vi) 88 cm
(vii) ≈37.71 cm
(viii) ≈37.71 cm
(ix) ≈62.86 cm
(ii) ≈35.43 cm
(iii) ≈62.86 cm
(iv) ≈56.57 cm
(v) 176 cm
(vi) 88 cm
(vii) ≈37.71 cm
(viii) ≈37.71 cm
(ix) ≈62.86 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).
(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: The centres of the arcs are the midpoints of the sides of the square. Fig. 6.15B: The centres of the arcs are the vertices of the hexagon.)
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 petal is bounded by 2 quarter-circle arcs of radius 7 cm, so the 4 petals have 8 quarter circles = 4 semicircles (diameter 14, radius 7)[each arc runs from a corner to the centre]
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
(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'?
Fig. 6.18 (redrawn from the textbook) — a ‘thin’ parallelogram ABCD: the foot of the perpendicular from C falls outside AD, on AD extended.
Fig. 6.19 — thin parallelogram ABCD with A′ and D′ marked so that AA′ = DD′; the shaded triangles BAA′ and CDD′ are congruent, and CF is the perpendicular from C to A′D′ (now falling on the base).
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
TRYou may wonder, like earlier, is 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? Please work out the answer to this question.
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. M, N, L are the midpoints of AB, AD, AC; the dashed line MNL is the cut. The shaded △AMN is rotated about M to form a parallelogram with base BD and height ½h.
Yes, it is possible[BD = DC and same height from A to BC]
Let h be the height of ΔABD. Cut ΔABD along the line MN joining the midpoints M of AB and N of AD[this line is parallel to BD; it gives a small top triangle and a trapezium]
Rotate the small triangle through 180° about the midpoint of AB[the two pieces now form a parallelogram]
Parallelogram: base BD, height \(\tfrac h2\)[area \(=BD\times\tfrac h2=\tfrac12\,BD\cdot h\), same as ΔABD]
Doing the same to ΔACD gives a parallelogram with base DC = BD and the same height \(\tfrac h2\)[BD = DC, same height from A]
Two parallelograms with equal base and equal height can be changed into each other by cutting off a triangle at one end and fixing it at the other end[as in Fig. 6.17 – 6.19; repeat if needed]
So the pieces of ΔABD can be rearranged to cover ΔACD exactly
Yes. Cut ΔABD along the line MN joining the midpoints M of AB and N of AD and rotate the small triangle through 180°; this gives a parallelogram of base BD and height ½h. ΔACD gives a parallelogram with the same base and height. One parallelogram can then be cut and rearranged into the other, so 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 out for familiar shapes, e.g.,
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 a conjecture of your own about this.
Think of various rectangles with perimeter 40 units (the sides do not have to be integers).
1. How many such rectangles are there?
2. Among them, is there one whose area is the largest? What are its dimensions?
3. Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?
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 \(A=x(20-x)=20x-x^2=100-(x-10)^2\)[area = product of sides; completing the square]
\((x-10)^2\ge0\), so \(A\le100\), with equality only when \(x=10\)[a square is never negative]
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?
Triangle → rectangle of the same area: the dashed midline (at height h/2) and altitude AH are the construction; the dashed rectangle has sides b and h/2.
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.
(Figure: ABCD is a rectangle with DC = 10 cm and BC = 8 cm; E is a point on BC.)
Fig. 6.31 — Triangle ADE always has base AD = 8 cm and height DC = 10 cm, whichever point E is on BC.
Base AD = BC = 8 cm[opposite sides of a rectangle]
Height of ΔADE = distance of E from AD = DC = 10 cm, wherever E lies on BC[BC ∥ AD, and DC ⊥ AD]
Area \(=\tfrac12\times8\times10=40\) 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 ABCD (AB = 20 cm, DC = 40 cm, legs 26 cm); AE and BF are the perpendiculars dropped from A and B to DC.
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 AB; the dashed perpendicular is the common height of △PCD and △QCD.
△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; h1, h2 are the perpendiculars from Q and S to PR (dashed).
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 (also known as a quadrilateral, but we prefer to call it 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. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)
Rough sketch — 4-gon ABCD with midpoints P, Q, R, S. Dashed lines: diagonals BD, AC and the median BS used in the proof. Shaded corner triangles: APS, CQR (orange) and BPQ, DRS (blue).
Given: P, Q, R, S = midpoints of AB, BC, CD, DA
To Prove: Area(PQRS) = ½ Area(ABCD)
Draw diagonals BD and AC[splits ABCD into △ABD, △CBD]
area(ABD) + area(CBD) = area(ABCD)[the two halves]
Join BS. In △ABD, S is the midpoint of AD, so BS is a median[S midpoint of AD]
area(ABS) \(=\tfrac12\)area(ABD)[a median divides a triangle into two triangles of equal area]
In △ABS, P is the midpoint of AB, so SP is a median: area(APS) \(=\tfrac12\)area(ABS)[median property again]
area(APS) \(=\tfrac12\times\tfrac12\)area(ABD)\(=\tfrac14\)area(ABD)
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 P inside. Red: △PAB and △PCD. Green: △PBC and △PDA. The dashed line through P (perpendicular to AB and CD) has parts h1 and h2.
Given: Square ABCD, P inside; PA,PB,PC,PD joined
To Find: ratio of red (PAB+PCD) to green (PBC+PDA)
Let side = a; draw the line through P perpendicular to AB and CD; let its parts be \(h_1\) (to AB) and \(h_2\) (to CD)[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 the midpoint of AB; CQ ∥ PD. The dashed line CD is the construction; the shaded triangle is BPQ.
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 (that subtends 90° at the centre), and
(ii) major sector (that subtends 270° at the centre).
(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
(ii) 235.5 sq cm
5A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
Rough sketch — OA, OB are the radii and AB the chord; the minor segment (teal) = sector OAB − △OAB.
Minor sector (60°): \(=3.14\times225\times\dfrac16=117.75\) sq cm[πr² × θ/360]
Join OA, OB and the chord AB. 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 of the circle. Show that the area of the corresponding minor segment of the circle is equal to \(\pi r^2\left(\frac{1}{6}-\frac{\sqrt{3}}{4}\right)\).
Note: the textbook prints the result as \(\pi r^2\left(\tfrac{1}{6}-\tfrac{\sqrt{3}}{4}\right)\), which is a misprint. The correct area of the minor segment is \(r^2\left(\tfrac{\pi}{6}-\tfrac{\sqrt{3}}{4}\right)\), and that is what we prove below.
Rough sketch — OA, OB are the radii and AB the chord subtending 60° at O; the minor segment (teal) = sector − △OAB.
Minor sector (60°) \(=\pi r^2\times\dfrac{60}{360}=\dfrac{\pi r^2}{6}\)[sector area formula]
Join OA, OB and the chord AB. 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 ABC inscribed in a circle of radius r; AD is the altitude (dashed), passing through the centre O, and OB is a radius.
Let the side of the triangle be \(a\). Draw the altitude AD (D is the midpoint of BC); it passes through the centre O, which is also the centroid, so \(r=OA=\tfrac23\times\) altitude[medians meet in the ratio 2 : 1]
Altitude \(=\dfrac{\sqrt3}{2}a\), so \(r=\tfrac23\cdot\dfrac{\sqrt3}{2}a=\dfrac{a}{\sqrt3}\Rightarrow a=r\sqrt3\)[Pythagoras in half 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 ABCD inscribed in a circle of radius r; the diagonals AC and BD (dashed) are diameters.
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 ABCDEF in a circle of radius r with centre O. Dashed: triangle ACE (Q8), and the rhombus OABC with its diagonal AC.
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 ACE joins alternate vertices of this hexagon ABCDEF (same circle, centre O)[radius r]
OABC is a rhombus made of 2 of the small equilateral triangles; diagonal AC cuts it into two equal halves, so area(ΔOAC) = 1 small triangle[a diagonal bisects a rhombus]
Similarly area(ΔOCE) = area(ΔOEA) = 1 small triangle, so area(ΔACE) = 3 small triangles
Hexagon = 6 small triangles = 2 × area(ΔACE)[from above]
So the hexagon : circle ratio is exactly double Q8's triangle : circle ratio[same circle, same area πr²]
Ratio = \(\dfrac{3\sqrt3}{2\pi}\) ≈ 0.827
It is exactly twice the Q8 answer, because the equilateral triangle of Q8 is formed by alternate vertices of this same hexagon and has area equal to 3 of the hexagon's 6 equal small triangles, i.e. exactly half the hexagon's area.
It is exactly twice the Q8 answer, because the equilateral triangle of Q8 is formed by alternate vertices of this same hexagon and has area equal to 3 of the hexagon's 6 equal small triangles, i.e. exactly half the hexagon's area.
End-of-Chapter Exercises
1Identities in algebra can sometimes be shown as area relationships. For example: the figure (Fig. 6.41: Area model of an identity) shown corresponds to the identity \((a+b)^2=a^2+2ab+b^2\). Do you see how?
Draw figures corresponding to the identities \((a+b)(a-b)=a^2-b^2\) and \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\).
Fig. 6.41 (redrawn from the textbook) — area model of (a + b)² = a² + 2ab + b².
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 along the dashed line into two rectangles: \(a\times(a-b)\) and \((a-b)\times b\)[see figure]
Turn the smaller rectangle through \(90^\circ\) and place it beside the larger one: they form a rectangle of length \(a+b\) and breadth \(a-b\)[same area, different shape]
So \((a+b)(a-b)=a^2-b^2\)[area of the rectangle = area of the L-shape]
(a + b)(a − b) = a² − b²: the square of side a with the corner square of side b removed is cut into two rectangles; the smaller one is turned through 90° and placed beside the larger one to make a rectangle of length a + b and breadth a − b.
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: a square of side a + b + c divided into 3 × 3 cells.
Draw a square of side \(a+b+c\); split each side into parts \(a\), \(b\), \(c\) and join the opposite points
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 are shown by area: \((a+b)(a-b)\) by cutting and rearranging the L-shape left after removing a square of side b from a square of side a, and \((a+b+c)^2\) by dividing a square of side a+b+c into 3 squares and 6 rectangles.
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]
Let the sides be \(x\) and \(y\) (\(x\ge y\)). Perimeter fixes \(x+y=S\) and area fixes \(xy=P\)[S = half the perimeter, P = area]
\((x-y)^2=(x+y)^2-4xy=S^2-4P\)[identity]
So \(x-y=\sqrt{S^2-4P}\), since \(x\ge y\)
\(x=\tfrac12\left(S+\sqrt{S^2-4P}\right),\quad y=\tfrac12\left(S-\sqrt{S^2-4P}\right)\)[adding and subtracting the two equations]
Length and breadth are completely fixed by S and P, so any two such rectangles have the same length and the same breadth — they are congruent
Yes — two rectangles with equal area and equal perimeter must have the same pair of side lengths, and are therefore always congruent.
11You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., \(\tfrac{1}{2}(a+b)h\).
(Fig. 6.42: Trapezium: sides a and b, height h)
Fig. 6.42 — Trapezium with parallel sides a and b and height h. The slanted line drawn from the top-right vertex, parallel to the left side, splits it into a parallelogram (purple) and a triangle (pink).
Given: Trapezium with parallel sides a (top) and b (bottom), b > a, and height h[Fig. 6.42]
To Prove: Area = ½(a+b)h
Construction: Through the top-right vertex, draw a line parallel to the left slanted side, meeting the base. This splits the trapezium into a parallelogram and a triangle[Fig. 6.42]
Parallelogram: base \(a\) (opposite sides are equal), height \(h\), so its area \(=ah\)[area of parallelogram = base × height]
Triangle: base \(=b-a\) (the rest of the bottom side), height \(h\), so its area \(=\tfrac{1}{2}(b-a)h\)[the triangle is half of a parallelogram of base \(b-a\) and height \(h\)]
Area of trapezium \(=ah+\tfrac{1}{2}(b-a)h=\tfrac{1}{2}h\,[2a+b-a]=\tfrac{1}{2}(a+b)h\)[adding the two areas, hence proved]
Area of trapezium = ½(a+b)h, obtained by splitting it into a parallelogram (area ah) and a triangle (area ½(b−a)h).
12By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).
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 identical trapeziums (the second rotated 180° about the midpoint of the shared slanted leg) form a parallelogram of base a + b and height h.
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)(Q15: Three problems about fitting congruent shapes together)
(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!
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)(ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c. Show that ∆PQR has 4 times the area of ∆ABC. Does this mean that 4 copies of ∆ABC will fit into ∆PQR? Check and see!
ΔPQR (sides doubled) split by its three midlines into 4 triangles, each congruent to ΔABC (sides a, b, c).
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)(iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a, 3b, 3c. Show that ∆PQR has 9 times the area of ∆ABC. Does this mean that 9 copies of ∆ABC will fit into ∆PQR? Check and see!
ΔPQR (sides tripled) split by lines parallel to its sides at one-third and two-thirds into 9 triangles, each congruent to ΔABC (sides a, b, c).
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*Fig. 6.43: What fraction of the triangle is shaded? Fig. 6.44: What fraction of the square is shaded?
Fig. 6.43 — P is the midpoint of AB; Q and R are the points of trisection of AC. The shaded region is the 4-gon BPQR. The dashed line BQ is the construction used in the solution (BR is already a side of the shaded region).
Fig. 6.43: In ΔABC, P is the midpoint of AB; Q and R are the points of trisection of AC, so AQ = QR = RC. The shaded region is the 4-gon BPQR[reading the tick marks]
Let area(ΔABC) \(=T\)
ΔABR and ΔABC have the same height from B, and \(AR=\tfrac23AC\), so area(ΔABR) \(=\tfrac23T\)[same height, bases in the ratio 2 : 3]
Similarly \(AQ=\tfrac13AC\), so area(ΔABQ) \(=\tfrac13T\)
In ΔABQ, P is the midpoint of AB, so QP is a median: area(ΔAPQ) \(=\tfrac12\times\tfrac13T=\tfrac16T\)[a median halves the area]
Shaded region = ΔABR − ΔAPQ \(=\tfrac23T-\tfrac16T=\tfrac12T\)[PQ cuts ΔABR into ΔAPQ and the 4-gon BPQR]
Fig. 6.44 — Lines are drawn from each vertex to the midpoint of a side (AR, PC, BS, DQ). X, Y are the points where AR meets DQ and BS; the right angles at X and Y are marked. The shaded region is the inner 4-gon.
Fig. 6.44: Let ABCD be the square of side 2 units (area 4 sq. units); P, Q, R, S are the midpoints of AB, BC, CD, DA. The lines AR, PC, BS, DQ are drawn[reading the tick marks]
Let AR meet DQ at X, and AR meet BS at Y
ΔADR ≅ ΔDCQ[SAS: AD = DC = 2, DR = CQ = 1, ∠ADR = ∠DCQ = 90°]
So ∠DAR = ∠CDQ, and \(\angle DAX+\angle ADX=\angle CDQ+\angle ADX=\angle ADC=90^\circ\)[X lies on DQ and on AR]
Hence ∠AXD = 90°, i.e. DX ⊥ AR[angle sum of ΔADX]
\(AR^2=AD^2+DR^2=4+1=5\Rightarrow AR=\sqrt5\)[Pythagoras]
Area(ΔADR) \(=\tfrac12\times AD\times DR=1\) and also \(=\tfrac12\times AR\times DX\), so \(DX=\dfrac{2}{\sqrt5}\)[area found in two ways]
\(AX^2=AD^2-DX^2=4-\tfrac45=\tfrac{16}{5}\Rightarrow AX=\dfrac{4}{\sqrt5}\)[Pythagoras in ΔADX]
In the same way (ΔBAS ≅ ΔADR), AY ⊥ BS and \(AY=\dfrac{AB\times AS}{BS}=\dfrac{2\times1}{\sqrt5}=\dfrac{2}{\sqrt5}\)[area of ΔABS found in two ways]
\(XY=AX-AY=\dfrac{4}{\sqrt5}-\dfrac{2}{\sqrt5}=\dfrac{2}{\sqrt5}\)
The four lines meet at right angles and the figure looks the same after a 90° turn about the centre, so the shaded 4-gon is a square of side \(\dfrac{2}{\sqrt5}\)
Area of the shaded square \(=\dfrac45\), so fraction \(=\dfrac{4/5}{4}=\dfrac15\)
Fraction shaded: \(\dfrac12\) of the triangle (Fig. 6.43); \(\dfrac15\) of the square (Fig. 6.44).
17Fig. 6.45: What fraction of the rectangle is covered by the circles? Fig. 6.46: What fraction of the rectangle is covered by the 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.
Fig. 6.47 — 4 rectangles in the top row (length L horizontal) and 5 in the bottom row (breadth B horizontal), both rows of the same width.
Let each small rectangle have length \(L\) and breadth \(B\), with \(L>B\)[from Fig. 6.47]
Top row: 4 rectangles with the length horizontal, so its width is \(4L\). Bottom row: 5 rectangles with the breadth horizontal, so its width is \(5B\)[reading Fig. 6.47]
Both rows have the same width as the large rectangle: \(4L=5B\), so \(L:B=5:4\)[same width]
Let \(L=5k\) and \(B=4k\) for some \(k>0\) (cm)[from the ratio 5 : 4]
Height of large rectangle \(=B+L=4k+5k=9k\) and width \(=4L=20k\)[one row of height B above one row of height L]
Area of large rectangle \(=20k\times9k=180k^2=72\)[given area]
\(k^2=\dfrac{72}{180}=\dfrac25\Rightarrow k=\sqrt{\dfrac25}=\dfrac{\sqrt{10}}{5}\)[k > 0]
\(L=5k=\sqrt{10}\approx3.16\) cm and \(B=4k=\dfrac{4\sqrt{10}}{5}\approx2.53\) cm
Check: \(9\times LB=9\times20k^2=180\times\dfrac25=72\) ✓[area of 9 rectangles]
Perimeter \(=2(L+B)=2(9k)=18k=18\times\dfrac{\sqrt{10}}{5}=\dfrac{18\sqrt{10}}{5}\approx11.38\) cm[perimeter of a rectangle = 2(l + b)]
Perimeter of each small rectangle \(=\dfrac{18\sqrt{10}}{5}\) cm \(\approx11.38\) cm (each rectangle is \(\sqrt{10}\) cm × \(\dfrac{4\sqrt{10}}{5}\) cm).
20*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 into some number of pieces and rearranging the pieces to cover the red triangle.
(Fig. 6.48: Lines from a vertex to the points of trisection of the opposite side)
Fig. 6.48 — BC is trisected at D, E. M, N are the midpoints of AB, AD and M′, N′ are the midpoints of AE, AC; the dashed lines MN and M′N′ are the cuts.
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]
Equal bases and the same height ⇒ equal areas[each = ⅓ of △ABC, hence proved]
Let h be the height of each triangle. Cut △ABD along the line MN joining the midpoints M of AB and N of AD[parallel to BD; gives a small triangle and a trapezium]
Rotate the small triangle through 180° about the midpoint of AB[forms a parallelogram of base BD and height \(\tfrac h2\)]
Do the same to △AEC with the line M′N′ joining the midpoints M′ of AE and N′ of AC: it gives a parallelogram of base EC (= BD) and height \(\tfrac h2\)[BD = EC, same height]
Two parallelograms with equal bases and equal heights can be changed into each other by cutting off a triangle at one end and fixing it at the other end[as in Fig. 6.17 – 6.19; repeat if needed]
So the pieces of the blue triangle can be rearranged to cover the red triangle exactly
Area(ABD) = Area(AEC), since their bases BD and EC are equal (each ⅓ of BC) and they have the same height. Cut the blue triangle along the line MN joining the midpoints M of AB and N of AD and rotate the small triangle through 180° to get a parallelogram of base BD and height ½h; the red triangle gives a parallelogram of the same base and height. One parallelogram can be cut and rearranged into the other, so the pieces of the blue triangle can be rearranged to exactly cover the red triangle.
21*The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.
(Fig. 6.49: A quarter circle and two semicircles)
Fig. 6.49 — Quarter circle centred at O (radius a) and two semicircles on the sides through O. A (orange) is the region common to the two semicircles; B (blue) is the region of the quarter circle outside both semicircles.
Given: Square of side \(a\); quarter circle with centre O (a vertex) and radius \(a\); two semicircles of radius \(\tfrac a2\) on the two sides through O[Fig. 6.49]
To Prove: Area(A) = Area(B)[A = region common to the two semicircles; B = region of the quarter circle outside both semicircles]
Quarter circle \(=\tfrac14\pi a^2\)[¼πr², r = a]
Each semicircle \(=\tfrac12\pi\left(\tfrac a2\right)^2=\dfrac{\pi a^2}{8}\)[½πr², r = a/2]
The two semicircles together \(=\dfrac{\pi a^2}{8}+\dfrac{\pi a^2}{8}=\dfrac{\pi a^2}{4}\) — equal to the quarter circle[key equality]
Both semicircles lie inside the quarter circle, and they overlap in A. So the area they cover together \(=\dfrac{\pi a^2}{4}-\) Area(A)[A would be counted twice in the sum]
Quarter circle = (area covered by the semicircles) + B[B is the rest of the quarter circle]
\(\dfrac{\pi a^2}{4}=\left[\dfrac{\pi a^2}{4}-\text{Area}(A)\right]+\text{Area}(B)\)
Area(A) = Area(B)[hence proved]
Area(A) = Area(B), because the quarter circle has exactly the same area as the two semicircles together. The overlap A is counted twice in the sum of the two semicircles, so the part of the quarter circle outside them (B) must be equal to A.
22*In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.
Fig. 6.50 — Four semicircles inside the square, centred at the midpoints of the sides (radius 1 unit), forming a 4-petalled flower. Dashed: square AMON (M, N midpoints of AB, AD; O the centre) and the chord AO of the petal at A.
Perimeter: each semicircle has a side of the square (2 units) as diameter, so its radius is 1 unit[centres at the midpoints of the sides]
Each petal is bounded by 2 quarter-circle arcs of radius 1[each arc runs from a corner to the centre of the square]
Length of one quarter-circle arc \(=\tfrac14\times2\pi(1)=\dfrac{\pi}{2}\)
4 petals have 8 such arcs: perimeter \(=8\times\dfrac{\pi}{2}=4\pi=4\times\dfrac{22}{7}=\dfrac{88}{7}\approx12.57\) units
Area: take the petal at corner A. M, N are the midpoints of AB, AD and O is the centre of the square[Fig. 6.50]
AMON is a square of side 1 unit[AM = MO = ON = NA = 1, angles 90°]
The quarter circle (centre M, radius 1) and the quarter circle (centre N, radius 1) together cover the square AMON, and the petal is the part covered twice
Petal \(=2\times\dfrac{\pi}{4}-\) area of square AMON \(=\dfrac{\pi}{2}-1\)[quarter circle = ¼π(1)² = π/4]
4 petals: area \(=4\left(\dfrac{\pi}{2}-1\right)=2\pi-4=\dfrac{44}{7}-4=\dfrac{16}{7}\approx2.29\) sq units
Perimeter of the flower \(=4\pi=\dfrac{88}{7}\approx12.57\) units. Area of the flower \(=2\pi-4=\dfrac{16}{7}\approx2.29\) sq units.
23*In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is \(\tfrac{1}{4}\pi l^2\).
Fig. 6.51 — Concentric circles with centre O; chord BC of the larger circle touches the smaller circle at A. The dashed lines OA (⊥ BC) and OB are the construction.
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*In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).
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 the three sides of a right triangle. The semicircle on the hypotenuse (dashed) passes through the right-angle vertex. A and B are the two lunes; C is the triangle.
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*Fig. 6.53 shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius r.
Fig. 6.53 — Circles centred at A and B, each through the other's centre, meeting at C and D. The dashed lines are the radii AC, AD, BC, BD and the common chord CD; the shaded lens is the overlap.
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]
ACBD is a rhombus of side \(r\), made of the two equilateral triangles ABC and ABD, each of area \(\tfrac{\sqrt3}{4}r^2\)[AC = CB = BD = DA = r]
Diagonal CD bisects the rhombus, so area(ΔACD) \(=\) area(ΔABC) \(=\tfrac{\sqrt3}{4}r^2\)[ΔACD is the triangle with ∠CAD = 120°]
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*In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is \(\dfrac{2(A+C)(B+C)}{C}\).
Fig. 6.54 — Rectangle with corner O; P on the top side, Q on the right side, R directly below P and level with Q. A = ΔOPR, B = ΔORQ, C = ΔPRQ.
Given: O is a corner of the rectangle; P lies on the top side and Q on the right side. R is the point where the vertical line through P meets the horizontal line through Q. A = ΔOPR, B = ΔORQ, C = ΔPRQ[Fig. 6.54]
To Prove: Area of rectangle \(=\dfrac{2(A+C)(B+C)}{C}\)
Let PR \(=h\), RQ \(=k\); let \(p\) = distance of O from line PR and \(q\) = distance of O from line RQ[as marked in the figure]
Length of rectangle \(=p+k\), breadth \(=q+h\)[from the figure]
\(A=\tfrac12\times PR\times p=\tfrac12hp\)[base PR, height = distance of O from PR]
\(B=\tfrac12\times RQ\times q=\tfrac12kq\)[base RQ, height = distance of O from RQ]
\(C=\tfrac12hk\)[right angle at R]
\(A+C=\tfrac12h(p+k)\) and \(B+C=\tfrac12k(q+h)\)[taking out the common factor]
\((A+C)(B+C)=\tfrac14\,hk\,(p+k)(q+h)=\tfrac12\,C\,(p+k)(q+h)\)[\(\tfrac14hk=\tfrac12C\)]
\(\dfrac{2(A+C)(B+C)}{C}=(p+k)(q+h)=\) length × breadth \(=\) area of the rectangle[hence proved]
Area of the rectangle \(=(p+k)(q+h)=\dfrac{2(A+C)(B+C)}{C}\), as required.
27*In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.
(Fig. 6.55) Show that the areas of the two shaded regions are equal.
Fig. 6.55 — Quarter circle AFB (centre O) and semicircle AEB on the chord AB (centre D). The shaded regions are the lune AEBF and the triangle AOB.
Given: O is the centre; OA = OB = r and ∠AOB = 90°. Arc AFB is a quarter circle (centre O). AEB is a semicircle on AB as diameter (centre D)[Fig. 6.55]
To Prove: area of the lune AEBF = area of ΔAOB[the two shaded regions]
\(AB^2=OA^2+OB^2=2r^2\)[Pythagoras in right ΔAOB]
Radius of the semicircle \(=\dfrac{AB}{2}\), so (radius)\(^2=\dfrac{2r^2}{4}=\dfrac{r^2}{2}\)
Area of semicircle AEB \(=\tfrac12\pi\cdot\dfrac{r^2}{2}=\dfrac{\pi r^2}{4}\)[½πr²]
Area of quarter circle OAFB \(=\tfrac14\pi r^2=\dfrac{\pi r^2}{4}\)
Segment AFB \(=\) quarter circle \(-\) ΔAOB \(=\dfrac{\pi r^2}{4}-\) area(ΔAOB)[segment = sector − triangle]
Lune AEBF = semicircle AEB − segment AFB[the segment lies inside the semicircle]
\(=\dfrac{\pi r^2}{4}-\left[\dfrac{\pi r^2}{4}-\text{area}(\triangle AOB)\right]=\text{area}(\triangle AOB)\)[hence proved]
Area of the lune AEBF = area of ΔAOB \(\left(=\tfrac12r^2\right)\), so the two shaded regions are equal.
Extra Practice Questions
Eight 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.
(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.