NCERT Solutions Class 9 Maths Chapter 6 End-of-Chapter Exercises — Measuring Space: Perimeter and Area

NCERT Class 9 Mathematics — Ganita Manjari Part I, Measuring Space: Perimeter and Area, End-of-Chapter Exercises. All 27 questions solved.

The longest exercise in the book, and it ranges widely. The tools:

$$\text{triangle} = \tfrac12 bh \qquad \text{Heron} = \sqrt{s(s-a)(s-b)(s-c)} \qquad \text{trapezium} = \tfrac12(a+b)h$$ $$\text{circle} = \pi r^2 \qquad \text{circumference} = 2\pi r \qquad \text{sector} = \tfrac{\theta}{360}\pi r^2$$

Use $\pi = \frac{22}{7}$ unless told otherwise. Sixteen questions carry an asterisk, marking them as enrichment that is not assessed; all are solved in full.

Key insight. The hardest-looking questions here — 16, 21, 24, 25, 27 — all yield to the same move: express each region as a sum or difference of shapes you can already measure, then let the common parts cancel. Nothing in this exercise needs a formula beyond the list above; what it needs is deciding what to add and what to subtract.

Question 1

Identities in algebra can sometimes be shown as area relationships. Draw figures corresponding to the identities $(a+b)(a-b) = a^2 – b^2$ and $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Solution. The worked example, Fig. 6.41, shows a square of side $a + b$ cut into four pieces of areas $a^2$, $ab$, $ab$ and $b^2$ — the whole square is $(a+b)^2$, and the pieces add to $a^2 + 2ab + b^2$.

For $(a+b)(a-b) = a^2 – b^2$. Start with a square of side $a$, area $a^2$, and cut from one corner a square of side $b$, leaving an L-shape of area $a^2 – b^2$.

Now cut that L-shape along the line separating its two arms. You get:

• a rectangle $b$ by $(a-b)$, and
• a rectangle $a$ by $(a-b)$.

Slide the first to sit alongside the second, making one rectangle of height $(a-b)$ and width $a + b$. Its area is $(a+b)(a-b)$, and nothing was added or removed — so

$$(a+b)(a-b) = a^2 – b^2$$

For $(a+b+c)^2$. Draw a square of side $a+b+c$ and divide each side into three parts of lengths $a$, $b$, $c$. Two horizontal and two vertical cuts give a $3 \times 3$ grid of nine rectangles, each of area (its row label) $\times$ (its column label):

	$a$	$b$	$c$
$a$	$a^2$	$ab$	$ac$
$b$	$ab$	$b^2$	$bc$
$c$	$ac$	$bc$	$c^2$

The three squares on the diagonal give $a^2 + b^2 + c^2$; the six rectangles pair up into $2ab + 2bc + 2ca$. Their total is the area of the whole square, $(a+b+c)^2$.

Question 2

An isosceles triangle has perimeter $40$ cm; the equal sides are $15$ cm each. Find the area of the triangle.

Solution. The base is what is left of the perimeter:

$$b = 40 – 15 – 15 = 10 \text{ cm}$$

In an isosceles triangle the height to the base bisects it, giving a right-angled triangle with hypotenuse $15$ and base $5$:

$$h = \sqrt{15^2 – 5^2} = \sqrt{225 – 25} = \sqrt{200} = 10\sqrt2$$

$$\text{Area} = \frac{1}{2} \times 10 \times 10\sqrt2 = 50\sqrt2 \approx 70.71 \text{ cm}^2$$

Question 3

An isosceles triangle has base $10$ cm, and its area is $60 \text{ cm}^2$. What are the lengths of the equal sides?

Solution. Find the height from the area first:

$$\frac{1}{2} \times 10 \times h = 60 \quad\Longrightarrow\quad h = 12 \text{ cm}$$

The height bisects the base, so each equal side is the hypotenuse of a right-angled triangle with legs $5$ and $12$:

$$\text{equal side} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$$

Question 4

The area of a right-angled triangle is $54$ sq. cm. One of its legs has length $12$ cm. Find its perimeter.

Solution. For a right-angled triangle the two legs are base and height:

$$\frac{1}{2} \times 12 \times b = 54 \quad\Longrightarrow\quad b = 9 \text{ cm}$$

The hypotenuse follows:

$$c = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \text{ cm}$$

$$\text{Perimeter} = 12 + 9 + 15 = 36 \text{ cm}$$

Question 5

The sides of a triangle are in the ratio $2 : 3 : 4$, and its perimeter is $45$ cm. Find its area.

Solution. Write the sides as $2k$, $3k$, $4k$:

$$9k = 45 \quad\Longrightarrow\quad k = 5$$

so the sides are $10$, $15$ and $20$ cm, and $s = 22.5$.

$$\text{Area} = \sqrt{22.5(22.5-10)(22.5-15)(22.5-20)} = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5}$$ $$= \sqrt{\frac{84375}{16}} = \frac{75\sqrt{15}}{4} \approx 72.62 \text{ cm}^2$$

Question 6

The sides of a triangle have lengths $7$ cm, $24$ cm, $25$ cm. Find the area of the triangle in two different ways.

Solution. First notice the triangle is right-angled:

$$7^2 + 24^2 = 49 + 576 = 625 = 25^2$$

so the angle between the $7$ cm and $24$ cm sides is a right angle, and those two are the legs.

Way 1 — half base times height.

$$\text{Area} = \frac{1}{2} \times 7 \times 24 = 84 \text{ cm}^2$$

Way 2 — Heron’s formula. $s = \frac{7 + 24 + 25}{2} = 28$:

$$\text{Area} = \sqrt{28(28-7)(28-24)(28-25)} = \sqrt{28 \times 21 \times 4 \times 3} = \sqrt{7056} = 84 \text{ cm}^2$$

The two agree — which is a useful check on both, and a reminder that Heron’s formula works for every triangle including right-angled ones.

Question 7

If the wheel of a bicycle has a diameter of $60$ cm, find how far a cyclist will have travelled after the wheel has rotated $100$ times.

Solution. One rotation covers one circumference:

$$C = \pi d = \frac{22}{7} \times 60 = \frac{1320}{7} \text{ cm}$$

$$\text{distance} = 100 \times \frac{1320}{7} = \frac{132000}{7} \approx 18857.14 \text{ cm} \approx 188.57 \text{ m}$$

Question 8

Find the area of a quadrant of a circle whose circumference is $66$ cm.

Solution. Find the radius first:

$$2 \times \frac{22}{7} \times r = 66 \quad\Longrightarrow\quad r = \frac{66 \times 7}{44} = \frac{21}{2} = 10.5 \text{ cm}$$

A quadrant is a quarter of the circle:

$$\frac{1}{4} \times \frac{22}{7} \times \left( \frac{21}{2} \right)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{441}{4} = \frac{693}{8} = 86.625 \text{ cm}^2$$

Question 9

The wheel of a car has an outer radius of $28$ cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of $1$ km.

Solution. One turn:

$$C = 2\pi r = 2 \times \frac{22}{7} \times 28 = 176 \text{ cm}$$

Turns in $1$ km. Convert: $1$ km $= 100{,}000$ cm.

$$\frac{100000}{176} = \frac{6250}{11} \approx 568.2$$

So the wheel turns about $568$ times.

Question 10

*Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Solution. Yes — for rectangles, equal area and equal perimeter do force congruence.

Let the first have sides $a$, $b$ and the second $c$, $d$. Then

$$ab = cd \qquad \text{and} \qquad 2(a+b) = 2(c+d), \text{ so } a + b = c + d$$

Two numbers are completely determined by their sum and their product: $a$ and $b$ are the two roots of

$$x^2 – (a+b)x + ab = 0$$

and $c$, $d$ are the roots of $x^2 – (c+d)x + cd = 0$. Since the sum and the product agree, the two quadratics are identical, so they have the same pair of roots. Hence ${a, b} = {c, d}$ and the rectangles have the same dimensions — congruent, possibly after a rotation.

A worked instance. Area $12$ and perimeter $14$ give $a+b = 7$, $ab = 12$, so $x^2 – 7x + 12 = 0$, whose roots are $3$ and $4$. No other rectangle has that area and perimeter.

Worth knowing: this is special to rectangles. Two triangles with the same area and perimeter need not be congruent, and nor need two general quadrilaterals — a square and a rhombus of the same side have equal perimeters but different areas, and shapes can be found matching on both counts without being congruent.

Question 11

You know that the area of a parallelogram is base $\times$ height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides $\times$ height, $\frac{1}{2}(a+b)h$.

Solution. Fig. 6.42 splits the trapezium by a line from one end of the shorter parallel side down to the longer one, cutting it into a parallelogram and a triangle.

Let the parallel sides be $a$ (shorter, on top) and $b$ (longer, below), with height $h$.

The parallelogram has base $a$ — it is bounded by the whole top side and an equal length of the bottom — and height $h$:

$$\text{area} = a h$$

The triangle is what remains of the bottom side, of length $b – a$, with the same height:

$$\text{area} = \frac{1}{2}(b – a)h$$

Add them:

$$ah + \frac{1}{2}(b-a)h = h\left[ a + \frac{b-a}{2} \right] = h \cdot \frac{2a + b – a}{2} = \frac{1}{2}(a+b)h \qquad \blacksquare$$

Question 12

By dividing a trapezium into two triangles show that its area is half the sum of the parallel sides multiplied by the height.

Solution. Draw a diagonal of the trapezium $ABCD$, where $AB = a$ and $DC = b$ are the parallel sides, $h$ apart. The diagonal $AC$ splits it into $\triangle ABC$ and $\triangle ACD$.

$\triangle ABC$ has base $AB = a$. Its apex $C$ lies on the line $DC$, which is parallel to $AB$ at distance $h$ — so its height is $h$:

$$\text{area} = \frac{1}{2}ah$$

$\triangle ACD$ has base $DC = b$ and apex $A$, again at distance $h$:

$$\text{area} = \frac{1}{2}bh$$

Add:

$$\frac{1}{2}ah + \frac{1}{2}bh = \frac{1}{2}(a+b)h \qquad \blacksquare$$

This proof is shorter than Question 11’s, and it explains the formula’s shape directly: the trapezium is two triangles of the same height, one on each parallel side.

Question 13

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Solution. The construction. Take the trapezium, make an identical copy, and rotate the copy through $180°$ about the midpoint of one of its slanted sides. Fit the two together along that side.

The rotated copy has its short side where the original has its long side, so the combined figure has:

• top side $= a + b$,
• bottom side $= b + a$,
• and height $h$ (unchanged by the rotation).

Both pairs of opposite sides are equal and parallel, so the result is a parallelogram of base $a + b$ and height $h$.

The formula. The parallelogram’s area is base $\times$ height:

$$(a+b)h$$

and it is made of exactly two copies of the trapezium. So one trapezium is half of it:

$$\text{trapezium} = \frac{1}{2}(a+b)h \qquad \blacksquare$$

This is the same idea used to derive the area of a triangle, where two copies make a parallelogram of base $b$ and height $h$.

Question 14

Show that the area of a kite is half the product of its diagonals: (i) using algebra, and (ii) using geometry.

Solution. A kite has one diagonal that is an axis of symmetry, and the diagonals cross at right angles.

(i) Using algebra

Let the diagonals be $d_1$ (the axis of symmetry) and $d_2$, crossing at $X$. The axis bisects the other diagonal, so $X$ splits $d_2$ into two halves of $\frac{d_2}{2}$ each, and splits $d_1$ into two parts $p$ and $q$ with $p + q = d_1$.

The kite is four right-angled triangles. Taking the two above the crossing and the two below:

$$\text{area} = 2 \times \frac{1}{2} \cdot p \cdot \frac{d_2}{2} + 2 \times \frac{1}{2} \cdot q \cdot \frac{d_2}{2} = \frac{d_2}{2}(p + q) = \frac{d_2}{2} \cdot d_1 = \frac{1}{2}d_1 d_2$$

(ii) Using geometry

Draw a rectangle around the kite, with its sides parallel to the two diagonals. Its width is $d_2$ and its height is $d_1$, so its area is $d_1 d_2$.

Now look at each of the four corner regions between the kite and the rectangle. Each is a right-angled triangle congruent to the piece of the kite beside it — because the diagonals are perpendicular and the rectangle’s sides are parallel to them, each corner region is a mirror image of the kite piece across a diagonal.

So the kite fills exactly half the rectangle:

$$\text{kite} = \frac{1}{2}d_1 d_2 \qquad \blacksquare$$

Question 15

Three problems about fitting congruent shapes together.

Solution.

(i) Rectangle $ABCD$ has sides $a$, $b$; rectangle $PQRS$ has sides $2a$, $2b$. Show $PQRS$ has $4$ times the area. Do $4$ copies of $ABCD$ fit into $PQRS$?

$$\text{area}(PQRS) = (2a)(2b) = 4ab = 4 \times \text{area}(ABCD)$$

And yes, four copies do fit — in a simple $2 \times 2$ arrangement, two across and two down. Doubling each side of a rectangle exactly doubles it in both directions.

(ii) $\triangle ABC$ has sides $a$, $b$, $c$; $\triangle PQR$ has sides $2a$, $2b$, $2c$. Show $\triangle PQR$ has $4$ times the area. Do $4$ copies fit?

The two triangles are similar, with every length doubled. Using Heron’s formula with every side doubled, $s$ doubles too, and each of the four factors doubles:

$$\sqrt{2s \cdot 2(s-a) \cdot 2(s-b) \cdot 2(s-c)} = \sqrt{16}\sqrt{s(s-a)(s-b)(s-c)} = 4 \times \text{area}(\triangle ABC)$$

And yes, four copies fit. Join the midpoints of the three sides of $\triangle PQR$: this cuts it into four triangles, each with sides $a$, $b$, $c$ — three the same way up and one inverted in the middle.

(iii) $\triangle ABC$ has sides $a$, $b$, $c$; $\triangle PQR$ has sides $3a$, $3b$, $3c$. Show $\triangle PQR$ has $9$ times the area. Do $9$ copies fit?

By the same Heron calculation with a factor of $3$:

$$\sqrt{81}\sqrt{s(s-a)(s-b)(s-c)} = 9 \times \text{area}(\triangle ABC)$$

And yes, nine copies fit. Divide each side of $\triangle PQR$ into three equal parts and draw the lines parallel to the sides through those points. This cuts it into nine small triangles, each with sides $a$, $b$, $c$ — six the same way up and three inverted.

The pattern. Scaling every length by $k$ multiplies the area by $k^2$, and the larger figure can always be tiled by $k^2$ copies of the original.

Question 16

*Fig. 6.43: what fraction of the triangle is shaded? Fig. 6.44: what fraction of the square is shaded?

Solution.

Fig. 6.43 — the triangle

The tick marks show that one side of the triangle is cut into two equal parts and another into three equal parts. Take the apex as $A$, and let $B$, $C$ be the other two vertices.

• $P$ is the midpoint of $AB$;
• $Q$ and $R$ divide $AC$ into three equal parts, with $Q$ nearer $A$.

The shaded quadrilateral is $BPQR$.

Working with position vectors from $A$: $P = \frac{1}{2}\vec{b}$, $Q = \frac{1}{3}\vec{c}$, $R = \frac{2}{3}\vec{c}$, and $B = \vec{b}$.

Adding the signed areas of the quadrilateral $B \to P \to Q \to R$:

$$\tfrac12 \left| \vec{b}\times\tfrac{\vec b}{2} + \tfrac{\vec b}{2}\times\tfrac{\vec c}{3} + \tfrac{\vec c}{3}\times\tfrac{2\vec c}{3} + \tfrac{2\vec c}{3}\times\vec b \right| = \tfrac12\left| 0 + \tfrac16 – 0 – \tfrac23 \right| |\vec b \times \vec c| = \tfrac14|\vec b \times \vec c|$$

and the whole triangle is $\frac12|\vec b \times \vec c|$. So the shaded part is exactly half.

(Check with coordinates $A(0,0)$, $B(-2,-6)$, $C(10,-6)$: the quadrilateral comes to $18$ and the triangle to $36$.)

Fig. 6.44 — the square

The midpoint of each side is marked, and four lines are drawn from each corner to the midpoint of a non-adjacent side, all turning the same way — a pinwheel. The shaded region is the small square they enclose in the middle.

Take the square $(0,0)$, $(4,0)$, $(4,4)$, $(0,4)$ and the four lines

$$y = \tfrac{x}{2}, \quad y = -2x+8, \quad y = \tfrac{x}{2}+2, \quad y = -2x+4$$

Their pairwise intersections are $(3.2, 1.6)$, $(2.4, 3.2)$, $(0.8, 2.4)$ and $(1.6, 0.8)$ — the corners of the inner square. Its side is $\sqrt{0.8^2 + 1.6^2} = \sqrt{3.2}$, so its area is $3.2$ against the big square’s $16$:

$$\frac{3.2}{16} = \frac{1}{5}$$

Question 17

Fig. 6.45: what fraction of the rectangle is covered by the circles? Fig. 6.46: the same for four circles.

Solution. In each figure the circles are all the same size, sit in a row, and each touches the rectangle’s top and bottom edges and its neighbours.

Let each circle have radius $r$. Then each has diameter $2r$, so:

• the rectangle’s height is $2r$;
• its length is $n$ diameters, $2nr$, for $n$ circles.

$$\text{rectangle} = 2nr \times 2r = 4nr^2$$ $$\text{circles} = n \times \pi r^2 = n\pi r^2$$ $$\text{fraction} = \frac{n\pi r^2}{4nr^2} = \frac{\pi}{4} \approx 0.785$$

The $n$ cancels — so the answer is the same for both figures: about $78.5\%$.

Question 18

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for $10$, $20$ and $50$ circles. Then prove it.

Solution. The conjecture. However many circles are fitted in this way, they always cover $\frac{\pi}{4} \approx 78.5\%$ of the rectangle — the fraction does not depend on the number of circles.

Testing it. With radius $r$ throughout:

Circles	Rectangle	Circles’ area	Fraction
$10$	$20r \times 2r = 40r^2$	$10\pi r^2$	$\frac{10\pi}{40} = \frac{\pi}{4}$
$20$	$40r \times 2r = 80r^2$	$20\pi r^2$	$\frac{20\pi}{80} = \frac{\pi}{4}$
$50$	$100r \times 2r = 200r^2$	$50\pi r^2$	$\frac{50\pi}{200} = \frac{\pi}{4}$

The proof. For $n$ circles of radius $r$, the rectangle is $2nr$ by $2r$ and so has area $4nr^2$; the circles have total area $n\pi r^2$. Hence

$$\frac{n\pi r^2}{4nr^2} = \frac{\pi}{4}$$

The $n$ and the $r^2$ both cancel, which is why neither the number of circles nor their size matters. $\blacksquare$

Put another way: each circle sits in its own $2r \times 2r$ square, filling $\frac{\pi r^2}{4r^2} = \frac{\pi}{4}$ of it — and the whole rectangle is just $n$ of those squares side by side.

Question 19

*The figure shows nine identical rectangles fitted together to make a large rectangle whose area is $72 \text{ cm}^2$. Find the perimeter of each small rectangle.

Solution. Read the arrangement off Fig. 6.47: the top row has four rectangles lying on their long sides, and the bottom row has five standing on their short sides.

Let each small rectangle be $\ell$ long and $w$ wide.

The widths must match, since both rows span the big rectangle:

$$4\ell = 5w \qquad \text{(i)}$$

The area gives a second equation. Nine identical rectangles:

$$9\ell w = 72 \quad\Longrightarrow\quad \ell w = 8 \qquad \text{(ii)}$$

Solve. From (i), $w = \frac{4\ell}{5}$. Substituting into (ii):

$$\ell \cdot \frac{4\ell}{5} = 8 \quad\Longrightarrow\quad 4\ell^2 = 40 \quad\Longrightarrow\quad \ell^2 = 10 \quad\Longrightarrow\quad \ell = \sqrt{10}$$

$$w = \frac{4\sqrt{10}}{5}$$

$$\text{Perimeter} = 2(\ell + w) = 2\left( \sqrt{10} + \frac{4\sqrt{10}}{5} \right) = 2 \cdot \frac{9\sqrt{10}}{5} = \frac{18\sqrt{10}}{5} \approx 11.38 \text{ cm}$$

Check: the big rectangle is $4\sqrt{10} \approx 12.65$ cm wide and $w + \ell = \frac{9\sqrt{10}}{5} \approx 5.69$ cm tall, giving an area of $72 \text{ cm}^2$ ✓

Question 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.

Solution. Fig. 6.48 shows lines drawn from one vertex to the two points of trisection of the opposite side.

Let the triangle be $ABC$, with $P$ and $Q$ trisecting $BC$, so

$$BP = PQ = QC = \frac{1}{3}BC$$

The three small triangles $ABP$, $APQ$ and $AQC$ all have:

• equal bases — each one third of $BC$;
• the same height — the perpendicular from $A$ to the line $BC$, common to all three.

Therefore

$$\text{area}(\triangle ABP) = \text{area}(\triangle APQ) = \text{area}(\triangle AQC) = \frac{1}{3}\text{area}(\triangle ABC)$$

The blue and red triangles are two of these three, so their areas are equal. $\blacksquare$

Cutting one to cover the other. Take the blue triangle $ABP$ and the red one $AQC$. They have equal bases ($BP = QC$) and the same height, but different shapes.

A two-piece dissection works:

1. Mark $M$, the midpoint of $AB$, and $N$, the midpoint of $AP$. Cut the blue triangle along $MN$.
2. The top piece is $\triangle AMN$; rotate it through $180°$ about $N$.

This turns the triangle into a parallelogram on base $BP$ with height half that of the triangle. Do the same to the red triangle and you get a parallelogram on base $QC$ with the same height. Since $BP = QC$, the two parallelograms are congruent after a slide, so one set of pieces covers the other.

Question 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.

Solution. Let the square have side $s$.

The quarter circle is centred at one vertex with radius $s$ (it passes through the two adjacent vertices):

$$\text{quarter circle} = \frac{1}{4}\pi s^2$$

The two semicircles each stand on a side of length $s$, so each has radius $\frac{s}{2}$:

$$\text{each} = \frac{1}{2}\pi\left( \frac{s}{2} \right)^2 = \frac{\pi s^2}{8}, \qquad \text{together} = \frac{\pi s^2}{4}$$

The two totals are equal:

$$\text{quarter circle} = \text{two semicircles} = \frac{\pi s^2}{4}$$

Now let $X$ be the region common to the quarter circle and the two semicircles. Then

• region $A$ (the part of the semicircles outside the quarter circle) $=$ (two semicircles) $- X$;
• region $B$ (the part of the quarter circle outside the semicircles) $=$ (quarter circle) $- X$.

Subtracting the same quantity $X$ from two equal quantities leaves equal results:

$$\text{area}(A) = \text{area}(B) \qquad \blacksquare$$

The shapes look nothing alike, and neither area is easy to compute on its own — which is exactly why the subtraction argument is the right one.

Question 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. Find the perimeter and the area of this flower.

Solution. Each semicircle is centred at the midpoint of a side with radius $1$ (half the side), so each passes through the centre of the square — the distance from a side’s midpoint to the centre is exactly $1$.

Where the arcs meet. Adjacent semicircles cross at a corner of the square and at the centre. So each semicircle is split by the centre into two quarter-circle arcs of radius $1$.

Perimeter. The flower has four petals, each bounded by two of those quarter arcs — eight arcs in all:

$$P = 8 \times \frac{1}{4}\left( 2\pi \times 1 \right) = 8 \times \frac{\pi}{2} = 4\pi \approx 12.57 \text{ units}$$

Area. Take one petal. It is bounded by two arcs on the chord from a corner to the centre, of length $\sqrt2$. For each arc, the chord subtends $90°$ at that semicircle’s centre, so the circular segment between chord and arc is

$$\text{sector} – \text{triangle} = \frac{1}{4}\pi(1)^2 – \frac{1}{2}(1)(1) = \frac{\pi}{4} – \frac{1}{2}$$

A petal is two such segments:

$$\text{petal} = 2\left( \frac{\pi}{4} – \frac{1}{2} \right) = \frac{\pi}{2} – 1$$

Four petals:

$$\text{flower} = 4\left( \frac{\pi}{2} – 1 \right) = 2\pi – 4 \approx 2.28 \text{ square units}$$

(With $\pi = \frac{22}{7}$: area $= \frac{44}{7} – 4 = \frac{16}{7} \approx 2.29$ and perimeter $= \frac{88}{7} \approx 12.57$.)

Question 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 $\frac{1}{4}\pi l^2$.

Solution. Let the larger circle have radius $R$ and the smaller radius $r$. The region between them — the annulus — has area

$$\pi R^2 – \pi r^2 = \pi\left( R^2 – r^2 \right)$$

So everything turns on finding $R^2 – r^2$, and the chord provides it.

Use the tangency. $BC$ touches the smaller circle at $A$, so $OA \perp BC$ and $OA = r$. But $OA$ is also the perpendicular from the centre $O$ to the chord $BC$ of the larger circle — and a perpendicular from the centre bisects its chord. So

$$BA = AC = \frac{l}{2}$$

Apply Pythagoras in the right-angled triangle $OAB$:

$$OB^2 = OA^2 + AB^2$$ $$R^2 = r^2 + \left( \frac{l}{2} \right)^2$$ $$R^2 – r^2 = \frac{l^2}{4}$$

Substitute:

$$\text{annulus} = \pi\left( R^2 – r^2 \right) = \pi \cdot \frac{l^2}{4} = \frac{1}{4}\pi l^2 \qquad \blacksquare$$

Neither $R$ nor $r$ appears in the answer — the chord length alone determines the area between the circles, however large or small the pair happens to be.

Question 24

*In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle. Show that Area $(A) +$ Area $(B) =$ Area $(C)$.

Solution. Let the right-angled triangle have legs $a$, $b$ and hypotenuse $c$, so that $a^2 + b^2 = c^2$.

A semicircle drawn on a segment of length $s$ as diameter has radius $\frac{s}{2}$ and area

$$\frac{1}{2}\pi\left( \frac{s}{2} \right)^2 = \frac{\pi s^2}{8}$$

So the three semicircles have areas

$$A = \frac{\pi a^2}{8}, \qquad B = \frac{\pi b^2}{8}, \qquad C = \frac{\pi c^2}{8}$$

Adding the first two:

$$A + B = \frac{\pi}{8}\left( a^2 + b^2 \right) = \frac{\pi}{8}c^2 = C \qquad \blacksquare$$

The key is that the same constant $\frac{\pi}{8}$ multiplies each square, so the Baudhāyana–Pythagoras relation between the squares carries straight over to the semicircles. The same works for equilateral triangles, or any family of similar shapes built on the three sides — the theorem is not really about squares at all.

Question 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 $r$.

Solution. The shaded region is the lens where the two circles overlap. Let their centres be $A$ and $B$, and let them cross at $C$ and $D$.

The key distances. Each circle passes through the other’s centre, so

$$AB = r$$

and since $AC = BC = r$ as well, $\triangle ABC$ is equilateral — giving $\angle CAB = 60°$. By symmetry $\angle DAB = 60°$ too, so

$$\angle CAD = 120°$$

Split the lens in half. The chord $CD$ divides the lens into two equal pieces, one inside each circle. Take the piece belonging to circle $A$: it is the circular segment cut off by $CD$.

$$\text{sector } CAD = \frac{120}{360}\pi r^2 = \frac{\pi r^2}{3}$$

For the triangle $ACD$: the chord $CD$ has length $2 \times r\sin 60° = r\sqrt3$, and $A$ is at distance $\frac{r}{2}$ from it, so

$$\text{area}(\triangle ACD) = \frac{1}{2} \times r\sqrt3 \times \frac{r}{2} = \frac{\sqrt3}{4}r^2$$

$$\text{segment} = \frac{\pi r^2}{3} – \frac{\sqrt3}{4}r^2$$

Double it for the whole lens:

$$\text{lens} = 2\left( \frac{\pi r^2}{3} – \frac{\sqrt3}{4}r^2 \right) = \frac{2\pi r^2}{3} – \frac{\sqrt3}{2}r^2 = r^2\left( \frac{2\pi}{3} – \frac{\sqrt3}{2} \right)$$

Numerically that is about $1.228\,r^2$.

Question 26

*In Fig. 6.54 we see three triangles within a rectangle, of areas $A$, $B$, $C$ as marked. Show that the area of the rectangle is $\dfrac{2(A+C)(B+C)}{C}$.

Solution. Set up coordinates on the rectangle: let it be $w$ wide and $h$ tall, with its bottom-left corner at the origin. The figure has a point $T$ on the top edge and a point $U$ directly below it inside, with a point on the right edge level with $U$. Write $T = (t, h)$ and $U = (t, u)$.

The three triangles.

$$A = \frac{1}{2}t(h-u) \qquad C = \frac{1}{2}(w-t)(h-u) \qquad B = \frac{1}{2}(w-t)u$$

Now form the two sums in the formula.

$$A + C = \frac{1}{2}(h-u)\big[ t + (w – t) \big] = \frac{1}{2}w(h-u)$$ $$B + C = \frac{1}{2}(w-t)\big[ u + (h-u) \big] = \frac{1}{2}h(w-t)$$

Each sum simplifies beautifully because the bracket collapses to a full side.

Divide as the formula says:

$$\frac{2(A+C)(B+C)}{C} = \frac{2 \cdot \frac{1}{2}w(h-u) \cdot \frac{1}{2}h(w-t)}{\frac{1}{2}(w-t)(h-u)}$$

$$= \frac{\frac{1}{2}wh(h-u)(w-t)}{\frac{1}{2}(w-t)(h-u)} = wh \qquad \blacksquare$$

Both $(h-u)$ and $(w-t)$ cancel, leaving exactly the rectangle’s area.

Question 27

*In Fig. 6.55 we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.

Solution. This is the lune of Hippocrates, one of the oldest results in mathematics — and the first curved region ever to have its area found exactly.

Set up the figure: the large semicircle has diameter $AC$ and centre $O$, with radius $R$. The point $B$ is the top of that semicircle, directly above $O$, so $OB = R$ and $OB \perp AC$. A smaller semicircle is drawn on $AB$ as its diameter, passing through $E$.

The two shaded regions are:

• the lune — between the small semicircle’s arc $AEB$ and the large semicircle’s arc $AB$;
• the triangle $AOB$.

Find the small semicircle. Since $\triangle AOB$ is right-angled at $O$ with legs $R$ and $R$,

$$AB = \sqrt{R^2 + R^2} = R\sqrt2$$

so the small semicircle has radius $\frac{R\sqrt2}{2} = \frac{R}{\sqrt2}$ and area

$$\frac{1}{2}\pi\left( \frac{R}{\sqrt2} \right)^2 = \frac{1}{2}\pi \cdot \frac{R^2}{2} = \frac{\pi R^2}{4}$$

Find the segment. The lune is the small semicircle minus the part of it that lies inside the big semicircle — which is exactly the circular segment of the big circle cut off by the chord $AB$. Since $\angle AOB = 90°$:

$$\text{segment} = \text{sector } AOB – \triangle AOB = \frac{1}{4}\pi R^2 – \frac{1}{2}R^2$$

Subtract.

$$\text{lune} = \frac{\pi R^2}{4} – \left( \frac{\pi R^2}{4} – \frac{R^2}{2} \right) = \frac{R^2}{2}$$

The $\pi$ terms cancel completely — which is the remarkable part. And

$$\text{area}(\triangle AOB) = \frac{1}{2} \times R \times R = \frac{R^2}{2}$$

So the two shaded regions are equal, each $\frac{R^2}{2}$. $\blacksquare$

What makes this famous is that a region bounded entirely by curves has been shown to have exactly the same area as a straight-sided triangle, with no $\pi$ left over at all.

Common mistakes

• Question 2, using $15$ cm as the height. $15$ cm is the slant side. The height comes from the right-angled triangle with hypotenuse $15$ and base $5$.
• Question 5, using the ratio numbers as the sides. $2$, $3$, $4$ have perimeter $9$, not $45$. The multiplier $k$ must be found first.
• Question 6, giving the same method twice. The question asks for two different ways; Heron’s formula and the right-angle shortcut are genuinely different routes.
• Questions 7 and 9, mixing units. The circumference is in centimetres, so the journey must be too: $1$ km $= 100{,}000$ cm.
• Question 8, treating $66$ cm as the radius or diameter. It is the circumference.
• Question 10, answering “no” by analogy with triangles. For rectangles the answer is yes, because two numbers are determined by their sum and product. The analogy with triangles is what misleads here.
• Question 15, assuming doubled sides double the area. Doubling every length quadruples the area, and tripling multiplies it by nine.
• Question 16, guessing from the picture. Neither fraction is obvious by eye — $\frac12$ and $\frac15$ both need the tick marks read and a calculation done.
• Question 17, thinking more circles cover more. The fraction $\frac{\pi}{4}$ is the same for three circles, four, or fifty, because each circle fills the same fraction of its own square.
• Question 19, assuming the small rectangles are squares. They are not: four along the top must match five along the bottom, which forces $4\ell = 5w$.
• Question 21, trying to compute $A$ and $B$ separately. Neither has a simple formula. The argument is that they are equal differences from two equal totals.
• Question 22, treating each petal as a semicircle. A petal is two circular segments, not a semicircle — a segment being a sector minus a triangle.
• Question 23, trying to find $R$ and $r$ individually. They cannot be found, and need not be: only the difference $R^2 – r^2$ appears, and the tangent chord supplies it.
• Question 25, giving the union of the two circles. The shaded region in Fig. 6.53 is the overlap — the lens where they cross.
• Question 27, expecting $\pi$ in the answer. The whole point is that the $\pi$ terms cancel, leaving a region bounded by curves with exactly the area of a triangle.

Practise next

• Chapter 7, The Mathematics of Maybe — the next chapter, an introduction to probability.
• Exercise Set 6.3 — worth revisiting alongside Questions 21 to 27, since every one of them is built from sectors and segments.
• Exercise Set 6.2, Questions 6 to 11 — the equal-area arguments that Questions 20 and 26 here extend.