Class 9 Mathematics Chapter 6: Measuring Space: Perimeter and Area

The perimeter is the distance once round the boundary of the shape, so it is a length and is measured in units such as centimetres or metres. The area is the amount of flat space the shape covers, and it is measured in square units such as square centimetres or square metres.

The fencing follows the perimeter, which for a rectangle is $2(l + b)$. Here that is $2(23 + 14) = 2 \times 37 = 74$ metres of fencing.

The number $\pi$ is the ratio of the circumference of any circle to its diameter, and it comes out the same for every circle. It is irrational, meaning it cannot be written as a fraction of two whole numbers at all, so $\dfrac{22}{7}$ can only be an approximation. In fact $\dfrac{22}{7} = 3.142857\ldots$, slightly above the true $3.14159\ldots$.

Using $C = 2\pi r$, the circumference is $2 \times \dfrac{22}{7} \times 21$. The 7 divides into the 21 to leave 3, so the answer is $2 \times 22 \times 3 = 132$ cm.

From $2\pi r = 132$ we get $r = \dfrac{132 \times 7}{2 \times 22} = \dfrac{924}{44} = 21$ metres.

All circles have the same shape and differ only in size, so enlarging one circle into another multiplies every length in the figure by the same factor. Doubling the radius, for instance, doubles the diameter and also doubles the distance once round. Since the circumference and the diameter are stretched by exactly the same factor, their ratio is left unchanged, and it therefore takes one fixed value for every circle in existence. That value is what we call $\pi$.

The whole circumference is $2 \times \dfrac{22}{7} \times 21 = 132$ cm. The arc covers $\dfrac{120}{360} = \dfrac{1}{3}$ of the full turn, so it is one third of the circle. Its length is therefore $\dfrac{132}{3} = 44$ cm.

The curved edge is half the circumference, which is $\dfrac{1}{2} \times 2 \times \dfrac{22}{7} \times 7 = 22$ cm. But a half-disc is not closed by its arc alone: the straight diameter runs across the flat side, and that measures $2 \times 7 = 14$ cm. The whole boundary is therefore $22 + 14 = 36$ cm. A piece of string laid round the shape would have to travel along the straight edge as well as the curve.

Take a square of side 6 and a rectangle measuring 10 by 2. The square has perimeter $4 \times 6 = 24$ and the rectangle has perimeter $2(10 + 2) = 24$, so the two boundaries are equal. But the square covers $6 \times 6 = 36$ square units while the rectangle covers only $10 \times 2 = 20$. The same fencing therefore encloses almost twice as much ground in one case as in the other, which shows that perimeter does not determine area.

Since $C = \pi d$, the diameter is $d = 176 \times \dfrac{7}{22} = 56$ cm, and the radius is half of that, namely 28 cm. A student who divides the circumference by $\pi$ and calls the answer the radius has forgotten that the division returns the diameter; one further halving is needed.

(i) The perimeter is the distance once round the boundary of a shape. Being a length, it is measured in units such as centimetres or metres.

(ii) A square of side $a$ has perimeter $4a$, and a rectangle of length $l$ and breadth $b$ has perimeter $2(l + b)$.

(iii) There is no formula. You follow the outside edge all the way round and add every edge you cross. A line drawn inside the figure to cut it into rectangles does not count, because the perimeter follows only the outside.

(iv) $\pi$ is the ratio of the circumference of a circle to its diameter, and that ratio comes out the same for every circle.

(v) Because $\pi$ is irrational: it cannot be written as a fraction of two whole numbers, and its decimal expansion never ends and never settles into a repeating block. So any fraction or terminating decimal used in its place, such as $\dfrac{22}{7}$ or $3.14$, must differ from it.

(i) $C = 2 \times \dfrac{22}{7} \times 14 = 88$ metres of edging.

(ii) From $2\pi r = 220$, $r = \dfrac{220 \times 7}{44} = 35$ metres.

(iii) A quarter of the edge is $\dfrac{88}{4} = 22$ metres.

(iv) The quarter piece is bounded by that arc and by two straight radii of 14 m each, so its boundary is $22 + 28 = 50$ metres.

(v) Part (iii) asks only for the curved edge, while part (iv) asks for the whole way round a closed region. The quarter piece is not closed by its arc alone; the two radii are needed to shut it, and they add 28 metres.

(i) The arc length is $\dfrac{\theta}{360} \times 2\pi r$.

(ii) A full turn is $360°$, so an arc that subtends $\theta$ at the centre covers $\dfrac{\theta}{360}$ of the whole way round. Its length is that same fraction of the circumference.

(iii) A semicircular arc is $\pi r$ and a quarter arc is $\dfrac{\pi r}{2}$.

(iv) The circumference is $2 \times \dfrac{22}{7} \times 42 = 264$ cm, and $\dfrac{30}{360} = \dfrac{1}{12}$ of that is 22 cm.

(v) The ratio is $1 : 2$. Arc length is a fixed fraction of one circumference, and that fraction is proportional to the angle at the centre, so doubling the angle doubles the arc.

(i) A square of side 6 has perimeter 24 and area 36. A rectangle of 10 by 2 also has perimeter $2(10 + 2) = 24$, and its area is 20.

(ii) The square encloses more, by $36 - 20 = 16$ square units.

(iii) The square. For a fixed perimeter, no other rectangle beats it.

(iv) The circle, which beats every polygon of the same perimeter.

(v) The area falls steadily towards zero. The perimeter is unchanged throughout, so this shows plainly that knowing the perimeter tells you nothing on its own about how much space a shape encloses.

The area of a rectangle is its length times its breadth, $l \times b$. The unit squares inside it lie in rows as long as the length, and there are as many rows as the breadth, so the count is the product of the two.

The area is $b \times h = 15 \times 8 = 120$ square cm. The height is the perpendicular distance between the base and the side parallel to it, not the length of the slanting side.

A copy of any triangle, turned through half a turn, fits alongside the original to make a parallelogram on the same base and of the same height. The triangle is therefore exactly half that parallelogram, whose area is $b \times h$, so the triangle covers $\tfrac{1}{2} \times b \times h$.

The semi-perimeter is half the distance once round the triangle, so $s = \dfrac{a+b+c}{2}$. Heron's formula then gives the area as $\sqrt{s(s-a)(s-b)(s-c)}$.

The semi-perimeter is $s = \dfrac{5+12+13}{2} = 15$. Heron's formula then gives $\sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30$ square cm.

Drop a perpendicular from one vertex to the base and cut along it, which removes a right-angled triangle from one end of the parallelogram. Slide that triangle across to the other end; because the opposite sides are equal and parallel, it fits there exactly. What remains is a rectangle standing on the same base and reaching the same height as the original. No part was added and none thrown away, so the two shapes cover the same amount of space. The area of the parallelogram is therefore the area of that rectangle, namely base times height.

Taking the 20 cm side as the base, the area is $20 \times 9 = 180$ square cm. The same parallelogram can equally be measured on the 15 cm base, and the area must come out the same, so $15 \times h = 180$ and the second height is $h = 12$ cm.

The theorem is that a median of a triangle divides it into two triangles of equal area. To prove it, let $AD$ be the median of triangle $ABC$, so $D$ is the midpoint of $BC$. The triangles $ABD$ and $ACD$ have equal bases, since $BD = DC$, and they have the same height, because both have their apex at $A$ and stand on the same line $BC$. Equal base and equal height give equal area. The result is worth remarking on because the two pieces are in general not congruent: they are different shapes, and yet they cover exactly the same amount of space.

The semi-perimeter is $s = \dfrac{8+15+17}{2} = 20$, so Heron gives $\sqrt{20 \times 12 \times 5 \times 3} = \sqrt{3600} = 60$ square metres. As a check, $8^2 + 15^2 = 64 + 225 = 289 = 17^2$, so the triangle is right-angled with legs 8 and 15. Half the product of the legs is $\tfrac{1}{2} \times 8 \times 15 = 60$, which agrees.

The area is $\dfrac{\sqrt{3}}{4} a^2$. For $a = 6$ this gives $\dfrac{\sqrt{3}}{4} \times 36 = 9\sqrt{3}$ square cm. The answer should be left as $9\sqrt{3}$ because $\sqrt{3}$ is irrational: any decimal written for it is only an approximation, and rounding it here and again later would lose accuracy for no gain.

(i) A rectangle of length $l$ and breadth $b$ has area $l \times b$. A parallelogram of base $b$ and perpendicular height $h$ has area $b \times h$.

(ii) The height is the perpendicular distance between the base and the side parallel to it. It is measured at right angles to the base, and it is not the length of the slanting side.

(iii) No. The 8 cm is the slanting side, not the perpendicular height, and the height is always shorter than it. Without being told the height, the area cannot be worked out at all.

(iv) A triangle of base $b$ and height $h$ has area $\tfrac{1}{2} \times b \times h$.

(v) Take a copy of the triangle and turn it through half a turn. The two fit together into a parallelogram on the same base and of the same height, so the triangle is exactly half of it.

(i) The semi-perimeter is $s = \dfrac{a+b+c}{2}$, half the way round. The area is $\sqrt{s(s-a)(s-b)(s-c)}$.

(ii) When all three sides are known but the height is not, so that the half-base-times-height formula cannot be used directly.

(iii) $s = \dfrac{13+14+15}{2} = 21$, so the area is $\sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$ square cm.

(iv) Any two sides of a triangle must add to more than the third, and $3 + 4 = 7$ falls short of 9. The two short sides could never meet, so no triangle closes up. Heron's formula would ask for the square root of a negative number.

(v) For a right-angled triangle the area is also half the product of the two legs, so the two methods must agree. Sides 6, 8 and 10 give $s = 12$ and $\sqrt{12 \times 6 \times 4 \times 2} = 24$, which is $\tfrac{1}{2} \times 6 \times 8$ exactly.

(i) The area is $24 \times 10 = 240$ square metres.

(ii) The same area must come out on the other base, so $30 \times h = 240$ and $h = 8$ metres.

(iii) The semi-perimeter is $s = 18$, so the area is $\sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54$ square metres. As a check, $9^2 + 12^2 = 225 = 15^2$, so the triangle is right-angled and $\tfrac{1}{2} \times 9 \times 12 = 54$ as well.

(iv) That path is a median, and a median divides a triangle into two of equal area, so each part is 27 square metres.

(v) $\dfrac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}$ square metres.

(i) A median of a triangle divides it into two triangles of equal area.

(ii) Let $AD$ be a median of triangle $ABC$, so that $D$ is the midpoint of $BC$. Triangles $ABD$ and $ACD$ have equal bases because $BD = DC$, and they have the same height, since both have their apex at $A$ and their bases along the same line. Equal base with equal height gives equal area.

(iii) Congruence would need all three sides to match, and in a triangle that is not isosceles about that median the two remaining sides $AB$ and $AC$ differ. The pieces therefore have different shapes.

(iv) A triangle with sides 5, 5 and 6 and one with sides 5, 5 and 8. Heron gives 12 for each, yet their shapes are quite different.

(v) Area counts how much flat space is covered, and space can be redistributed by cutting and rearranging without any of it being lost. Two figures may therefore cover the same amount while looking nothing alike, which is why equal area never by itself proves congruence.

To square a shape meant to construct, with straight edge and compasses alone, a square covering exactly the same area as the shape. A rectangle can be squared, and Baudhāyana gave a construction for it around 800 BCE. A circle cannot be squared at all, as was finally proved after more than two thousand years of attempts.

The identity is $\left(\dfrac{a+b}{2}\right)^2 - \left(\dfrac{a-b}{2}\right)^2 = ab$. For $a = 9$ and $b = 4$ the two halves are $6.5$ and $2.5$, and $6.5^2 - 2.5^2 = 42.25 - 6.25 = 36$, which is indeed $9 \times 4$.

Using $A = \pi r^2$, the area is $\dfrac{22}{7} \times 21 \times 21$. The 7 divides into one of the 21s to leave 3, so the answer is $22 \times 3 \times 21 = 1386$ square cm.

From $\dfrac{22}{7} r^2 = 616$ we get $r^2 = \dfrac{616 \times 7}{22} = 196$, and taking the square root gives $r = 14$ metres.

The circumference is $C = 2\pi r$ and the area is $A = \pi r^2$. The circumference is a length and is measured in units such as centimetres or metres; the area is measured in square units such as square centimetres or square metres.

Cut the disc into a large number of equal sectors and lay them out in a row, turning every second one upside down so that they interlock. Nothing has been added and nothing thrown away, so the strip covers exactly the same area as the disc did. The strip is very nearly a rectangle: it is $r$ high, since every slice reaches from the centre out to the rim. It is $\pi r$ wide, since half of the curved edges lie along the top and half along the bottom, making the width half the circumference. A rectangle $\pi r$ by $r$ has area $\pi r^2$, and the more slices are taken the closer the strip is to a true rectangle.

The area of the rectangle is $ab$, and the identity $\left(\dfrac{a+b}{2}\right)^2 - \left(\dfrac{a-b}{2}\right)^2 = ab$ writes that area as a difference of two squares. A right-angled triangle supplies exactly such a difference, because the square on one leg equals the square on the hypotenuse minus the square on the other leg. The construction therefore builds a right-angled triangle whose hypotenuse is $\dfrac{a+b}{2}$ and one of whose legs is $\dfrac{a-b}{2}$. The remaining leg is then $\sqrt{ab}$, and the square on it matches the rectangle.

The inner radius is 7 m and the outer radius is $7 + 7 = 14$ m. The path is the ring between the two circles, so its area is $\pi(R^2 - r^2) = \dfrac{22}{7}(196 - 49) = \dfrac{22}{7} \times 147 = 462$ square metres.

The whole disc covers $\dfrac{22}{7} \times 441 = 1386$ square cm. The sector covers $\dfrac{120}{360} = \dfrac{1}{3}$ of that, which is 462 square cm. The rest of the disc is therefore $1386 - 462 = 924$ square cm.

The area is $\pi r^2$, so doubling the radius replaces $r^2$ by $(2r)^2 = 4r^2$ and multiplies the area by four. The common mistake this explains is putting the diameter into the formula in place of the radius: the diameter is twice the radius, so the answer comes out four times too large. Writing $r = d \div 2$ on a line of its own prevents it.

(i) It means constructing, with straight edge and compasses alone, a square that covers exactly the same area as the given rectangle.

(ii) $\left(\dfrac{a+b}{2}\right)^2 - \left(\dfrac{a-b}{2}\right)^2 = ab$.

(iii) With $a = 16$ and $b = 9$ the two halves are $12.5$ and $3.5$, and $12.5^2 - 3.5^2 = 156.25 - 12.25 = 144$, which is $16 \times 9$. The square therefore has side $\sqrt{144} = 12$.

(iv) The identity writes the area as a difference of two squares, and the theorem of the right triangle produces exactly such a difference: the square on one leg is the square on the hypotenuse minus the square on the other leg.

(v) No. Squaring the circle is impossible with straight edge and compasses, as was finally proved after more than two thousand years.

(i) For any one shape, the ratio of the square of the perimeter to the area is a fixed number that does not depend on the size. For a square of side $a$ it is $\dfrac{16a^2}{a^2} = 16$. Applying the same reasoning to circles, and using $C = 2\pi r$, forces the area to be some fixed multiple of $r^2$.

(ii) Cut the disc into many equal sectors and lay them in a row with every second one turned over so that they interlock. Nothing is added or thrown away, so the area is unchanged.

(iii) The height is $r$, because every slice reaches from the centre to the rim. The width is $\pi r$, because half the curved edges lie along the top and half along the bottom, making the width half the circumference.

(iv) $A = \pi r \times r = \pi r^2$.

(v) With few slices the top and bottom edges are visibly bumpy and the strip is only roughly a rectangle. The more slices are taken, the straighter those edges become, so the strip approaches a true rectangle and the area approaches $\pi r^2$ exactly.

(i) $\dfrac{22}{7} \times 196 = 616$ square metres.

(ii) The outer radius is the inner radius plus the width of the paving, $14 + 7 = 21$ metres.

(iii) The ring covers $\dfrac{22}{7}(441 - 196) = \dfrac{22}{7} \times 245 = 770$ square metres.

(iv) A quarter of 616 is 154 square metres.

(v) The 14 m is the radius of the fountain itself, which is the inner edge of the paving. Using it as the outer radius would make the ring have zero width, giving an area of nothing at all. The outer radius must reach right across the paving as well.

(i) $C = 2\pi r$ and $A = \pi r^2$.

(ii) The circumference is a length and is measured in centimetres or metres. The area is measured in square centimetres or square metres, because it counts unit squares rather than unit lengths.

(iii) $r^2 = \dfrac{1386 \times 7}{22} = 441$, so $r = 21$ cm.

(iv) The value $r^2$ is the square of the radius, not the radius itself. Stopping there gives 441 in place of 21, an answer more than twenty times too large, and in the wrong units besides.

(v) The area depends on the square of the radius, so trebling $r$ multiplies the area by $3^2 = 9$.

The perimeter is the distance once round the boundary of the shape, so it is a length and is measured in units such as centimetres or metres. The area is the amount of flat space the shape covers, and it is measured in square units such as square centimetres or square metres.

The fencing follows the perimeter, which for a rectangle is $2(l + b)$. Here that is $2(23 + 14) = 2 \times 37 = 74$ metres of fencing.

The number $\pi$ is the ratio of the circumference of any circle to its diameter, and it comes out the same for every circle. It is irrational, meaning it cannot be written as a fraction of two whole numbers at all, so $\dfrac{22}{7}$ can only be an approximation. In fact $\dfrac{22}{7} = 3.142857\ldots$, slightly above the true $3.14159\ldots$.

The area of a rectangle is its length times its breadth, $l \times b$. The unit squares inside it lie in rows as long as the length, and there are as many rows as the breadth, so the count is the product of the two.

The area is $b \times h = 15 \times 8 = 120$ square cm. The height is the perpendicular distance between the base and the side parallel to it, not the length of the slanting side.

A copy of any triangle, turned through half a turn, fits alongside the original to make a parallelogram on the same base and of the same height. The triangle is therefore exactly half that parallelogram, whose area is $b \times h$, so the triangle covers $\tfrac{1}{2} \times b \times h$.

To square a shape meant to construct, with straight edge and compasses alone, a square covering exactly the same area as the shape. A rectangle can be squared, and Baudhāyana gave a construction for it around 800 BCE. A circle cannot be squared at all, as was finally proved after more than two thousand years of attempts.

The identity is $\left(\dfrac{a+b}{2}\right)^2 - \left(\dfrac{a-b}{2}\right)^2 = ab$. For $a = 9$ and $b = 4$ the two halves are $6.5$ and $2.5$, and $6.5^2 - 2.5^2 = 42.25 - 6.25 = 36$, which is indeed $9 \times 4$.

Using $A = \pi r^2$, the area is $\dfrac{22}{7} \times 21 \times 21$. The 7 divides into one of the 21s to leave 3, so the answer is $22 \times 3 \times 21 = 1386$ square cm.