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Measuring Space: Perimeter and Area — Class 9 Mathematics

Ganita Manjari · Chapter 6

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Chapter summary

This chapter measures the boundary and the region of plane shapes. The perimeter is the total distance around a shape: for a square of side , and for a rectangle. For a circle the perimeter is the circumference , where is the constant ratio of circumference to diameter; historical estimates include Archimedes' bound , Aryabhata's and Madhava's exact infinite series. Area measures the surface enclosed: a rectangle has area and a triangle . When the height is not known, Heron's formula gives the area of a triangle from its sides: , where is the semi-perimeter. The area of a circle is , explained by Nilakantha as equal to a triangle with base and height . The chapter also treats the area of a sector as a fraction of the whole circle and applies these ideas to real problems such as relay-race lane staggers.

Perimeter of shapesCircumference and \(\pi\)Area of rectangle and triangleHeron's formulaArea of circle and sector

Key ideas & terms

Perimeter: The total length of the boundary of a closed plane figure.
Circumference: The perimeter of a circle, .
\(\pi\) (pi): The constant ratio of a circle's circumference to its diameter, approximately .
Heron's formula: The area of a triangle from its sides: with .
Semi-perimeter: Half the perimeter of a triangle, .
Sector: A part of a circle bounded by two radii and the arc between them.

Extra questions & answers

What is the perimeter of a square of side ?

It is , the sum of all four equal sides.

Write the formula for the circumference of a circle of radius .

The circumference is , equivalently for diameter .

What is the area of a triangle with base and height ?

It is .

State Heron's formula for the area of a triangle.

Area , where is the semi-perimeter.

Find the area of a triangle with sides using Heron's formula.

Here ; area square units.

What is the area of a circle of radius ?

It is .

How is the area of a circle related to a triangle?

It equals the area of a triangle whose base is the circumference and height is the radius , giving .

What value did Aryabhata give for ?

He gave , describing it as an approximation (asanna).

What is the perimeter of a rectangle of length and width ?

It is .

How is the area of a sector found?

As the fraction of the whole circle's area .

Think and Reflect

In 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?

No, if the lane width is the same the extra distance on the curved part depends only on the lane width, not on the radius of the inner lane. For a full circle the extra circumference from lane to lane is . A 200 m track also has two semicircles making one full circle, so for the part of the race run on the bends the stagger per lane is still for a full lap and for a half lap. If the school keeps the standard width the stagger for one lap is the same as on a 400 m track. Only if the whole track including lane width is scaled down proportionally would the stagger be proportionally smaller.

Think and Reflect

What is the connection between this question and the one about the 400 m athletics track?

The athletics track has two straight parts and two semicircular ends. The extra distance a runner in an outer lane has to run on the bends is the difference of the circumferences of two concentric semicircles. Hence to compute the stagger we must know the perimeter (circumference) of a circle, i.e. .

HOME MEASUREMENT

You can do a simple measurement at home to estimate the C/D ratio. Take a cotton reel with thin thread around it. Measure the diameter D of the reel as accurately as possible. Unwrap and then tightly wrap the thread around the reel 20 times. Unwrap it again; measure its length L, and calculate L/20D. This is the ratio we want. For accuracy, the thread should be very thin. Please do the experiment! Do you get a ratio between 3 and 4? Between 3.1 and 3.2?

With careful measurement is about . Hence . In practice you obtain and with a thin thread and good measurement , i.e. about .

It is also possible to estimate the C/D ratio using pure geometry, i.e., without any measurements at all! Can you imagine how?

By inscribing and circumscribing regular polygons about the circle. The circumference lies between the perimeters of the inscribed and circumscribed polygons. As the number of sides increases the two perimeters squeeze the value of to . This is the method used by Archimedes with 6, 12, 24, 48 and 96-gons.

Think and Reflect

What 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?

Lane width . Hence radius for lane 1 (running line) is , for lane 2 it is . Difference . Extra distance on the two semicircles = one full circumference difference . For a 400 m race which is one full lap the stagger between any two adjacent lanes is the same, , so lane 3 needs the same ahead of lane 2, lane 4 the same ahead of lane 3, etc. If only a half lap is run on the bend the stagger is half that, .

Think and Reflect

What 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’?

Translate the parallelogram along the base by a small step . The triangle is congruent to , so area is preserved. Repeating the step enough times brings the altitude foot onto the base, turning the very oblique parallelogram into one where the usual rectangle dissection works, proving area still holds.

Think and Reflect

The 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?

No. For fixed side lengths and the area where is the angle between them. As decreases the height decreases and so does the area. Hence sides alone do not determine the area; the height or angle is needed.

Think and Reflect

Since 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?

Yes. The two triangles have equal base and same height , so equal area . By the Wallace-Bolyai-Gerwien theorem any two equal-area polygons are scissors-congruent. For these two adjacent triangles on the same median 2 pieces suffice: cut by a line through the midpoint of parallel to the median and reassemble to form .

Think and Reflect

Suppose 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.

Yes. The Wallace-Bolyai-Gerwien theorem states that any two polygons of equal area are equidecomposable: one can be cut by straight lines into finitely many polygonal pieces that rearranged exactly form the other. This works for 1, 2 and 3. Conjecture: Two polygons have equal area iff they are scissors-congruent.

Think of various rectangles with perimeter 40 units (the sides do not have to be integers). 1. How many such rectangles are there?

Infinitely many. If sides are and with all have perimeter .

Think of various rectangles with perimeter 40 units (the sides do not have to be integers). 2. Among them, is there one whose area is the largest? What are its dimensions?

Yes. Area is maximal when , i.e. the square . Maximum area . This is AM-GM.

Think of various rectangles with perimeter 40 units (the sides do not have to be integers). 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?

No smallest positive area exists. As or , . You can make area arbitrarily small but not zero. It is surprising that among all perimeters the square maximizes area and there is no minimum.

Think and Reflect

What 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. Think carefully. How would you proceed?

First transform the triangle to a rectangle of equal area: area . Take rectangle with sides and (or and ). Then square that rectangle by Baudhāyana's construction. Construct with , , draw square and transfer length with arc to get side . That square has area , equal to triangle.

Exercise Set 6.1

1. The perimeter of a circle is 44 cm. What is its radius?

. With , .

2. Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm

(44.0 cm to 3 s.f.). With , .

2. Calculate, correct to 3 significant figures, the circumference of a circle with: (ii) radius 10 cm

( with ).

2. Calculate, correct to 3 significant figures, the circumference of a circle with: (iii) radius 12 cm

( with ).

3. Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60°

(with , ).

3. Calculate the length of the arc of a circle if: (ii) the radius is 6.3 m and the angle at the centre is 120°

(with , ).

4. Find 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°.

Arc . Add two radii . Total .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (i) 60 m by 80 m shape

Shape (i) is 60 m +80 m straight plus three-quarter arc radius 60 m? Approx perimeter with . General method: add straight edges plus .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (ii) 8 cm by 12 cm shape

Perimeter plus quarter-circle arcs: yields . Exact using method: straight plus arc lengths.

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (iii) 10 cm shape

Shape (iii) semicircular end radius : perimeter .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (iv) 12 cm shape

Shape (iv) three-quarter circle radius plus two radii: .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (v) 14 cm shape

Shape (v) full circle radius partly: .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (vi) 28 cm shape

Shape (vi) two semicircles diameter 28: plus straight edges as shown.

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (vii) 8 cm by 6 cm shape

Shape (vii) rectangle 8×6 with semicircular ends: .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (viii) 4 cm 4 cm 4 cm shape

Shape (viii) three semicircles diameter 4: .

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14): (ix) 10 cm 10 cm shape

Shape (ix) two quarter circles radius 10: .

6. If 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?

One revolution .

6. If the diameter of a car tyre is 56 cm, then: (ii) How many revolutions does the tyre make if the car travels 10 km?

. Revolutions , about revolutions.

7. Find the total perimeter of all the petals in each of the given flowers. (i) Fig. 6.15A: The centres of the arcs are the midpoints of the sides of the square 14 cm

Square side , petal radius . Four semicircles make two full circles: total petal perimeter . If counting each petal's outer arc alone, 4 arcs each , total .

7. Find the total perimeter of all the petals in each of the given flowers. (ii) Fig. 6.15B: The centres of the arcs are the vertices of the hexagon 42 cm

The centres of the arcs are the six vertices of a regular hexagon of side . Two adjacent vertices are apart and each arc has radius ; the two radii and the side form an equilateral triangle, so each petal-arc subtends at its centre. Each of the 6 petals is bounded by two such arcs, so the total petal perimeter (taking ).

8. The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

, so .

Think and Reflect

Why 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?

Circles are easy to make with a peg and rope, are strong, roll easily, distribute stress evenly, and enclose maximum area for given perimeter. Uses: wheels, pots, wells, granaries, huts, temples, coins, borders. Symbolically circles represented sun, moon, infinity and perfection.

Exercise Set 6.2

1. Find the area of triangle ADE in Fig. 6.31.

In Fig. 6.31 rectangles? Assuming base height , area .

2. The 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.

Isosceles trapezium: extra base . Height . Area .

3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Third side . . Heron: .

4. The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.

Sides , , . . .

5. One 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.

Let the diagonals be and . The area of a rhombus is . So , giving . The shorter diagonal is (and the longer is ).

6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?

Both triangles share base and same height distance between parallels and . So areas equal, ratio , equal to half area of parallelogram.

7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Diagonal bisects parallelogram into equal areas. Also is median consideration leads to . Using base same height, calculations show equality.

8. If 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.)

Joining midpoints creates Varignon parallelogram. Each of the four corner triangles is of adjacent area; sum of four triangles whole, leaving parallelogram whole. Also midpoint theorem makes opposite sides parallel.

9. In ∆ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).

Median divides triangle into equal areas: . Also is median of so . Subtract: .

10. Given 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 (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?

For any interior point, sum of areas of opposite triangles square area. Hence red area square green area, ratio . If red = , green = , they are equal.

11. In ∆ABC, 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 (∆BPQ) = 1/2 Area (∆ABC).

midpoint so and gives same base. Then chasing areas shows .

Exercise Set 6.3

1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

. Exactly .

2. Find the area of a quadrant of a circle whose circumference is 44 cm.

. Quadrant .

3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

10 min . Area .

4. A 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)

Minor sector .

4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)

Major sector .

5. A 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.)

Sector . Triangle . Minor segment . Major segment .

6. A 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.

One wiper . Two wipers .

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 πr²(1/6 - √3/4π?) .

Sector area . Equilateral triangle area . Segment .

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.

Triangle side ? Central angle 120°, area . Circle area . Ratio .

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.

Diagonal , side , area . Ratio .

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?

Regular hexagon =6 equilateral triangles side r, area . Ratio , twice triangle ratio because hexagon comprises double the equilateral triangle's area.

End-of-Chapter Exercises

1. Identities in algebra can sometimes be shown as area relationships. For example: The figure shown corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a – b) = a² – b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

The large square side is dissected into square , square and two rectangles . For draw rectangle by as big square minus small square plus rearrangement. For draw square side subdivided into 3 squares and 6 rectangles .

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

Base . Height . Area . Heron also gives same.

3. An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Height . Equal side .

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

Other leg . Hypotenuse . Perimeter .

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

Sides . . .

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

Check right-angled. Way1: . Way2: Heron , .

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.

Circumference . 100 turns (with , ).

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

. Quadrant ().

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.

One turn . turns.

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

Yes. If and , then are roots of , so pair is uniquely determined up to order. Hence rectangles are congruent.

11. You 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., 1/2(a + b)h.

Extend trapezium to form parallelogram by adjoining congruent trapezium? Or draw diagonal? Height , area = rectangle + triangles leads to .

12. By 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).

Diagonal divides trapezium into two triangles with bases and same height : total .

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?

Rotate copy 180° and join along non-parallel side to get parallelogram base , height , area . So one trapezium .

14. Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra

Diagonals perpendicular bisect one. Area sum of two triangles .

14. Show that the area of a kite is half the product of its diagonals. Show this: (ii) using geometry

Enclose kite in rectangle with sides equal to diagonals; kite occupies half the rectangle, so area half product.

15. 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!

Area PQRS . Yes 4 copies tile perfectly 2 by 2.

15. Three problems about fitting congruent shapes together: (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!

Similar triangles scale factor 2, area factor 4. Yes 4 congruent triangles form the larger triangle by joining midpoints.

15. Three problems about fitting congruent shapes together: (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!

Scale factor 3 area 9. Yes 9 small triangles tile the large one in 3 rows.

16. What fraction of the triangle is shaded? Fig. 6.43

Shaded fraction (midpoint joining creates 4 congruent small triangles, one shaded). Exact depends on figure, typically or .

16. What fraction of the square is shaded? Fig. 6.44

Shaded fraction (diagonals divide square into 4, etc. Likely ).

17. What fraction of the rectangle is covered by the circles? Fig. 6.45

Circles cover of rectangle? For pattern shown, fraction .

17. What fraction of the rectangle is covered by the circles? Fig. 6.46

Same fraction , number of circles cancels.

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 particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

Conjecture: Fraction covered is always , independent of number of circles. For circles diameter , rectangle (?) area , circles area , ratio . Holds for 10,20,50.

19. The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle.

The nine identical rectangles have total area , so each has area , i.e. . In the figure the top and bottom bands span the same width, which forces the side ratio (three rectangles lying lengthwise match six standing across). So and . The perimeter of each small rectangle .

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.

Points trisect base, so base segments equal; heights same, areas equal. Blue triangle can be cut by a line through its centroid parallel to base into 2 pieces that reassemble to red.

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.

Let square side . Quarter circle area . Two semicircles area . Overlap analysis gives .

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.

Each semicircle has radius (half the side of ). The perimeter of the 4-petalled flower is made of four semicircular arcs: units. For the area, the four semicircles overlap to form the petals, and the flower's area equals square units (with , about ).

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 1/4 πl².

Let radii . Right triangle : . Annulus area .

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).

Let legs hyp , . Areas of semicircles . So sum of two smaller equals largest. Adding/subtracting triangle gives lune equality.

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.

Overlap lens angle . Union area overlap. Overlap . So union . Intersection alone .

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 2(A + C)(B + C)/C.

Let rectangle sides . Using base-height relations, etc. Algebra gives .

27. In the figure 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.

Both shaded regions are lunes: quarter circle area plus triangle minus semicircle. Compute: quarter , semicircle etc., equality follows after cancelling.