Area and Perimeter, titled Measuring Space: Perimeter and Area in the new NCERT book, starts with the stagger on a running track and builds the constant π, the ratio of circumference to diameter that is the same for every circle. It covers arc length, the areas of rectangles, parallelograms and triangles, the median theorem, Heron’s formula, Brahmagupta’s formula for a cyclic 4-gon, and the areas of circles, sectors and segments. These formulas carry straight into Class 10 mensuration, so accurate working here saves marks for years.
Key Concepts
1. Perimeter and the C/D Ratio (the Constant π)
The perimeter of a shape is the total length around its border. The textbook pictures a tiny insect walking once around the border, never turning back, until it returns to where it started. The distance it walks is the perimeter.
| Shape | Perimeter | Ratio perimeter : side |
|---|---|---|
| Square of side a | 4a | 4 : 1 for every square |
| Equilateral triangle of side a | 3a | 3 : 1 for every equilateral triangle |
| Rectangle a by b | 2(a + b) | Depends on the shape of the rectangle |
The square formula is a special case of the rectangle formula: put b = a in 2(a + b) and you get 4a.
Double the side of a square from 2 units to 4 units and the perimeter doubles from 8 units to 16 units. The ratio of perimeter to side stays fixed at 4 : 1 whatever the size. The same idea works for circles. The perimeter of a circle is called its circumference (C). People noticed in ancient times that C ÷ D, the circumference divided by the diameter, is the same for every circle, a coin or a giant wheel. The textbook calls this the “C/D ratio”. Today we call it π (said “pie”).
π = C/D, so C = πD = 2πr, because the diameter is twice the radius.
Home measurement from the textbook. Wrap a very thin thread tightly around a cotton reel 20 times. Unwrap it and measure its length L. Measure the diameter D of the reel. Then L/(20D) estimates the C/D ratio. You should get a number between 3.1 and 3.2.
2. The Journey of π and Why π Is Irrational
The chapter traces how mathematicians in many regions pinned down π. You are expected to list the main approximations and who found them.
| Who and when | Value or method |
|---|---|
| Mesopotamia (c. 1900 BCE) | A circle is slightly longer than the hexagon inside it, so π > 3. They used 3 + 1/8 = 3.125 |
| Archimedes (by 250 BCE) | Trapped π between inscribed and circumscribed polygons. Hexagons gave 3 < π < 2√3 ≈ 3.46. With 96-sided polygons: 3 10/71 < π < 3 1/7 |
| Ptolemy (c. 150 CE) | 377/120 ≈ 3.14167 |
| Liu Hui (263 CE), Zu Chongzhi (480 CE) | Circle-cutting with up to 24,576 sides. Zu gave 22/7 ≈ 3.1428 (Yuelü) and 355/113 ≈ 3.1415929 (Miü), the best value in the world for over 800 years |
| Āryabhaṭa (499 CE) | 62832/20000 = 3.1416, called asanna, meaning “approaching” or approximate |
| Brahmagupta (628 CE) | √10 ≈ 3.1622, chosen because it is easy to work with |
| Mādhava of Sangamagrāma | First exact formula: π/4 = 1 − 1/3 + 1/5 − 1/7 + …, an infinite series. He used series to get π to 11 decimal places, 3.14159265358 |
In 1706 William Jones used the Greek letter π, the first letter of perimetros (perimeter), and Euler made the symbol popular.
π is irrational. A rational number can be written as a/b, where a and b are integers and b ≠ 0. Its decimal expansion ends or repeats in a pattern: 1/3 = 0.333…, 1/11 = 0.0909…, 1/7 = 0.142857 142857…. The digits of π go on forever with no repeating pattern, and π cannot be written as a ratio of two integers. Lambert proved this in 1761. Āryabhaṭa and Zu Chongzhi seem to have regarded π as irrational long before.
Since π is irrational, there is no “best fraction” for it: for any fraction close to π, another fraction is closer. So we write π ≈ 22/7 and also π ≠ 22/7, just as √2 ≈ 1.414 and √2 ≠ 1.414.
3. Length of an Arc and the Athletics Track
A diameter AB splits a circle into two semicircles. Reflect the circle in AB (or rotate it through 180° about the centre O) and the two semicircles swap places, so they are equal. Each has length 2πr ÷ 2 = πr. Rotating through 90° shows the four quarter circles are equal, so each is 2πr ÷ 4 = πr/2.
Written as a share of the full turn: semicircle = 2πr × 180/360 and quarter circle = 2πr × 90/360. So an arc AB that subtends an angle θ° at the centre has
length of arc = 2πr × θ/360
A sector is the region bounded by an arc and the two radii to its ends. Its perimeter is the arc plus two radii: perimeter of sector = 2πr × θ/360 + 2r.
The 400 m track. A standard track has two straight sections of 84.39 m each and two semicircular ends with a common centre. The innermost semicircle has radius 36.5 m, and each lane is 1.22 m wide. An athlete runs 0.3 m from the inner border of lane 1:
- Straights: 2 × 84.39 = 168.78 m.
- The two semicircles of radius 36.5 + 0.3 = 36.8 m make one full circle: 2 × 3.1416 × 36.8 = 231.22 m.
- Total: 168.78 + 231.22 = 400 m.
On the straights, all lanes run the same distance. On the bends, a runner in an outer lane runs a semicircle of larger radius, so she covers more ground. The stagger, the distance between the starting points of adjacent lanes, makes up for this. The textbook leaves the size of the stagger as a question. A simple estimate assumes every athlete runs the same distance (0.3 m) from the inner edge of her own lane. Then the running radius in lane 2 is 1.22 m larger than in lane 1. Over the two semicircles (one full circle) the extra distance is 2π × 1.22 ≈ 7.67 m, so the lane 2 runner starts about 7.67 m ahead. Under this assumption the extra length depends only on the difference in radius, so each pair of neighbouring lanes needs the same stagger. Official track rules measure the outer lanes slightly differently, so real staggers differ a little from this estimate.
4. Areas of Rectangles, Parallelograms, Triangles and the Median Theorem
Area is the amount of space a flat region takes up. The unit is a 1 × 1 square, whose area is 1 unit2 (1 sq. unit). A rectangle with sides a and b has area ab, and a square of side a has area a2.
Parallelogram. Cut a right triangle off one end of parallelogram ABCD and slide it to the other end. The pieces form a rectangle with the same base b and the same height h. So area of parallelogram = base × height = bh. When the parallelogram is very “thin” and the perpendicular does not land on the opposite side, the textbook replaces it step by step with less slanted parallelograms of the same base and height (using congruent triangles) until the cut works.
Side lengths are not enough for a parallelogram. Keep the four sides fixed and change the angle between them: the height changes, so the area changes. A rhombus of side 3 can have area 9 (a square), 8.01 or 5.41, depending on how much it is pushed over.
Triangle. Two congruent copies of a triangle fit together to make a parallelogram with the same base and height. So
area of triangle = ½ × base × height = ½bh
Enclosing the triangle in a rectangle gives the same result: the triangle fills half the rectangle.
Median theorem. A median joins a vertex to the midpoint of the opposite side. In ΔABC, let AD be a median with BD = DC = a, and let h be the height from A to BC. Triangles ABD and ACD have equal bases a and the same height h, so each has area ah/2.
Theorem: A median of a triangle divides it into two triangles with equal area. The two triangles are usually different in shape (not congruent), yet their areas are equal.
Other shapes built from these:
- Trapezium with parallel sides a and b and height h: area = ½(a + b)h.
- Kite (and rhombus): area = ½ × d1 × d2, half the product of the diagonals.
5. Heron’s Formula and Other Triangle Area Formulas
Heron of Alexandria gave a way to find the area of a triangle from its three sides alone, without the height. For ΔABC with BC = a, CA = b, AB = c:
semi-perimeter s = (a + b + c)/2
Area = √[s(s − a)(s − b)(s − c)]
The textbook tests the formula against ½ × base × height in three cases.
| Triangle | Heron’s formula gives | Check with ½bh |
|---|---|---|
| Equilateral, side a | s = 3a/2, each (s − a) = a/2, area = √(3a4/16) = (√3/4)a2 | h = (√3/2)a, ½ × a × (√3/2)a = (√3/4)a2 |
| Isosceles, equal sides a, base 2b | s = a + b, area = √[(a + b)(b)(b)(a − b)] = b√(a2 − b2) | h = √(a2 − b2), ½ × 2b × h = b√(a2 − b2) |
| Sides 3, 4, 5 | s = 6, area = √(6 × 3 × 2 × 1) = 6 | 32 + 42 = 52, so right-angled: ½ × 3 × 4 = 6 |
The height of an equilateral triangle comes from the Baudhāyana-Pythagoras theorem: h2 = a2 − (a/2)2 = 3a2/4, ∴ h = (√3/2)a. One proof of Heron’s formula uses this theorem and the identity a2 − b2 = (a − b)(a + b); it comes in Grade 10.
Two circle formulas for area. Every triangle has exactly one circumcircle (passing through all three vertices, radius R) and one incircle (touching all three sides, radius r). Then
- Area of ΔABC = abc/(4R)
- Area of ΔABC = r(a + b + c)/2 = rs
6. Brahmagupta’s Formula, Special Cases and Squaring a Rectangle
Can we find the area of a 4-gon (quadrilateral) from its four sides? No. The rhombus with all sides 3 already showed three different areas. We need one more fact, such as an angle, a diagonal, or a geometric property. One such property is being cyclic: all four vertices lie on one circle.
Brahmagupta’s formula (628 CE). For a cyclic 4-gon with sides a, b, c, d and s = (a + b + c + d)/2:
Area = √[(s − a)(s − b)(s − c)(s − d)]
- Rectangle (all rectangles are cyclic), sides a, b, a, b: s = a + b, area = √(b × a × b × a) = √(ab × ab) = ab. Correct.
- Isosceles trapezium (all isosceles trapezia are cyclic) with parallel sides 2a and 2b and equal legs c: s = a + b + c, and the formula simplifies to (a + b)√[c2 − (b − a)2]. The height is h = √[c2 − (b − a)2], so this is (a + b)h, which matches ½ × (2a + 2b) × h.
Heron is a special case of Brahmagupta. Treat a triangle with sides a, b, c as a 4-gon whose fourth side d = 0 (two vertices coincide). Every triangle is cyclic, since a circle passes through any three points not on a line. With d = 0, s = (a + b + c)/2 and s − d = s, so Brahmagupta’s formula becomes √[s(s − a)(s − b)(s − c)], which is Heron’s formula. We say Brahmagupta’s formula generalises Heron’s formula.
Special case and generalisation. A special case comes from a general result by adding an extra condition. Examples from the chapter:
- A square is a special case of a rectangle: A = ab with b = a gives A = a2.
- An isosceles right triangle is a special case of a right triangle: a2 + b2 = c2 with a = b gives c = a√2.
- (a + b)2 = a2 + b2 + 2ab is (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca with c = 0.
Squaring a rectangle (Baudhāyana, Śulbasūtra, 800 BCE). To “square” a shape means to construct a square with the same area. For rectangle ABCD with AD = a, AB = b and a > b:
- Mark E on AD with AE = AB = b.
- Mark F, the midpoint of ED.
- Draw square AFGH with side AF, with H on AB produced.
- With centre H, draw arc AG. Let it cut side BC at K.
- Through K draw a line parallel to AH, cutting GH at P.
- Square HPQS on side HP has the same area as ABCD.
Why it works. AF = (AE + AD)/2 = (a + b)/2, so HK = HG = (a + b)/2 (radius of the arc). BH = AH − AB = (a + b)/2 − b = (a − b)/2. In right triangle HKP, KP = BH, so HP2 = HK2 − BH2 = [(a + b)/2]2 − [(a − b)/2]2 = (a2 + 2ab + b2)/4 − (a2 − 2ab + b2)/4 = ab. The construction is geometry for the identity [(a + b)/2]2 − [(a − b)/2]2 = ab.
7. Area of a Circle, Sector and Segment
Why area grows with the square of the size. For every square, P2 : A = (4a)2 : a2 = 16 : 1. For every equilateral triangle, P2 : A = (3a)2 : (√3/4)a2 = 36 : √3. The ratio depends only on the shape, not its size. So for circles too, C2 : A must be a fixed constant.
- The Babylonians (well before 1500 BCE) measured C2 : A ≈ 12 : 1, so A ≈ C2/12.
- The Egyptians (around 1500 BCE) used A ≈ (8d/9)2 = (256/81)r2. The same formula appears in the Baudhāyana Śulbasūtra.
- Archimedes (c. 250 BCE) proved the constant A/r2 is exactly π: A = πr2.
Archimedes put it this way: the area of a circle equals the area of a right triangle whose two shorter sides are the radius and the circumference. So A = ½ × 2πr × r = πr2.
Nīlakaṇṭha’s slices (c. 1500). Cut a circle into many thin slices like a pizza and lay them alternately point-up and point-down. As the slices get thinner, the shape becomes almost a parallelogram with base = half the circumference = πr and height = r. Area = πr × r = πr2.
Sector. By reflection symmetry a semicircular disc is 180/360 = ½ of the circle, ½πr2. By quarter-turn symmetry a quadrant is 90/360 = ¼ of the circle, ¼πr2. In general, a sector whose arc subtends θ° at the centre has
area of sector = πr2 × θ/360
Segment. A segment is the region bounded by an arc and the chord joining its ends. The minor segment is the part of the sector left after removing the triangle formed by the chord and the two radii:
area of minor segment = area of sector − area of triangle OAB
area of major segment = area of circle − area of minor segment
Two triangles come up again and again. For θ = 90°, triangle OAB is right-angled with area ½r2. For θ = 60°, OA = OB and the angle between them is 60°, so triangle OAB is equilateral with area (√3/4)r2.
8. How Length and Area Change with Scale
If every length in a figure is multiplied by k, every perimeter is multiplied by k and every area by k2.
- A rectangle with sides 2a and 2b has area 4ab, four times the area of an a by b rectangle, and four copies really do fit inside it.
- A triangle with sides 2a, 2b, 2c has 4 times the area of one with sides a, b, c. With sides 3a, 3b, 3c it has 9 times the area.
- For circles, C = 2πr is proportional to r. If circumferences are in the ratio 5 : 4, the radii are also 5 : 4, and the areas are 25 : 16.
Formula and Theorem Sheet
| Result | Formula or statement | Notes |
|---|---|---|
| Perimeter of square, triangle, rectangle | 4a, 3a (equilateral), 2(a + b) | Square is the case b = a |
| Definition of π | π = C/D | Same for all circles; irrational |
| Circumference | C = πd = 2πr | Semicircle arc πr, quarter arc πr/2 |
| Arc length | l = 2πr × θ/360 | θ = angle at centre in degrees |
| Parallelogram | base × height | Sides alone do not fix the area |
| Triangle | ½ × base × height | Half of a parallelogram |
| Median theorem | A median splits a triangle into two triangles of equal area | Equal bases, same height |
| Heron’s formula | √[s(s − a)(s − b)(s − c)], s = (a + b + c)/2 | Any triangle, three sides known |
| Brahmagupta’s formula | √[(s − a)(s − b)(s − c)(s − d)], s = (a + b + c + d)/2 | Cyclic 4-gon only; d = 0 gives Heron |
| Trapezium | ½(a + b)h | a, b parallel |
| Kite, rhombus | ½d1d2 | Diagonals perpendicular |
| Squaring a rectangle | [(a + b)/2]2 − [(a − b)/2]2 = ab | Baudhāyana’s construction |
| Circle | A = πr2 = ½ × C × r | Archimedes |
| Sector | πr2 × θ/360 | Semicircle ½πr2, quadrant ¼πr2 |
| Segment | Sector − triangle OAB | Major = circle − minor |
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Important Definitions
- Perimeter: the total length around the border of a shape.
- π (pi): the constant ratio of the circumference of a circle to its diameter, the same for all circles. π ≈ 22/7 ≈ 3.14.
- Irrational number: a number that cannot be written as a ratio of two integers. π and √2 are irrational.
- Stagger: the distance between the starting points of adjacent lanes on a running track.
- Arc: a part of a circle between two points on it. It subtends an angle at the centre.
- Sector: the region bounded by an arc and the two radii containing the endpoints of the arc.
- Segment: the region bounded by an arc of a circle and the chord joining the endpoints of the arc.
- Median of a triangle: the line segment joining a vertex to the midpoint of the opposite side.
- Circumcircle: the one circle that passes through all three vertices of a triangle.
- Incircle: the one circle that fits tightly inside a triangle, touching all three sides.
- Cyclic 4-gon: a 4-gon (quadrilateral) whose four vertices all lie on one circle. Rectangles and isosceles trapezia are cyclic.
- Squaring a shape: constructing a square with the same area as the given shape.
Solved Examples (NCERT-Based)
Use π = 22/7 unless another value is stated.
Example 1 (Exercise Set 6.1, Q1). The perimeter of a circle is 44 cm. What is its radius?
Solution: 2πr = 44 ⇒ 2 × (22/7) × r = 44 ⇒ (44/7)r = 44 ⇒ r = 7 cm.
Example 2 (Exercise Set 6.1, Q2). Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.
Solution:
- (i) C = 2 × (22/7) × 7 = 44 cm = 44.0 cm.
- (ii) C = 2 × (22/7) × 10 = 440/7 = 62.857… ≈ 62.9 cm.
- (iii) C = 2 × (22/7) × 12 = 528/7 = 75.428… ≈ 75.4 cm.
Example 3 (Exercise Set 6.1, Q3). 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°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.
Solution:
- (i) Full circumference = 2 × (22/7) × 3.5 = 22 cm. Arc = 22 × 60/360 = 22/6 = 11/3 ≈ 3.67 cm.
- (ii) Full circumference = 2 × (22/7) × 6.3 = 39.6 m. Arc = 39.6 × 120/360 = 39.6/3 = 13.2 m.
Example 4 (Exercise Set 6.1, Q4). 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°.
Solution: Circumference = 2 × (22/7) × 14 = 88 cm. Arc = 88 × 75/360 = 6600/360 = 55/3 ≈ 18.33 cm. Perimeter = arc + 2r = 18.33 + 28 = 46.33 cm (exactly 139/3 cm).
Example 5 (Textbook Example 1). Two circles of equal radius r pass through each other’s centres A and B and meet at C and D. Find the perimeter of the shape formed by the two circles (the outer boundary only).
Solution: AB = AC = BC = r, so ΔABC is equilateral and ∠CAB = 60°. Similarly ∠BAD = 60°, so ∠CAD = 120°. The arc CD of circle A that lies inside circle B subtends 120° at A, so it is 120/360 = 1/3 of that circle. The outer part of each circle is the other 2/3. Total perimeter = 2 × (2/3) × 2πr = 8πr/3 units.
Example 6 (Textbook Example 2). Points P and Q are joined by two paths. Path 1 is one semicircle a on PQ as diameter. Path 2 is made of three semicircles b, c, d whose diameters lie along PQ and together make up PQ. Which path is longer?
Solution: Let the radii be a′, b′, c′, d′. A semicircle of radius r has length πr. Path 1 = πa′. Path 2 = π(b′ + c′ + d′). Since PQ = 2a′ = 2b′ + 2c′ + 2d′, we get a′ = b′ + c′ + d′. The two paths have equal length.
Example 7 (Exercise Set 6.2, Q3). Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Solution: Third side = 32 − 8 − 11 = 13 cm. s = 32/2 = 16 cm. s − a = 8, s − b = 5, s − c = 3.
Area = √(16 × 8 × 5 × 3) = √1920 = √(64 × 30) = 8√30 ≈ 8 × 5.477 ≈ 43.8 cm2.
Example 8 (Exercise Set 6.2, Q2). 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.
Solution: Drop perpendiculars from the ends of the 20 cm side to the 40 cm side. They cut off two equal pieces of (40 − 20)/2 = 10 cm. Each slant side is the hypotenuse of a right triangle with base 10 cm, so height h = √(262 − 102) = √(676 − 100) = √576 = 24 cm.
Area = ½(40 + 20) × 24 = 30 × 24 = 720 cm2.
Check with Brahmagupta (an isosceles trapezium is cyclic): s = (40 + 20 + 26 + 26)/2 = 56. Area = √[(56 − 40)(56 − 20)(56 − 26)(56 − 26)] = √(16 × 36 × 30 × 30) = 4 × 6 × 30 = 720 cm2.
Example 9 (End-of-Chapter, Q6). The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Solution: Method 1: 72 + 242 = 49 + 576 = 625 = 252, so the triangle is right-angled with legs 7 cm and 24 cm. Area = ½ × 7 × 24 = 84 cm2.
Method 2 (Heron): s = (7 + 24 + 25)/2 = 28. Area = √[28 × 21 × 4 × 3] = √7056 = 84 cm2.
Example 10 (Exercise Set 6.3, Q1). Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Solution: Area of circle = (22/7) × 49 = 154 cm2. Sector = 154 × 60/360 = 154/6 = 77/3 ≈ 25.67 cm2.
Example 11 (Exercise Set 6.3, Q5). 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.)
Solution:
- Area of sector = 3.14 × 225 × 60/360 = 706.5/6 = 117.75 cm2.
- OA = OB = 15 cm and ∠AOB = 60°, so ΔOAB is equilateral. Area = (√3/4) × 225 = 1.73 × 225/4 = 97.3125 cm2.
- Minor segment = 117.75 − 97.3125 = 20.4375 ≈ 20.44 cm2.
- Area of circle = 3.14 × 225 = 706.5 cm2. Major segment = 706.5 − 20.4375 = 686.0625 ≈ 686.06 cm2.
Example 12 (Exercise Set 6.3, Q6). 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.
Solution: Each wiper sweeps a sector of radius 28 cm and angle 120°. Area = (22/7) × 28 × 28 × 120/360 = 2464 × 1/3 = 2464/3 cm2. Two wipers: 2 × 2464/3 = 4928/3 ≈ 1642.67 cm2.
Competency-Based Questions (with answers)
Q1. Case study: the athletics track. A 400 m track has two straight sections of 84.39 m each and two semicircular ends with a common centre. The innermost semicircle has radius 36.5 m and each lane is 1.22 m wide. An athlete in lane 1 runs 0.3 m from the inner border. Use π = 3.1416.
- (i) What is the radius of the path she runs on the bends?
- (ii) Find the distance she runs on the two bends together.
- (iii) Show that one lap is 400 m.
- (iv) Why does a runner in lane 2 start ahead of her?
Answer: (i) 36.5 + 0.3 = 36.8 m. (ii) The two semicircles form one circle: 2 × 3.1416 × 36.8 = 231.22 m. (iii) 2 × 84.39 + 231.22 = 168.78 + 231.22 = 400 m. (iv) On the bends the lane 2 runner follows a semicircle of larger radius, so she runs further. The stagger puts her start further ahead so that both runners cover the same total distance to the common finish line.
Q2. Assertion-reason. Assertion (A): The area of a parallelogram cannot be found from the lengths of its sides alone. Reason (R): If the sides are kept fixed and the angle between them is changed, the height changes.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (a). Area = base × height, and the height depends on the angle. A rhombus of side 3 can have area 9, 8.01 or 5.41.
Q3. Assertion-reason. Assertion (A): π = 22/7. Reason (R): π is an irrational number.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (d). 22/7 is a fraction of two integers, so it is rational. π is irrational, so π ≠ 22/7. We only write π ≈ 22/7.
Q4. Assertion-reason. Assertion (A): A median of a triangle divides it into two triangles of equal area. Reason (R): The two triangles formed by a median are always congruent.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (c). The two triangles have equal bases and the same height, so their areas are equal. They are usually different in shape, so R is false.
Q5. Assertion-reason. Assertion (A): Heron’s formula is a special case of Brahmagupta’s formula. Reason (R): Every triangle is cyclic, so it can be treated as a cyclic 4-gon with the fourth side d = 0.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (a). Putting d = 0 makes s − d = s, and Brahmagupta’s formula becomes √[s(s − a)(s − b)(s − c)].
Q6. Source-based: the story of π. Read: “In 499 CE, Āryabhaṭa provided a value of 62832/20000 = 3.1416 for π. Crucially, he described this value as asanna, i.e., ‘approaching’ or ‘approximate’.”
- (i) Why is the word asanna important?
- (ii) Which approximation stayed the most accurate value for π for over 800 years, and who found it?
- (iii) Which mathematician gave the first exact formula for π, and what is it?
Answer: (i) It shows Āryabhaṭa understood that 3.1416 is only close to π and that π cannot be written exactly as one simple fraction, which fits π being irrational. (ii) 355/113, found by Zu Chongzhi. (iii) Mādhava of Sangamagrāma: π/4 = 1 − 1/3 + 1/5 − 1/7 + ….
Q7. Case study: the clock. The minute hand of a wall clock is 7 cm long.
- (i) Through what angle does the minute hand turn in 10 minutes?
- (ii) Find the area swept by the minute hand in 10 minutes.
- (iii) How far does the tip of the hand travel in the same time?
Answer: (i) 60 minutes = 360°, so 10 minutes = 60°. (ii) Sector area = (22/7) × 49 × 60/360 = 154/6 = 77/3 ≈ 25.67 cm2. (iii) Arc = 2 × (22/7) × 7 × 60/360 = 44/6 = 22/3 ≈ 7.33 cm.
Q8. Case study: the garden plot. A triangular garden has sides in the ratio 3 : 5 : 7 and a perimeter of 300 m. A gardener wants to lay grass on the whole plot.
- (i) Find the three sides.
- (ii) Find the semi-perimeter.
- (iii) Find the area of grass needed.
Answer: (i) Sum of ratio parts = 15, so one part = 300/15 = 20 m. Sides: 60 m, 100 m, 140 m. (ii) s = 150 m. (iii) Area = √(150 × 90 × 50 × 10) = √6750000 = 1500√3 ≈ 2598 m2.
Important Questions for Exams
The grouping below is a practice ladder from quick recall to full working. Use π = 22/7 unless stated.
1-Mark Questions
Q1. Define π in terms of a circle.
Answer: π is the ratio of the circumference of any circle to its diameter, C/D.
Q2. Is π equal to 22/7? Give a reason.
Answer: No. 22/7 is rational and π is irrational, so π ≈ 22/7 and π ≠ 22/7.
Q3. Find the semi-perimeter of a triangle with sides 7 cm, 8 cm and 9 cm.
Answer: s = (7 + 8 + 9)/2 = 12 cm.
Q4. Find the area of a quadrant of a circle of radius 14 cm.
Answer: ¼ × (22/7) × 196 = 154 cm2.
2-Mark Questions
Q5 (Exercise Set 6.1, Q8). The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Answer: 2πr1 : 2πr2 = 5 : 4. Cancel 2π: r1 : r2 = 5 : 4.
Q6 (Exercise Set 6.2, Q5). 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.
Answer: Let the diagonals be d and 2d. Area = ½ × d × 2d = d2 = 128, so d = √128 = 8√2 ≈ 11.31 cm.
Q7 (End-of-Chapter, Q3). An isosceles triangle has base 10 cm, and its area is 60 cm2. What are the lengths of the equal sides?
Answer: ½ × 10 × h = 60, so h = 12 cm. The height bisects the base into 5 cm pieces. Equal side = √(122 + 52) = √169 = 13 cm.
Q8 (End-of-Chapter, Q4). The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Answer: ½ × 12 × x = 54, so the other leg x = 9 cm. Hypotenuse = √(144 + 81) = √225 = 15 cm. Perimeter = 12 + 9 + 15 = 36 cm.
3-Mark Questions
Q9 (End-of-Chapter, Q2). An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Answer: Base = 40 − 30 = 10 cm. s = 20. Area = √[20 × (20 − 15) × (20 − 15) × (20 − 10)] = √(20 × 5 × 5 × 10) = √5000 = 50√2 ≈ 70.71 cm2. Check: h = √(225 − 25) = √200 = 10√2, and ½ × 10 × 10√2 = 50√2.
Q10 (End-of-Chapter, Q5). The sides of a triangle are in the ratio 2: 3: 4, and its perimeter is 45 cm. Find its area.
Answer: One part = 45/9 = 5, so the sides are 10, 15, 20 cm. s = 22.5. s − a = 12.5, s − b = 7.5, s − c = 2.5. Area = √(22.5 × 12.5 × 7.5 × 2.5) = √5273.4375 = (75√15)/4 ≈ 72.62 cm2.
Q11 (Exercise Set 6.3, Q4). 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), and (ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)
Answer: Circle = 3.14 × 100 = 314 cm2. (i) Minor sector = 314 × 90/360 = 78.5 cm2. (ii) Major sector = 314 × 270/360 = 235.5 cm2.
Q12 (End-of-Chapter, Q9). 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.
Answer: One turn = 2 × (22/7) × 28 = 176 cm. 1 km = 100000 cm. Number of turns = 100000/176 ≈ 568.2, so the wheel makes about 568 complete turns.
5-Mark Questions
Q13 (Exercise Set 6.2, Q9). In ΔABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).
Answer: AD is a median of ΔABC, so by the median theorem, area(ΔABD) = area(ΔACD) … (1). In ΔPBC, PD joins P to the midpoint D of BC, so PD is a median of ΔPBC and area(ΔPBD) = area(ΔPCD) … (2). Subtract (2) from (1): area(ΔABD) − area(ΔPBD) = area(ΔACD) − area(ΔPCD). Since P lies on AD, ΔABD is made of ΔABP and ΔPBD, and ΔACD is made of ΔACP and ΔPCD. ∴ area(ΔABP) = area(ΔACP).
Q14 (Exercise Set 6.3, Q8). 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 3√3/(4π) ≈ 0.413.
Answer: Triangle: join the centre O to the three vertices. This makes three equal triangles, each with two sides r and angle 120° at O. The foot of the perpendicular from O to a side bisects it, and the angle at O is split into two 60° angles, so the perpendicular is r/2 and half the side is √(r2 − r2/4) = (√3/2)r. So the side is √3 r. Area = (√3/4)(√3 r)2 = (3√3/4)r2. Ratio = (3√3/4)r2 ÷ πr2 = 3√3/(4π) ≈ 5.196/12.566 ≈ 0.413.
Q15 (End-of-Chapter, Q24). Semicircles are drawn outward on all three sides of a right-angled triangle. Show that Area (A) + Area (B) = Area (C), where A and B are the semicircles on the two shorter sides and C is the semicircle on the hypotenuse.
Answer: Let the legs be a, b and the hypotenuse c. A semicircle on a side of length x has radius x/2, so area ½π(x/2)2 = πx2/8. Area(A) + Area(B) = πa2/8 + πb2/8 = π(a2 + b2)/8. By the Baudhāyana-Pythagoras theorem a2 + b2 = c2, so the sum is πc2/8 = Area(C).
Q16 (End-of-Chapter, Q23). Two concentric circles have common centre O. A chord BC of the larger circle touches the smaller circle at A, and BC = l. Show that the area of the region between the two circles is ¼πl2.
Answer: Let the radii be R (larger) and r (smaller). BC touches the smaller circle only at A, so every other point of BC lies outside the smaller circle and is further than r from O. So OA is the shortest distance from O to BC, which means OA ⊥ BC. In right triangles OAB and OAC, the hypotenuses OB = OC = R and the side OA is common, so the triangles are congruent (RHS) and AB = AC = l/2. (This is the circle fact that the perpendicular from the centre to a chord bisects the chord.) In right triangle OAB, R2 = r2 + (l/2)2, so R2 − r2 = l2/4. Area between the circles = πR2 − πr2 = π(R2 − r2) = ¼πl2.
Common Mistakes and Examiner Tips
- Using the diameter in C = 2πr. If the question gives the diameter, halve it first or use C = πd. A 56 cm tyre has circumference (22/7) × 56 = 176 cm, not 352 cm.
- Writing π = 22/7. π is irrational. Write π ≈ 22/7 or “taking π = 22/7”. In a true/false or assertion question, “π = 22/7” is false.
- Forgetting the two radii in a sector’s perimeter. Perimeter of a sector = arc + 2r. The arc alone is the arc length.
- Mixing arc length and sector area. Arc uses 2πr × θ/360 (units cm). Sector uses πr2 × θ/360 (units cm2). Check the unit of your answer.
- Using the slant side as the height. In a parallelogram or trapezium, the height is the perpendicular distance between the parallel sides. Find it with the Baudhāyana-Pythagoras theorem when only the slant side is given.
- Using the perimeter for s in Heron’s formula. s is half the perimeter. Check that s − a, s − b and s − c are all positive and add up to s.
- Applying Brahmagupta’s formula to any 4-gon. It works only for a cyclic 4-gon. For a general 4-gon the four sides do not fix the area.
- Segment = sector. A segment is the sector minus the triangle on the chord. Show the triangle area as a separate step.
- Wrong triangle in a segment problem. For 60° the triangle is equilateral, area (√3/4)r2. For 90° it is right-angled, area ½r2. Do not use ½r2 for 60°.
- Unit conversion in wheel problems. Convert km to cm (1 km = 100000 cm) or the circumference to metres before dividing. Round the number of complete turns sensibly.
- Assuming a median gives congruent triangles. It gives equal areas only. Quote the reason: equal bases and the same height.
Quick Revision Points
- Square 4a, equilateral triangle 3a, rectangle 2(a + b).
- For every circle, C/D is the same constant, π.
- C = πd = 2πr. Semicircle arc = πr, quarter arc = πr/2.
- Arc length = 2πr × θ/360. Sector perimeter = arc + 2r.
- Mesopotamia 3.125; Archimedes 3 10/71 < π < 3 1/7; Ptolemy 377/120; Zu Chongzhi 22/7 and 355/113; Āryabhaṭa 3.1416; Brahmagupta √10.
- Mādhava: π/4 = 1 − 1/3 + 1/5 − 1/7 + …, the first exact formula.
- π is irrational (Lambert, 1761); π ≈ 22/7 ≈ 3.14, π ≠ 22/7.
- Rectangle ab; parallelogram base × height; triangle ½ × base × height.
- Sides alone do not fix the area of a parallelogram or a general 4-gon.
- A median divides a triangle into two triangles of equal area (not always congruent).
- Heron: s = (a + b + c)/2, area = √[s(s − a)(s − b)(s − c)].
- Brahmagupta (cyclic 4-gon): √[(s − a)(s − b)(s − c)(s − d)]; put d = 0 to get Heron.
- Trapezium ½(a + b)h; kite and rhombus ½d1d2.
- Baudhāyana squared a rectangle using [(a + b)/2]2 − [(a − b)/2]2 = ab.
- Babylonians A ≈ C2/12; Egyptians A ≈ (8d/9)2; Archimedes A = πr2.
- A = ½ × circumference × radius = πr2; Nīlakaṇṭha’s slices form a near-parallelogram of base πr and height r.
- Sector area = πr2 × θ/360; semicircle ½πr2, quadrant ¼πr2.
- Minor segment = sector − triangle; major segment = circle − minor segment.
- Scale lengths by k: perimeters × k, areas × k2.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, Unit V, Mensuration, carries 14 marks out of the 80-mark theory paper. The unit has two chapters, Area and Perimeter and Surface Area and Volume, so the 14 marks are shared between them. The unit is allotted 27 teaching periods. The remaining 20 marks of the year are internal assessment: pen-paper tests and multiple assessment (10), portfolio (5) and lab practical (5).
The course structure expects students to compute circumference, arc length and the areas of circles, sectors and segments, use Heron’s formula, explain the median theorem and the squaring of a shape, list historical approximations to π, and explain why Heron’s formula is a special case of Brahmagupta’s. The 1, 2, 3 and 5-mark grouping above is a practice ladder; the exact question types in a given school paper are set by that paper.
The CBSE Class IX question paper design weights questions by thinking skill: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating.
This chapter is the foundation for Class 10, where arc length, sector and segment areas and triangle areas return in harder problems.
Class 9 Maths · Chapter 6 – swipe through all 9 cards to understand the whole chapter.
π, the circle constant
For every circle, circumference divided by diameter gives the same number, π.
π ≈ 22/7 ≈ 3.14, and π ≠ 22/7
- The perimeter of a circle is its circumference
- Square: perimeter : side is always 4 : 1
- π is irrational, proved by Lambert in 1761
Approximations through the ages
Mathematicians in many regions pinned π down more and more closely.
Mādhava gave the first exact formula: π/4 = 1 − 1/3 + 1/5 − 1/7 + …
- Mesopotamia 3.125, Ptolemy 377/120
- Zu Chongzhi 22/7 and 355/113
- Āryabhaṭa 3.1416, called asanna (approximate)
Arcs, sectors and the running track
An arc is the fraction θ/360 of the full circumference.
Perimeter of a sector = arc + 2r
- Semicircle πr, quarter circle πr/2
- Outer lanes run bigger semicircles on the bends
- The stagger evens out the total distance
Rectangle, parallelogram, triangle
A parallelogram rearranges into a rectangle; a triangle is half a parallelogram.
Four side lengths alone do not fix the area of a parallelogram
- Rectangle ab, parallelogram bh
- Trapezium ½(a + b)h
- Kite and rhombus ½ d1d2
A median halves the area
A median joins a vertex to the midpoint of the opposite side.
Equal areas, but the two triangles are usually not congruent
- Equal bases BD = DC
- Same height from A
- Use it twice to prove area(ΔABP) = area(ΔACP)
Area from three sides
Heron’s formula needs only the side lengths, no height.
Also: area = abc/(4R) = r(a + b + c)/2
- Sides 3, 4, 5: s = 6, area = 6
- Equilateral side a: area (√3/4)a2
- Check that s − a, s − b, s − c are all positive
Brahmagupta’s formula
For a 4-gon whose vertices lie on a circle, the four sides fix the area.
Put d = 0 to get Heron’s formula, a special case
- s = (a + b + c + d)/2
- Rectangles and isosceles trapezia are cyclic
- Rectangle a by b gives √(ab · ab) = ab
Baudhāyana squares a rectangle
To square a shape is to construct a square of the same area.
From the Śulbasūtra, about 800 BCE
- Mark AE = b on AD, then the midpoint F of ED
- Square on AF, arc with centre H cuts BC at K
- HP2 = HK2 − BH2 = ab
Circle, sector and segment
Archimedes proved the area of a circle is πr2, half the circumference times the radius.
Minor segment = sector − triangle OAB
- Semicircle ½πr2, quadrant ¼πr2
- 60° chord: the triangle is equilateral, (√3/4)r2
- 90° chord: the triangle has area ½r2
📝 Practice Area and Perimeter - 10 questions
Concept-check questions · with answers & explanations
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Why D: Step 1 (add the first three terms): 1 – 1/3 + 1/5 = 15/15 – 5/15 + 3/15 = 13/15. Step 2 (the series equals π/4, so multiply by 4): π ≈ 4 × 13/15 = 52/15 ≈ 3.47. The estimate is above π; adding the next term, -1/7, would bring it below, and the estimates close in on π as more terms are used.
Why not A: 8/3 = 4 × (1 – 1/3) uses only two terms
Why not B: 13/15 is the value of π/4 from three terms; it still has to be multiplied by 4
Why not C: 304/105 = 4 × (1 – 1/3 + 1/5 – 1/7) uses four terms
Remember: Mādhava’s series gives π/4, so always multiply the sum by 4.
Why B: Step 1 (equilateral triangle): the two centres A, B and an intersection point C are all r apart, so triangle ABC is equilateral and angle CAB = 60°. By symmetry, the angle at A between the two intersection points is 120°. Step 2 (arc length = 2πr × θ/360): the inner arc on each circle is 120°, so the outer arc is 240°, i.e. 2/3 of the circle. Total = 2 × (2/3) × 2πr = 8πr/3.
Why not A: 4πr adds both full circumferences, including the parts hidden inside the overlap
Why not C: 4πr/3 adds the two inner 120° arcs, which are the hidden parts, not the boundary
Why not D: 2πr is one full circle; the outer boundary is longer because each circle keeps 2/3 of its length
Remember: Equal circles through each other’s centres: 120° inside, 240° outside.
Why B: Both triangles have the same base CD. Their third vertices P and Q both lie on AB, which is parallel to CD, so both triangles have the same height: the distance between the parallel lines AB and CD. By Area = ½ × base × height, the areas are equal, whatever the positions of P and Q. Each is half the area of the parallelogram.
Why not A: moving a vertex along a line parallel to the base changes the shape but not the height
Why not C: there is no reason one should be twice the other; they share base and height
Why not D: the distance of the vertex from A does not enter the area; only base and height matter
Remember: Same base, same parallels, same area.
Why D: Step 1 (share the perimeter in the ratio): 3 + 5 + 7 = 15 parts, 300/15 = 20 m per part, so the sides are 60, 100 and 140 m. Step 2 (semi-perimeter): s = 150, so s – a, s – b, s – c = 90, 50, 10. Step 3 (Heron’s formula): area = √(150 × 90 × 50 × 10) = √6750000 = √(2250000 × 3) = 1500√3 ≈ 2598 m².
Why not A: 750√3 is half the correct area, as if Heron’s result needed an extra ½
Why not B: 3000 = ½ × 60 × 100 assumes a right angle, but 60² + 100² = 13600 is not 140² = 19600
Why not C: 6750000 is the product under the root; the square root has not been taken
Remember: Ratio → actual sides → s → Heron → square root.
Why A: Step 1 (the diagonal of an inscribed square is a diameter): diagonal = 2r. Step 2 (area of a square from its diagonal d, like a rhombus: ½d²): ½ × (2r)² = 2r². Step 3 (compare with A = πr²): 2r² / πr² = 2/π ≈ 0.637.
Why not B: π/2 ≈ 1.57 is greater than 1, which would mean the square is bigger than the circle around it
Why not C: 4/π uses a square of side 2r, which is the square drawn around the circle, not inside it
Why not D: 1/π uses a square of side r; the side of the inscribed square is r√2
Remember: Inscribed square: diagonal = diameter, area 2r², share 2/π ≈ 64%.
Why B: A is true: the C/D ratio does not change when a circle is scaled up or down, so a coin and the equator share the same ratio π. R is also true: π cannot be written as a ratio of two integers. But irrationality says nothing about why the ratio is the same for all sizes. A would be true even if the constant were a simple number, just as perimeter : side is 4 : 1 for every square.
Why not A: R is true, but being irrational is not the reason the ratio is size-independent; the square has a constant ratio 4 : 1 that is rational
Why not C: R is a correct fact: π is irrational, proved by Lambert in 1761
Why not D: A is correct: the C/D ratio is the same for circles of all sizes
Remember: Constant (same for all circles) and irrational (not a/b) are two different facts about π.
Why C: Step 1 (arc length = 2πr × θ/360): 2 × 22/7 × 14 × 90/360 = 88 × 1/4 = 22 cm. Step 2 (a sector’s boundary has two straight radii as well): perimeter = 22 + 14 + 14 = 50 cm.
Why not A: 22 cm is only the curved part; the two straight radii are also part of the boundary
Why not B: 36 cm adds only one radius; a sector is closed by two radii
Why not D: 154 is the area of the quadrant in cm², 22/7 × 14² × 1/4
Remember: Sector perimeter = arc + r + r.
Why A: Step 1 (symmetry of an isosceles trapezium): dropping perpendiculars from the ends of the 20 cm side splits the 40 cm side into 10 + 20 + 10. Step 2 (Baudhāyana-Pythagoras theorem): height h = √(26² – 10²) = √(676 – 100) = √576 = 24 cm. Step 3 (trapezium area = ½(a + b)h): ½ × (40 + 20) × 24 = 30 × 24 = 720 cm².
Why not B: 1560 = (40 + 20) × 26 uses the slant side as the height and leaves out the ½
Why not C: 780 = ½ × 60 × 26 uses the slanting 26 cm side instead of the perpendicular height
Why not D: 1440 = 60 × 24 forgets the ½ in the trapezium formula
Remember: Trapezium = ½ × (sum of parallel sides) × perpendicular height; find h first.
Why A: Step 1 (perimeter): third side = 32 – 8 – 11 = 13 cm. Step 2 (semi-perimeter): s = 32/2 = 16, so s – a, s – b, s – c = 8, 5, 3. Step 3 (Heron’s formula): area = √(16 × 8 × 5 × 3) = √1920 = √(64 × 30) = 8√30 ≈ 43.8 cm².
Why not B: 44 = ½ × 8 × 11 treats the two given sides as base and height, but they are not known to be perpendicular
Why not C: 1920 is the product under the root; the square root has not been taken
Why not D: 4√30 is half the correct area, as if Heron’s result needed an extra ½
Remember: Heron already contains everything; no extra ½ and no guessed height.
Why D: Step 1 (sector area = πr² × θ/360): 3.14 × 100 × 90/360 = 78.5 cm². Step 2 (triangle formed by two perpendicular radii, ½ × base × height): ½ × 10 × 10 = 50 cm². Step 3 (segment = sector – triangle): 78.5 – 50 = 28.5 cm².
Why not A: 78.5 cm² is the minor sector; the triangle under the chord has not been removed
Why not B: 235.5 cm² is the major sector, for the 270° angle
Why not C: 128.5 adds the triangle to the sector instead of subtracting it
Remember: Minor segment = sector minus triangle.
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Frequently Asked Questions
π is the ratio of the circumference of a circle to its diameter, C/D. If a circle is enlarged, its circumference and diameter grow by the same factor, so their ratio does not change. People noticed this in ancient times. π is roughly 22/7 or 3.14, and the circumference of a circle is C = πd = 2πr.
No. 22/7 is a ratio of two integers, so it is rational, while π is irrational, as Lambert proved in 1761. The digits of π go on forever without a repeating pattern. So we write π ≈ 22/7 and π ≠ 22/7. The value 355/113 found by Zu Chongzhi is even closer, but it is still only an approximation.
An arc that subtends an angle θ° at the centre of a circle of radius r has length 2πr × θ/360. It is the fraction θ/360 of the full circumference. A semicircle (180°) has length πr and a quarter circle (90°) has length πr/2. The perimeter of a sector is this arc length plus the two radii, 2r.
Add the three sides and halve the total to get the semi-perimeter s = (a + b + c)/2. Then work out s − a, s − b and s − c. The area is √[s(s − a)(s − b)(s − c)]. For sides 3, 4 and 5, s = 6 and the area is √(6 × 3 × 2 × 1) = 6 square units.
For a cyclic 4-gon, one whose four vertices lie on a circle, with sides a, b, c, d and s = (a + b + c + d)/2, the area is √[(s − a)(s − b)(s − c)(s − d)]. It cannot be used for a general 4-gon, because four sides alone do not fix the area. Putting d = 0 turns it into Heron’s formula.
A sector is the region bounded by an arc and the two radii to its ends, like a slice of pizza. Its area is πr² × θ/360. A segment is the region bounded by an arc and the chord joining its ends. The minor segment equals the sector minus the triangle formed by the chord and the two radii.
A median joins a vertex to the midpoint of the opposite side, so the two smaller triangles have equal bases. They also share the same height, the perpendicular from that vertex to the opposite side. Since area is half of base times height, their areas are equal, even though the two triangles usually have different shapes.
