Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Shape | Perimeter / length | Area |
|---|---|---|
| Rectangle | 2(a + b) | ab |
| Parallelogram | 2(a + b) | bh |
| Triangle | a + b + c | ½bh; Heron √(s(s−a)(s−b)(s−c)) |
| Trapezium | sum of sides | ½(a + b)h |
| Kite / rhombus | — | ½ d₁d₂ |
| Circle | 2πr | πr² |
| Arc / sector | 2πr·θ/360 | πr²·θ/360 |
| Cyclic 4-gon | — | √((s−a)(s−b)(s−c)(s−d)) |
Try each problem first, then open the solution to check your working.
Find the circumference of a circle of radius 14 cm.
C = 2πr = 2 × 22/7 × 14 = 88.
Answer: 88 cm
The circumference of a circle is 132 cm. Find its radius.
2πr = 132, so r = 132 × 7/(2 × 22) = 21.
Answer: 21 cm
Find the length of an arc of angle 120° in a circle of radius 21 cm.
l = 2πr × 120/360 = 2 × 22/7 × 21 × 1/3 = 44.
Answer: 44 cm
Find the perimeter of a quadrant of a circle of radius 7 cm.
Arc = πr/2 = 11 cm.
Perimeter = 11 + 7 + 7 = 25.
Answer: 25 cm
A wheel has diameter 70 cm. How many complete turns does it make in 1.1 km?
Circumference = 22/7 × 70 = 220 cm.
1.1 km = 110000 cm; turns = 110000/220 = 500.
Answer: 500
Lane width is 1.22 m. Find the extra distance covered by a runner in the next lane over two semicircular bends (π = 3.1416).
Extra = 2π × 1.22 = 7.665.
Answer: about 7.67 m
Find the area of a triangle with sides 13, 14, 15 cm and the height on the side 14.
s = 21; area = √(21 × 8 × 7 × 6) = √7056 = 84.
Height = 2 × 84/14 = 12 cm.
Answer: 84 cm²; 12 cm
Find the area of a triangle with sides 9, 10, 17 cm.
s = 18; area = √(18 × 9 × 8 × 1) = √1296 = 36.
Answer: 36 cm²
The area of ΔABC is 60 cm² and AD is a median. Find the area of ΔABD. If P is on AD, compare the areas of ΔPBD and ΔPCD.
A median bisects the area: ΔABD = 30 cm².
ΔPBD and ΔPCD have equal bases (BD = DC) and the same height from P, so they are equal.
Answer: 30 cm²; equal
A parallelogram has sides 12 cm and 9 cm and the height on the 12 cm side is 7 cm. Find its area and the height on the 9 cm side.
Area = 12 × 7 = 84.
Height on 9 cm = 84/9 = 28/3.
Answer: 84 cm²; 28/3 cm
Find the area of a trapezium with parallel sides 10 cm and 16 cm and height 6 cm.
½(10 + 16) × 6 = 78.
Answer: 78 cm²
The diagonals of a rhombus are 16 cm and 12 cm. Find the area and the perimeter.
Area = ½ × 16 × 12 = 96.
Side = √(8² + 6²) = 10, so perimeter = 40.
Answer: 96 cm²; 40 cm
Find the area of a cyclic quadrilateral with sides 25, 39, 52, 60 cm.
s = 88; s − a = 63, 49, 36, 28.
Area = √(63 × 49 × 36 × 28) = √3111696 = 1764.
Answer: 1764 cm²
Find the area of a sector of angle 90° in a circle of radius 14 cm.
(1/4) × 22/7 × 14 × 14 = 154.
Answer: 154 cm²
Two concentric circles have radii 10 cm and 6 cm. Find the area between them. A chord of the larger circle touches the smaller circle; find its length.
Area = π(10² − 6²) = 64π ≈ 201.1.
Half-chord = √(100 − 36) = 8, so chord = 16 cm. (Check: π l²/4 = π × 256/4 = 64π.)
Answer: 64π ≈ 201 cm²; 16 cm
A circular park has radius 14 m and a path 7 m wide runs around it. Find the area of the path.
π(21² − 14²) = 22/7 × 245 = 770.
Answer: 770 m²
A chord of a circle of radius 6 cm subtends 60° at the centre. Find the area of the minor segment (π = 3.14, √3 = 1.73).
Sector = 3.14 × 36/6 = 18.84.
Triangle = (√3/4) × 36 = 9 × 1.73 = 15.57.
Segment = 18.84 − 15.57 = 3.27.
Answer: about 3.27 cm²
Show that a circle’s C² : A is the same for all circles and find it.
C² = 4π²r² and A = πr², so C²/A = 4π ≈ 12.57 (the Babylonians used 12).
Answer: 4π
n equal circles of diameter d are placed in a row in a rectangle that just contains them. What fraction of the rectangle do they cover?
Rectangle: nd × d = nd². Circles: n × πd²/4.
Fraction = π/4 ≈ 0.785, whatever n is.
Answer: π/4
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No. The extra distance run in an outer lane over a bend is 2π × (lane width) per full circle of bends; it depends on the lane width and the number of bends, not on the radius of the inner lane. A 200 m track has smaller bends but the same 400 m race needs twice as many laps, so with the same lane width the stagger is the same per lap (and larger in total), not smaller.
The curved ends of the track are semicircles. Their lengths (πr each) need the circumference formula, and the stagger comes from comparing semicircles of different radii.
Radius difference = lane width = 1.22 m. Over two bends (one full circle) the extra distance is 2π × 1.22 ≈ 7.67 m, which is the stagger. The third lane is again 1.22 m wider than the second, so the same 7.67 m stagger is needed.
A regular hexagon inscribed in a circle of radius 1 has six sides of length 1 (equilateral triangles with the centre), so its perimeter is 6. The circle is longer than the hexagon, so 2π > 6 and π > 3.
Circumscribed hexagon: the tangent at distance 1 from the centre; half-side = tan 30° = 1/√3 (from the right triangle with the Baudhāyana–Pythagoras theorem: a² = 1 + (a/2)² gives a = 2/√3). Perimeter = 6 × 2/√3 = 4√3 ≈ 6.93. The circle lies between the two hexagons: 6 < 2π < 4√3, so 3 < π < 2√3 ≈ 3.46.
Wind 20 turns, measure L and D. The ratio L/(20D) is the C/D ratio. A careful measurement with thin thread gives about 3.1–3.2 (close to 3.14).
2πr = 44, r = 44 × 7/(2 × 22) = 7 cm.
(i) 2 × 22/7 × 7 = 44.0 cm. (ii) 2 × 22/7 × 10 = 62.857 ≈ 62.9 cm. (iii) 2 × 22/7 × 12 = 75.43 ≈ 75.4 cm.
(i) 2 × 22/7 × 3.5 × 60/360 = 22/6 = 11/3 ≈ 3.67 cm. (ii) 2 × 22/7 × 6.3 × 1/3 = 39.6/3 = 13.2 m.
Arc = 2 × 22/7 × 14 × 75/360 = 88 × 5/24 = 55/3 ≈ 18.33. Perimeter = 18.33 + 14 + 14 = 46.33 cm (139/3).
2 × 80 + π × 60 = 160 + 188.57 = 348.57 m.
Outer arc π × 6 = 18.86; inner arc π × 4 = 12.57; two flat ends of (12 − 8)/2 = 2 cm each = 4. Perimeter = 18.86 + 12.57 + 4 = 35.43 cm.
4 semicircles of diameter 10: 4 × π × 5 = 20π = 62.86 cm.
3 × π × 6 = 18π = 56.57 cm.
Four quarter circles = one full circle: 2π × 14 = 88. Four semicircles: 4 × π × 7 = 88. Perimeter = 176 cm (4πs with s = 14). The figure is read as a grid of 14 cm cells.
Large arc π × 14 = 44; four small arcs 4 × π × 3.5 = 44. Perimeter = 88 cm.
Hypotenuse 10. Perimeter = π(10 + 8 + 6)/2 = 12π = 37.71 cm.
6π + 3 × 2π = 12π = 37.71 cm.
10π + 5π + 5π = 20π = 62.86 cm. (Figure read as: one outer semicircle of radius 10, an inner semicircle of radius 5 above, and one of radius 5 below the other half of the base.)
(i) π d = 22/7 × 56 = 176 cm. (ii) 10 km = 1,000,000 cm; 1000000/176 ≈ 5682 revolutions (5681.8).
(i) Eight quarter-circle arcs of radius 7: 8 × (2π × 7/4) = 28π = 88 cm. (ii) Twelve arcs of 60° with radius 42: 12 × (2π × 42/6) = 528 cm.
C = 2πr, so r ∝ C: 5 : 4.
Pick D′ on DA and A′ on DA extended with A′A = D′D. Then ΔCDD′ ≅ ΔBAA′, so A′BCD′ has the same area as ABCD and is closer to a rectangle. Repeat until the perpendicular foot lands on the base.
No. With the sides fixed, changing the angle changes the height h, so the area bh changes. (A rhombus of side 3 can have area 9, 8.01 or 5.41.)
Enclose the triangle in a rectangle after extending the base. The area is (big right triangle with base b + b′) − (small right triangle with base b′) = ½(b + b′)h − ½b′h = ½bh.
Yes. Equal-area polygons can always be cut into finitely many pieces and rearranged into each other (the Wallace–Bolyai–Gerwien theorem). For the two halves of a triangle, one can cut each into pieces by lines parallel to the base, with a suitable shear.
The conjecture is true: any two polygons of equal area can be cut into a finite number of pieces that rearrange to cover each other. A triangle becomes a rectangle (base b, height h/2), and a rectangle can be squared by Baudhāyana’s method (or cut and sheared to any other rectangle of the same area).
Sides a and 20 − a with 0 < a < 20: infinitely many. Area = a(20 − a) = 100 − (a − 10)², largest when a = 10: the 10 × 10 square, area 100. There is no smallest area: it can be made as close to 0 as we like (very thin rectangles) but never reaches 0.
Open-ended: a circle is easy to draw with a peg and a rope, has the same width in all directions, rolls (wheel), distributes stress evenly, and holds the most area for a given perimeter. It also has religious and aesthetic symmetry.
Make a rectangle with the same area: base b and height h/2. Then square the rectangle by Baudhāyana’s construction. The side of the square is √(bh/2).
The angles at C′ and C (and at B′ and B) are alternate angles. Since ∠B′C′A′ = ∠BCA, the opposite sides are parallel, so the figure is a parallelogram. Its area is bh, so each triangle has area ½bh.
Base AD = 10, and E lies on BC, which is 8 cm from AD: area = ½ × 10 × 8 = 40 cm².
Overhang each side = (40 − 20)/2 = 10; height = √(26² − 10²) = √576 = 24. Area = ½(40 + 20) × 24 = 720 cm².
Third side = 32 − 19 = 13; s = 16. Area = √(16 × 8 × 5 × 3) = √1920 = 8√30 ≈ 43.8 cm².
Sides 60, 100, 140 m; s = 150. Area = √(150 × 90 × 50 × 10) = √6,750,000 = 1500√3 ≈ 2598 m².
Diagonals d and 2d: area = ½ × d × 2d = d² = 128, so d = 8√2 ≈ 11.3 cm.
Both triangles have base CD and the same height (the distance between the parallel lines AB and CD). The ratio is 1 : 1.
The diagonals meet at M, the centre; reflection in M swaps Q and S and maps the line PR to itself, so Q and S are at the same distance from PR. ΔPSO and ΔPQO have the common base PO and equal heights, so their areas are equal.
Let P, Q, R, S be the midpoints of AB, BC, CD, DA. ΔAPS has AP = ½AB and AS = ½AD with the same angle A, so its area is ¼ of ΔABD; similarly ΔCQR = ¼ ΔCBD. Together ¼ of the 4-gon. In the same way ΔBPQ + ΔDRS = ¼ of the 4-gon. The four corner triangles total ½ the area, so the parallelogram PQRS has area ½ of ABCD.
ΔABD = ΔACD (median bisects the area) and ΔPBD = ΔPCD (equal bases BD = DC, same height). Subtract: ΔABP = ΔABD − ΔPBD = ΔACD − ΔPCD = ΔACP.
ΔPAB + ΔPCD = ½ × AB × (distances of P from AB and CD) = ½ × s × s = ½s², because the two heights add up to s. Likewise the green area is ½s². Ratio 1 : 1.
ΔPDC and ΔPDQ lie on the same base PD between the parallels PD and CQ, so they are equal. Then ΔBPQ = ΔBPD + ΔPDQ = ΔBPD + ΔPDC = ΔBDC = ½ ΔABC (CD is a median).
22/7 × 49 × 60/360 = 154/6 = 77/3 ≈ 25.67 cm².
r = 7 cm. Area = ¼ × 22/7 × 49 = 38.5 cm².
10 minutes = 60°. Area = 77/3 ≈ 25.67 cm².
(i) ¼ × 3.14 × 100 = 78.5 cm². (ii) 270°: ¾ × 314 = 235.5 cm².
Sector = 3.14 × 225/6 = 117.75. Triangle (equilateral, side 15) = (√3/4) × 225 = 97.31. Minor segment = 117.75 − 97.31 = 20.44 cm². Circle = 3.14 × 225 = 706.5; major segment = 706.5 − 20.44 = 686.06 cm².
One wiper: 22/7 × 784 × 1/3 = 821.33. Two wipers: 1642.67 cm².
Sector = πr²/6. The triangle is equilateral with side r: area (√3/4)r². Segment = r²(π/6 − √3/4) = (r²/12)(2π − 3√3).
Side = r√3 (chord subtending 120°), area = (√3/4) × 3r² = (3√3/4)r². Ratio = (3√3/4)r²/(πr²) = 3√3/(4π) ≈ 0.413.
The diagonal is 2r, so the area is ½(2r)(2r) = 2r². Ratio = 2r²/(πr²) = 2/π ≈ 0.637.
The hexagon is six equilateral triangles of side r: area 6 × (√3/4)r² = (3√3/2)r². Ratio 3√3/(2π) ≈ 0.827. The inscribed equilateral triangle (every second vertex) has area (3√3/4)r², exactly half of the hexagon, so the hexagon’s ratio is twice that of Q8.
AF = (a + b)/2 so HG = HK = (a + b)/2; BH = AH − AB = (a − b)/2. In the right triangle HKP, HP² = HK² − KP²... with KP = BH: HP² = ((a + b)/2)² − ((a − b)/2)² = ab. So the square HPQS has area ab.
(a + b)(a − b): draw a square of side a and remove a square of side b from a corner. The L-shape has area a² − b². Cut it along the dotted line and rearrange into a rectangle of sides (a + b) and (a − b). (a + b + c)²: draw a square of side a + b + c split into a 3 × 3 grid of rectangles; the diagonal cells are a², b², c² and the six other cells are ab, ab, bc, bc, ca, ca.
Base = 40 − 30 = 10. Height = √(15² − 5²) = √200 = 10√2. Area = ½ × 10 × 10√2 = 50√2 ≈ 70.7 cm².
Height = 2 × 60/10 = 12. Equal side = √(12² + 5²) = 13 cm.
Other leg = 108/12 = 9; hypotenuse = 15. Perimeter = 12 + 9 + 15 = 36 cm.
Sides 10, 15, 20; s = 22.5. Area = √(22.5 × 12.5 × 7.5 × 2.5) = √5273.4375 = 75√15/4 ≈ 72.6 cm².
Since 7² + 24² = 25², the triangle is right-angled: ½ × 7 × 24 = 84 cm². Heron: s = 28, √(28 × 21 × 4 × 3) = √7056 = 84 cm².
Circumference = 22/7 × 60 = 188.57 cm. 100 turns: 18857 cm ≈ 188.6 m.
r = 66 × 7/44 = 10.5 cm. Area = ¼ × 22/7 × 110.25 = 86.625 cm².
2πr = 176 cm in one turn. 1 km = 100000 cm: 100000/176 ≈ 568 turns (568.2).
Yes. If sides a, b and c, d have a + b = c + d and ab = cd, then the pairs are the roots of the same quadratic x² − (a + b)x + ab = 0, so {a, b} = {c, d}.
Draw a line from an end of the shorter side parallel to the slanted leg. It splits the trapezium into a parallelogram (base a, height h, area ah) and a triangle (base b − a, height h, area ½(b − a)h). Total = ah + ½(b − a)h = ½(a + b)h.
A diagonal splits it into two triangles with bases a and b and the same height h: area = ½ah + ½bh = ½(a + b)h.
Rotate a copy by 180° about the midpoint of a slanted leg and join. The result is a parallelogram with base a + b and height h, area (a + b)h. The trapezium is half of it: ½(a + b)h.
(i) Algebra: the diagonal AC is the axis; BD ⟂ AC at O. Area = ½ × AC × OB + ½ × AC × OD = ½ × AC × (OB + OD) = ½ × AC × BD. (ii) Geometry: enclose the kite in the rectangle with sides AC and BD; the diagonals divide it into four rectangles and the kite takes exactly half of each, so area = ½ × AC × BD.
Area PQRS = 4ab = 4 × ab. Yes: cut PQRS by the lines through the midpoints of its sides into four congruent a × b rectangles.
Heron: s becomes 2s and each of s − a, s − b, s − c doubles, so area = √(2s × 2(s−a) × 2(s−b) × 2(s−c)) = 4 × area. Yes: joining the midpoints of the sides of ΔPQR cuts it into 4 triangles congruent to ΔABC.
Heron gives 3² = 9 times the area. Yes: trisect each side and draw lines parallel to the sides through the points; this makes 9 triangles congruent to ΔABC.
Fig. 6.43: the shaded quadrilateral is ΔPBQ minus ΔPMN, where PQ = ⅔ of the side PC. ΔPBQ = ⅔ of the triangle; ΔPMN = ¼ of ΔPBQ = 1/6. Shaded = 2/3 − 1/6 = ½ of the triangle. Fig. 6.44: take the square with corners (0,0), (2,0), (2,2), (0,2). The four lines are y = 2 − 2x, y = 4 − 2x, y = x/2 and y = 1 + x/2. They meet at (0.8, 0.4), (1.6, 0.8), (1.2, 1.6), (0.4, 1.2): a tilted square of side √0.8, area 0.8. Fraction = 0.8/4 = 1/5.
With n circles of diameter d in a row the rectangle is nd × d. Fraction = nπd²/4 ÷ nd² = π/4 ≈ 0.785 in both figures.
Conjecture: the circles cover exactly π/4 of the rectangle, whatever the number. 10, 20, 50 circles all give π/4. Proof: n circles of diameter d have area nπd²/4 and the rectangle nd² (width nd, height d).
Let each be l × w (l > w). The top row of 4 lies flat (length 4l) and the bottom row of 5 stands (width 5w): 4l = 5w, so l : w = 5 : 4. Let l = 5k, w = 4k. Each has area 8 (72/9): 20k² = 8, k² = 2/5. Perimeter = 2(l + w) = 18k = 18√(2/5) = 18√10/5 ≈ 11.4 cm.
The blue and red triangles have equal bases (each ⅓ of the base) and the same height, so their areas are equal. Cutting: cut the blue triangle along the line through the midpoints of its two slanting sides, giving a small top triangle and a trapezium. Rotate the top triangle 180° about the midpoint of a slanted side to form a parallelogram (base b, height h/2). Slide a small triangle from one end of this parallelogram to the other to move it to the red triangle’s position, then do the first step in reverse to reach the red triangle.
Let the side be a. Quarter disc (centre at the corner) has area πa²/4. Each semicircle (diameter a) has area πa²/8, and their union plus region B makes the quarter disc. Union = S₁ + S₂ − A (A is the overlap). So πa²/4 = πa²/8 + πa²/8 − A + B, giving A = B.
Each petal has two quarter-circle arcs of radius 1: length 2 × π/2 = π. Four petals: perimeter = 4π ≈ 12.57 units. Area: the four semicircles total 4 × π/2 = 2π and cover the square once with the petals counted twice, so petals = 2π − 4 ≈ 2.28 sq. units.
OA ⟂ BC so A is the midpoint of BC: AB = l/2. With radii R and r: R² − r² = (l/2)² (Baudhāyana–Pythagoras in ΔOAB). Area of the ring = π(R² − r²) = π l²/4.
Let the legs be a, b and hypotenuse c: a² + b² = c². The semicircle on the hypotenuse passes through the right-angle vertex and contains the triangle and the two caps outside the legs. Area A + B = (semicircle on a) + (semicircle on b) − (the two caps) = π(a² + b²)/8 − (πc²/8 − C) = π(a² + b² − c²)/8 + C = C.
The overlap is two segments, each a 120° sector minus a triangle. 120° sector = πr²/3; triangle ACD (sides r, r, angle 120°) = ½r² sin 120° = (√3/4)r². Overlap = 2(πr²/3 − (√3/4)r²) = (2π/3 − √3/2) r² ≈ 1.23 r².
Let the rectangle be W × H, and use the notation of the figure: the right triangle C has legs u (vertical) and v (horizontal): C = ½uv. A = ½u·t where t is the horizontal distance from the left side to the vertical leg of C, and B = ½v·ρ, where ρ is the height of the right end above the base. Then A + C = ½u(t + v) = ½uW and B + C = ½v(ρ + u) = ½vH. So (A + C)(B + C) = ¼ uvWH, and 2(A + C)(B + C)/C = 2 × ¼uvWH/(½uv) = WH, the area of the rectangle.
Take OA = OB = r, ∠AOB = 90°, and let the shaded regions be the lune between the semicircle on AB (diameter AB = r√2) and the arc AFB, and the triangle AOB. Semicircle on AB: ½π(r/√2)² = πr²/4. Quarter sector OAB = πr²/4. Segment = sector − triangle = πr²/4 − T. Lune = semicircle − segment = πr²/4 − (πr²/4 − T) = T = ½r², the area of the triangle. So the two shaded regions are equal.