Lessons / Quantitative aptitude
Quantitative aptitude · Day 7

Mensuration

Areas scale with the square of lengths, volumes with the cube. A 10% longer side gives 21% more area and 33.1% more volume.

Why this topic matters

How much paint for a wall, how much water in a tank, how much wire around a field? Mensuration is measuring. Every question asks one of three things: the length around (perimeter), the flat space inside (area), or the space a solid fills (volume).

Placement papers ask 1 or 2 mensuration questions: areas of triangles, rectangles, circles and trapeziums, volumes of cylinders and cones, and 'melt one solid into another'. Once you can picture the shape, each question is one formula. So this lesson draws every shape first, and says what each letter in the formula stands for.

Perimeter, area, volume: what they measure

Perimeter is the distance around a flat shape. Walk around a field and count your steps. It is a length, measured in cm or m.

Area is how much flat surface a shape covers. Count how many 1 cm × 1 cm squares fit inside. It is measured in square units: cm2, m2.

Volume is how much space a solid takes up. Count how many 1 cm cubes fit inside. It is measured in cubic units: cm3, m3. Capacity (how much liquid it holds) is volume too: 1000 cm3 = 1 litre.

Surface area is the total area of all the faces of a solid: the paint you would need to cover a box. 'Curved surface area' (CSA) counts only the curved part; 'total surface area' (TSA) adds the flat ends as well.

6 cm4 cm6 cm4 cmdiagonal
A 6 cm × 4 cm rectangle: perimeter 20 cm (walk around), area 24 cm2 (24 unit squares fit inside)
MeasureUnitsThink of it as
Perimetercm, mfence around a field
Areacm2, m2tiles on a floor
Volumecm3, m3, litreswater in a tank
Surface areacm2, m2paint on a box
Example 1A rectangular field is 50 m long and 30 m wide. How much fencing goes around it, and how much grass is inside?
  1. Perimeter = 2 × (50 + 30) = 160 m.Two lengths and two breadths.
  2. Area = 50 × 30 = 1500 m2.Length times breadth counts the unit squares.
  3. Fencing 160 m, grass 1500 m2.
Exam tipCheck the units in the options first. If the question asks for area and an option says 'm', that option is wrong.

Triangles: the six types

Triangles are named in two ways. By their sides: equilateral (all three sides equal), isosceles (two sides equal), scalene (all different). By their biggest angle: acute (every angle under 90 degrees), right-angled (one angle exactly 90), obtuse (one angle over 90).

The three angles of every triangle add up to 180 degrees. So an equilateral triangle has three 60-degree angles, and an isosceles triangle has its two base angles equal.

A triangle can be both kinds at once: a right-angled isosceles triangle has angles 90, 45, 45.

aaa60°60°60°
Equilateral: all 3 sides equal, every angle 60°
baseaaxx
Isosceles: 2 sides equal, the 2 base angles equal
cab
Scalene: all 3 sides different
Acute-angled: every angle below 90°
basehypotenuseheight
Right-angled: one angle is 90°
>90°
Obtuse-angled: one angle above 90°
Angle sum = 180°
Any triangle.
Any two sides together are longer than the third
3, 4, 8 cannot make a triangle because 3 + 4 < 8.
Biggest side faces the biggest angle
In a right triangle the side opposite 90° (hypotenuse) is the longest.
Example 1Two angles of a triangle are 48° and 67°. Find the third angle and say what type the triangle is by its angles.
  1. Third angle = 180 - 48 - 67 = 65°.The three angles always add to 180.
  2. All three angles (48, 67, 65) are below 90.The type by angle depends on the biggest angle.
  3. 65°, and the triangle is acute-angled.
Exam tipIn a 'which type' question, look only at the biggest angle: under 90 acute, exactly 90 right, over 90 obtuse.

Area of a triangle, and right triangles

Area of any triangle = 1/2 × base × height. The height is the perpendicular (straight-down) distance from the top corner to the base, not a slanting side. A triangle is exactly half of a rectangle with the same base and height; that is where the 1/2 comes from.

If you only know the three sides, use Heron's formula: find s = half the perimeter, then area = √(s(s - a)(s - b)(s - c)).

In a right-angled triangle, Pythagoras connects the sides: hypotenuse2 = base2 + height2. Some whole-number triples come up again and again. Learn them and many questions need no square roots at all.

Equilateral triangle with side a: height = (√(3)/2) × a, area = (√(3)/4) × a2.

base bh
Area = 1/2 × base × height. The height is the perpendicular, not a side
basehypotenuseheight
Right triangle: hypotenuse2 = base2 + height2
Pythagorean triple (a, b, hypotenuse)Multiples also work
3, 4, 56, 8, 10 · 9, 12, 15
5, 12, 1310, 24, 26 · 15, 36, 39
8, 15, 1716, 30, 34 · 24, 45, 51
7, 24, 2514, 48, 50 · 21, 72, 75
20, 21, 2940, 42, 58 · 60, 63, 87
12, 35, 3724, 70, 74 · 36, 105, 111
9, 40, 4118, 80, 82 · 27, 120, 123
Area = 1/2 × base × height
Height must be perpendicular to the base.
Heron: s = (a + b + c)/2, area = √(s(s-a)(s-b)(s-c))
When only the three sides are given.
Equilateral: area = (√(3)/4) a2, height = (√(3)/2) a
One side is enough.
Right triangle: area = 1/2 × (the two shorter sides)
The two legs are base and height of each other.
Isosceles with equal sides a and base b: area = (b/4) √(4a2 - b2)
Drop the height to the middle of the base.
Example 1Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.
  1. s = (13 + 14 + 15)/2 = 21.Heron needs half the perimeter.
  2. s - a = 8, s - b = 7, s - c = 6.Subtract each side from s.
  3. Area = √(21 × 8 × 7 × 6) = √(7056) = 84 cm2.Pair factors: 21 × 8 × 7 × 6 = (7 × 3)(2 × 4)(7)(2 × 3) = 72 × 24 × 32, whose root is 7 × 4 × 3 = 84.
Example 2The hypotenuse of a right triangle is 25 cm and one side is 7 cm. Find its area.
  1. Other side = √(252 - 72) = √(576) = 24.Or spot the triple 7, 24, 25.
  2. Area = 1/2 × 7 × 24 = 84 cm2.The two shorter sides are base and height.
Example 3An equilateral triangle has side 8 cm. Find its area.
  1. Area = (√(3)/4) × 82 = (√(3)/4) × 64.Formula for equilateral triangles.
  2. = 16 √(3) cm2 (about 27.7 cm2).
Exam tipSee 5, 12, 13 or 8, 15, 17 hiding in the numbers? It is a right triangle; skip the square root.

Square and rectangle

A rectangle has opposite sides equal and all angles 90 degrees. Area = length × breadth. Perimeter = 2(length + breadth). The diagonal cuts it into two right triangles, so diagonal = √(l2 + b2).

A square is a rectangle with all four sides equal. Area = a2. Perimeter = 4a. Diagonal = a × √(2). If you know only the diagonal d, the area is d2 / 2.

lblbdiagonal
Rectangle: area = l × b, perimeter = 2(l + b), diagonal = √(l2 + b2)
aaaaa√2
Square: area = a2 = (diagonal2)/2, perimeter = 4a, diagonal = a × √(2)
Rectangle: area lb, perimeter 2(l + b), diagonal √(l2 + b2)
Square: area a2 = d2/2, perimeter 4a, diagonal a √(2)
Path of width w around a rectangle (outside)
Area of path = 2w(l + b + 2w)
Example 1The diagonal of a square is 12 cm. Find its area.
  1. Area = d2 / 2 = 144 / 2 = 72 cm2.A square is two right isosceles triangles on the diagonal; this formula skips finding the side.
Example 2A garden 40 m × 30 m has a path 2 m wide all around it on the outside. Find the area of the path.
  1. Outer rectangle: (40 + 4) × (30 + 4) = 44 × 34 = 1496 m2.The path adds 2 m on each side, so 4 m to each dimension.
  2. Garden: 40 × 30 = 1200 m2.The inner part.
  3. Path = 1496 - 1200 = 296 m2.

Parallelogram

A parallelogram is a 'pushed-over' rectangle: opposite sides are parallel and equal, but the corners are not 90 degrees.

Area = base × height, where the height is the perpendicular distance between the two parallel sides. Cut the triangle off one end and slide it to the other and you get a rectangle with the same base and height. That is why.

The slanted side is NOT the height. Using it is the most common mistake here.

base bsideh
Parallelogram: area = base × height (the height, NOT the slanted side)
Area = base × height
Perpendicular height, not the slanted side.
Perimeter = 2(a + b)
Two pairs of equal sides.
Diagonals bisect each other
Each diagonal cuts the other in half.
Example 1A parallelogram has base 15 cm and the perpendicular height to it is 8 cm. Its slanted side is 10 cm. Find its area and perimeter.
  1. Area = 15 × 8 = 120 cm2.Use the perpendicular height; ignore the 10 cm side for area.
  2. Perimeter = 2(15 + 10) = 50 cm.Perimeter uses the actual sides.

Rhombus

A rhombus is a parallelogram with all four sides equal: a diamond. Its two diagonals cut each other at right angles and in half.

Area = 1/2 × d1 × d2 (half the product of the diagonals). The diagonals split the rhombus into 4 right triangles, and together those 4 triangles make half of the rectangle d1 × d2.

Because the diagonals meet at 90 degrees, side2 = (d1/2)2 + (d2/2)2. That lets you get the side from the diagonals.

aaaad1/2d2/2
Rhombus: 4 equal sides, diagonals cut at 90°. Area = 1/2 × d1 × d2
Area = 1/2 × d1 × d2
Also = side × height, like any parallelogram.
Side = √((d1/2)2 + (d2/2)2)
Pythagoras on one of the four small right triangles.
Perimeter = 4 × side
Example 1The diagonals of a rhombus are 16 cm and 12 cm. Find its area, side and perimeter.
  1. Area = 1/2 × 16 × 12 = 96 cm2.Half the product of the diagonals.
  2. Half-diagonals are 8 and 6, so side = √(64 + 36) = 10 cm.The diagonals meet at 90°: a 6-8-10 right triangle.
  3. Perimeter = 4 × 10 = 40 cm.All four sides equal.

Trapezium

A trapezium has exactly one pair of parallel sides (the top a and bottom b), with a perpendicular height h between them.

Area = 1/2 × (a + b) × h: the average of the two parallel sides times the height. Put two copies of the trapezium together, one upside down, and they make a parallelogram with base (a + b) and height h. Half of that is one trapezium.

If the non-parallel sides are equal, it is an isosceles trapezium; drop two heights and the bottom splits into a, and two equal pieces of (b - a)/2.

b (bottom)a (top)h
Trapezium: one pair of parallel sides. Area = 1/2 × (a + b) × h
Area = 1/2 × (sum of parallel sides) × height
(a + b)/2 is the 'average width'.
Isosceles trapezium: each overhang = (b - a)/2
Then height = √(slant2 - overhang2).
Example 1The parallel sides of a trapezium are 12 cm and 20 cm, and the distance between them is 7 cm. Find its area.
  1. Area = 1/2 × (12 + 20) × 7.Add the parallel sides, halve, times height.
  2. = 16 × 7 = 112 cm2.
Example 2An isosceles trapezium has parallel sides 10 cm and 22 cm, and each slanted side is 10 cm. Find its area.
  1. Overhang on each side = (22 - 10)/2 = 6 cm.The longer base sticks out equally on both sides.
  2. Height = √(102 - 62) = 8 cm.Each slanted side is the hypotenuse of a 6-8-10 triangle.
  3. Area = 1/2 × (10 + 22) × 8 = 128 cm2.
Exam tipTrapezium area = average of the parallel sides × height. Say it that way and you will not forget the 1/2.

Circle, semicircle, sector and ring

Every point on a circle is the same distance r (radius) from the centre. The diameter d = 2r goes all the way across.

Circumference (perimeter) = 2 π r. Area = π r2. Use π = 22/7 when the radius is a multiple of 7, otherwise 3.14.

A sector is a slice, like a pizza slice, with angle t at the centre. It is t/360 of the whole circle, so its area is (t/360) × π r2 and its curved edge (arc) is (t/360) × 2 π r.

A semicircle is a sector of 180 degrees. Its perimeter is the curved half (π r) PLUS the diameter (2r). Forgetting the diameter is a classic trap.

A ring (a circular path) is a big circle minus a small one: π (R2 - r2).

rsectortchord
Circle: area = π r2, circumference = 2 π r. Sector with angle t: (t/360) × π r2
Rr
Ring (annulus): area = π (R2 - r2) = π (R + r)(R - r)
Circumference = 2 π r = π d
Area = π r2
Sector area = (t/360) π r2, arc = (t/360) 2 π r
t = angle at the centre.
Semicircle perimeter = π r + 2r
Curved part plus the straight diameter.
Ring area = π (R2 - r2) = π (R + r)(R - r)
The factorised form avoids big squares.
A wheel of radius r covers 2 π r in one turn
Distance = turns × circumference.
Example 1The circumference of a circle is 44 cm. Find its area (π = 22/7).
  1. 2 × 22/7 × r = 44, so r = 7 cm.Get the radius first; everything about a circle comes from r.
  2. Area = 22/7 × 7 × 7 = 154 cm2.
Example 2Find the area of a sector of angle 90° in a circle of radius 14 cm (π = 22/7).
  1. A 90° sector is 90/360 = 1/4 of the circle.Sectors are fractions of the full circle.
  2. Area = 1/4 × 22/7 × 14 × 14 = 154 cm2.
Example 3A circular park of radius 20 m has a path 7 m wide around it on the outside. Find the area of the path (π = 22/7).
  1. Outer radius R = 27, inner r = 20.The path adds 7 m to the radius.
  2. Area = 22/7 × (27 + 20)(27 - 20) = 22/7 × 47 × 7.Ring formula in factorised form; the 7 cancels.
  3. = 22 × 47 = 1034 m2.
Example 4A wheel of radius 35 cm makes how many turns to cover 1.1 km (π = 22/7)?
  1. One turn = 2 × 22/7 × 35 = 220 cm.Circumference is the distance per turn.
  2. 1.1 km = 110000 cm.Same units before dividing.
  3. Turns = 110000 / 220 = 500.

Polygons

A polygon is any closed shape with straight sides: pentagon (5), hexagon (6), octagon (8). 'Regular' means all sides and all angles are equal.

From one corner you can cut an n-sided polygon into (n - 2) triangles. Each triangle has 180 degrees, so the interior angles add up to (n - 2) × 180. In a regular polygon each interior angle is that divided by n.

The exterior angles (the turn you make at each corner walking around) always add to 360, so each exterior angle of a regular polygon is 360/n.

A regular hexagon is 6 equilateral triangles, so its area is 6 × (√(3)/4) a2 = (3 √(3)/2) a2.

5 sides: cut into 3 triangles from one corner, so angles add to 540°
6 sides: cut into 4 triangles from one corner, so angles add to 720°
Sides nAngle sumEach interior angle (regular)Each exterior angleDiagonals
3180601200
436090902
5540108725
6720120609
810801354520
1014401443635
1218001503054
Sum of interior angles = (n - 2) × 180
Each exterior angle (regular) = 360/n
Interior + exterior = 180.
Diagonals = n(n - 3)/2
Each corner joins n - 3 others; each diagonal counted twice.
Regular hexagon area = (3 √(3)/2) a2
Six equilateral triangles.
Example 1Each interior angle of a regular polygon is 140°. How many sides does it have?
  1. Each exterior angle = 180 - 140 = 40°.Interior and exterior angles at a corner make a straight line.
  2. n = 360 / 40 = 9.Exterior angles add to 360.
Exam tipFor 'how many sides' questions go through the exterior angle: n = 360 / (180 - interior).

Cuboid and cube

A cuboid is a box with length l, breadth b and height h. Volume = l × b × h: the number of 1 cm cubes that fill it.

It has 6 rectangular faces in 3 matching pairs (top/bottom, front/back, left/right). Total surface area = 2(lb + bh + hl). The four walls of a room without floor and ceiling (lateral surface) = 2h(l + b).

The longest rod that fits inside goes corner to corner: diagonal = √(l2 + b2 + h2).

A cube has l = b = h = a: volume a3, surface 6a2, diagonal a √(3).

lbh
Cuboid: volume = l × b × h, total surface = 2(lb + bh + hl), diagonal = √(l2 + b2 + h2)
Cuboid: volume lbh, TSA 2(lb + bh + hl), walls 2h(l + b), diagonal √(l2 + b2 + h2)
Cube: volume a3, TSA 6a2, diagonal a √(3)
Number of small cubes = big volume / small volume
Only if the sides divide exactly; otherwise count along each edge.
Example 1A room is 10 m long, 8 m wide and 4 m high. Find the cost of painting its four walls at Rs 20 per m2.
  1. Area of four walls = 2 × 4 × (10 + 8) = 144 m2.Walls only: no floor, no ceiling.
  2. Cost = 144 × 20 = Rs 2880.
Example 2How many 2 cm cubes can be cut from a cuboid 10 cm × 8 cm × 6 cm?
  1. Along each edge: 10/2 = 5, 8/2 = 4, 6/2 = 3.Count how many fit on each edge.
  2. 5 × 4 × 3 = 60 cubes.

Cylinder

A cylinder is a circle pulled straight up to height h: a pipe, a tin, a drum. Volume = base area × height = π r2 h.

Unroll the curved side and it becomes a rectangle: one side is the circumference 2 π r, the other is h. So curved surface area = 2 π r h. Add the two circular ends for the total: 2 π r h + 2 π r2 = 2 π r (r + h).

A hollow pipe with outer radius R and inner radius r has volume of material π (R2 - r2) h.

rh
Cylinder: volume = π r2 h, curved surface = 2 π r h, total = 2 π r (r + h)
Volume = π r2 h
Area of the circle × height.
CSA = 2 π r h
The label on a tin.
TSA = 2 π r (r + h)
Label plus the two lids.
Hollow: π (R2 - r2) h
Metal in a pipe.
Example 1A cylindrical tank has radius 7 m and height 10 m. How many litres does it hold (π = 22/7)?
  1. Volume = 22/7 × 7 × 7 × 10 = 1540 m3.π r2 h.
  2. 1 m3 = 1000 litres.Convert at the end.
  3. 1540 × 1000 = 15,40,000 litres.

Cone

A cone is a circle that narrows to a point: an ice-cream cone, a tent. Its height h goes straight up the middle; its slant height l runs down the side.

h, r and l make a right triangle, so l = √(r2 + h2).

A cone holds exactly one third of a cylinder with the same base and height: volume = (1/3) π r2 h. (Fill a cone with water and pour it into the matching cylinder: it takes three cones to fill it.)

Curved surface area = π r l. Total = π r (l + r).

hrl
Cone: volume = 1/3 π r2 h, slant l = √(r2 + h2), curved surface = π r l
Volume = (1/3) π r2 h
One third of the matching cylinder.
Slant l = √(r2 + h2)
CSA = π r l, TSA = π r (l + r)
Surface uses the slant height, volume uses the vertical height.
Example 1A cone has radius 6 cm and height 8 cm. Find its curved surface area, in terms of π.
  1. Slant l = √(36 + 64) = 10 cm.A 6-8-10 right triangle.
  2. CSA = π × 6 × 10 = 60 π cm2.
Example 2A conical tent has radius 7 m and height 24 m. Find the canvas needed (π = 22/7).
  1. Slant = √(49 + 576) = 25 m.The 7-24-25 triple.
  2. Canvas = CSA = 22/7 × 7 × 25 = 550 m2.A tent has no floor, so only the curved surface.

Sphere and hemisphere

A sphere is a perfect ball: every point on it is r from the centre. Volume = (4/3) π r3. Surface area = 4 π r2 (exactly four circles of the same radius).

A hemisphere is half a ball: a bowl. Volume = (2/3) π r3. Its curved surface is 2 π r2; a solid hemisphere also has a flat top, so total surface = 3 π r2.

r
Sphere: volume = 4/3 π r3, surface = 4 π r2. Hemisphere: 2/3 π r3, total surface 3 π r2
Sphere: volume (4/3) π r3, surface 4 π r2
Hemisphere: volume (2/3) π r3, CSA 2 π r2, TSA 3 π r2
TSA adds the flat circle on top.
Example 1Find the volume of a sphere of radius 21 cm (π = 22/7).
  1. (4/3) × (22/7) × 21 × 21 × 21.Formula for a sphere.
  2. Cancel first: one 21 divided by 7 is 3, another 21 divided by 3 is 7. Left: 4 × 22 × 3 × 7 × 21.Cancelling before multiplying keeps the numbers small.
  3. 4 × 22 = 88 and 3 × 7 × 21 = 441, so 88 × 441 = 38,808 cm3.

Frustum, prism and pyramid

Cut the top off a cone with a slice parallel to the base and the bottom piece is a frustum: a bucket or a glass. With top radius r, bottom radius R and height h, volume = (1/3) π h (R2 + r2 + Rr).

A prism has the same shape all the way through (a triangular chocolate box, a pencil). Volume = area of the base shape × length.

A pyramid rises from a base shape to a point. Like a cone, it is one third of the matching prism: volume = (1/3) × base area × height.

rRh
Frustum (bucket): volume = 1/3 π h (R2 + r2 + Rr)
prismpyramid
Prism: volume = base area × height. Pyramid: 1/3 × base area × height
Frustum volume = (1/3) π h (R2 + r2 + Rr)
Big cone minus small cone, done once.
Prism volume = base area × height
Any constant cross-section.
Pyramid volume = (1/3) × base area × height
One third of the prism.
Example 1A bucket is a frustum with top radius 14 cm, bottom radius 7 cm and height 15 cm. Find its capacity in cm3 (π = 22/7).
  1. R2 + r2 + Rr = 196 + 49 + 98 = 343.Work out the bracket first.
  2. Volume = 1/3 × 22/7 × 15 × 343.Put it into the formula.
  3. = 22 × 5 × 49 = 5390 cm3.15/3 = 5 and 343/7 = 49 cancel cleanly.
Example 2A pyramid has a square base of side 6 m and height 10 m. Find its volume.
  1. Base area = 36 m2.Square base.
  2. Volume = 1/3 × 36 × 10 = 120 m3.

Scaling, and melting one solid into another

If every length of a shape grows by a factor k, its area grows by k2 and its volume by k3. Double the radius of a ball and it needs 4 times the paint but holds 8 times the air.

As percentages: if each side rises by x%, the area rises by (2x + x2/100)%. A 10% longer side gives 21% more area and 33.1% more volume.

When one solid is melted and recast into another, nothing is lost: the volume stays the same. Write 'volume before = volume after', cancel π, and solve.

Lengths × k → area × k2, volume × k3
Each side up x% → area up (2x + x2/100)%
Successive percentage change applied twice.
Melt and recast: volume stays the same
Set the two volume formulas equal.
Number of small solids = big volume / small volume
Example 1A metal sphere of radius 6 cm is melted and drawn into a wire of radius 0.2 cm. Find the length of the wire in metres.
  1. Sphere volume = (4/3) π × 216 = 288 π.Keep π; it cancels.
  2. Wire is a cylinder: π × 0.04 × L.r = 0.2, so r2 = 0.04.
  3. 288 π = 0.04 π L, so L = 7200 cm.Volume is conserved; divide.
  4. = 72 m.
Example 2Each side of a rectangle is increased by 20%. By what percent does its area increase?
  1. Take sides 10 and 10: area 100. New sides 12 and 12: area 144.The 'take 100' method from Easy Maths.
  2. Increase = 44%. (Formula: 20 + 20 + 400/100 = 44.)
Exam tipMelting questions: write both volume formulas side by side and cancel π and any 1/3 or 4/3 before multiplying anything.

Formula summary

Rectangle
Area lb, perimeter 2(l + b), diagonal √(l2 + b2).
Triangle (Heron)
s = (a + b + c)/2, area √(s(s - a)(s - b)(s - c)).
Equilateral
√(3)/4 a2.
Circle
π r2, circumference 2 π r. Sector: (theta/360) π r2.
Cube / cuboid
a3, 6a2 / lbh, 2(lb + bh + hl). Cube diagonal a √(3).
Cylinder
Volume π r2 h, CSA 2 π r h, TSA 2 π r(r + h).
Cone
Volume (1/3) π r2 h, slant l = √(r2 + h2), CSA π r l.
Sphere / hemisphere
(4/3) π r3, 4 π r2 / (2/3) π r3, 3 π r2.
% change of area
Each side changes x% → area changes 2x + x2/100 %.

Shortcuts and tricks

Melt and recast

Use it when: One solid melted into one or many others.

  1. Volume is conserved: set the volume formulas equal.
  2. Cancel π and fractions before multiplying.
ExampleA hemispherical bowl of radius 9 cm is full of liquid, poured into cylinders of radius 1.5 cm and height 4 cm. How many cylinders?
  1. Bowl = (2/3) π × 729 = 486 π.Hemisphere volume.
  2. Cylinder = π × 2.25 × 4 = 9 π.π r2 h.
  3. 486 / 9 = 54 cylinders.

Spot the triple

Use it when: Any right triangle, cone slant, rhombus side or ladder question.

  1. Check whether the numbers are a multiple of 3-4-5, 5-12-13, 8-15-17 or 7-24-25 before taking a square root.
ExampleA ladder 13 m long reaches a window 12 m high. How far is its foot from the wall?
  1. 5, 12, 13 is a triple.Ladder = hypotenuse, wall = one side.
  2. 5 m.

Percent change of area

Use it when: Sides change by x% and y%.

  1. Change in area = x + y + xy/100 (use minus signs for decreases).
ExampleLength +20%, breadth -20%. Change in area?
  1. 20 - 20 + (20 × -20)/100 = -4.Successive-change formula.
  2. 4% decrease.

Common mistakes

Using the slanted side as the height of a parallelogram or trapezium.
Height is always the perpendicular distance between the parallel sides.
Semicircle perimeter = π r only.
Add the diameter: π r + 2r.
Using the vertical height for the cone's curved surface.
Surface uses the slant height l = √(r2 + h2).
Mixing cm and m in one calculation.
Convert everything to one unit before you start; 1 m3 = 1000 litres.
Thinking a 20% longer side gives 20% more area.
Area scales with the square: 1.22 = 1.44, so 44% more.

Before you move on, you should be able to...

  • say whether a question wants perimeter, area or volume, and check the units in the options
  • name a triangle by its sides and by its angles, and find its area three ways
  • find the area of a parallelogram, rhombus and trapezium, and explain why each formula works
  • use sector = (angle/360) of the circle, and remember the diameter in a semicircle's perimeter
  • find volume and surface of a cuboid, cylinder, cone and sphere, and solve a melt-and-recast question

Practice questions (27)

Try each one before opening the solution. Or practise them one by one so your score is saved.

easy Find the area of a triangle with base 12 cm and height 9 cm.
  1. 1/2 × 12 × 9.Half of base times height.
  2. = 54 cm2.

Answer: 54 sq cm

easy The parallel sides of a trapezium are 9 cm and 15 cm and the height is 6 cm. Find its area.
  1. 1/2 × (9 + 15) × 6.Average of the parallel sides times height.
  2. = 12 × 6 = 72.

Answer: 72 sq cm

easy The diagonals of a rhombus are 10 cm and 24 cm. Find its area.
  1. 1/2 × 10 × 24.Half the product of the diagonals.
  2. = 120.

Answer: 120 sq cm

easy A circle has circumference 44 cm (π = 22/7). Find its area.
  1. 2 π r = 44 gives r = 7.Radius first.
  2. 22/7 × 49 = 154.

Answer: 154 sq cm

easy A cylinder has radius 7 cm and height 10 cm (π = 22/7). Find its volume.
  1. 22/7 × 49 × 10.π r2 h.
  2. = 1540.

Answer: 1540 cu cm

easy Find the area of a triangle with sides 13, 14 and 15.
  1. s = 21.Half the perimeter.
  2. √(21 × 8 × 7 × 6) = 84.

Answer: 84

easy A cube has a diagonal of 6 √(3) cm. Find its volume.
  1. Diagonal = a √(3), so a = 6.Cube diagonal formula.
  2. 63 = 216.

Answer: 216

easy How many 2 cm cubes can be cut from a cuboid 10 × 8 × 6 cm?
  1. 5 × 4 × 3.Count along each edge.
  2. = 60.

Answer: 60

easy Two angles of a triangle are 35° and 55°. What kind of triangle is it by its angles?
  1. Third angle = 180 - 35 - 55 = 90.Angles add to 180.
  2. One angle is 90, so it is right-angled.

Answer: Right-angled

easy Each side of a rectangle is increased by 20%. By what percent does its area increase?
  1. 20 + 20 + 400/100.x + y + xy/100.
  2. = 44%.

Answer: 44%

medium A parallelogram has sides 12 cm and 8 cm, and the height on the 12 cm side is 5 cm. Find its area.
  1. Area = base × perpendicular height = 12 × 5.The 8 cm side is slanted; it is not the height.
  2. = 60.

Answer: 60 sq cm

medium The diagonals of a rhombus are 30 cm and 16 cm. Find its perimeter.
  1. Side = √(152 + 82) = 17.Half-diagonals make a right triangle (8-15-17).
  2. Perimeter = 4 × 17 = 68.

Answer: 68 cm

medium Each interior angle of a regular polygon is 150°. How many sides does it have?
  1. Exterior angle = 30.180 - 150.
  2. 360 / 30 = 12.

Answer: 12

medium How many diagonals does a regular octagon have?
  1. n(n - 3)/2 = 8 × 5 / 2.Each corner joins 5 non-neighbours.
  2. = 20.

Answer: 20

medium Find the perimeter of a semicircular plate of radius 14 cm (π = 22/7).
  1. Curved part = 22/7 × 14 = 44.Half of 2 π r.
  2. Add the diameter 28.The straight edge is part of the boundary.
  3. 44 + 28 = 72.

Answer: 72 cm

medium Find the area of a sector of angle 60° in a circle of radius 21 cm (π = 22/7).
  1. 60/360 = 1/6 of the circle.Sector as a fraction.
  2. 1/6 × 22/7 × 441 = 231.

Answer: 231 sq cm

medium A sphere has radius 21 cm (π = 22/7). Find its volume.
  1. 4/3 × 22/7 × 9261.Sphere formula; 213 = 9261.
  2. = 38,808.

Answer: 38,808 cu cm

medium A cone has radius 6 and height 8. Find its curved surface area in terms of π.
  1. l = 10.6-8-10.
  2. π × 6 × 10 = 60 π.

Answer: 60 π

medium A wire bent into a square of area 121 sq cm is re-bent into a circle. Find the area of the circle (π = 22/7).
  1. Side 11, so the wire is 44 cm long.Perimeter of the square = length of wire.
  2. 2 π r = 44 gives r = 7.Same wire, now the circumference.
  3. Area = 154.

Answer: 154 sq cm

hard An isosceles trapezium has parallel sides 6 cm and 18 cm, and each slanted side is 10 cm. Find its area.
  1. Overhang each side = (18 - 6)/2 = 6.The longer base sticks out equally.
  2. Height = √(100 - 36) = 8.6-8-10 triangle.
  3. 1/2 × (6 + 18) × 8 = 96.

Answer: 96 sq cm

hard A hemispherical bowl of radius 9 cm is full of liquid, which is poured into cylinders of radius 1.5 cm and height 4 cm. How many cylinders are filled?
  1. Bowl: (2/3) π × 729 = 486 π. Cylinder: π × 2.25 × 4 = 9 π.Volumes, keeping π.
  2. 486 / 9 = 54.

Answer: 54

hard A rectangle's length rises 20% and its breadth falls 20%. What is the change in area?
  1. 1.2 × 0.8 = 0.96.Multiply the scale factors.
  2. A 4% fall.

Answer: -4%

hard A bucket (frustum) has radii 14 cm and 7 cm and height 15 cm. Find its capacity (π = 22/7).
  1. R2 + r2 + Rr = 196 + 49 + 98 = 343.The bracket.
  2. 1/3 × 22/7 × 15 × 343 = 5390.

Answer: 5390 cu cm

hard A circular path 7 m wide runs around the outside of a park of radius 35 m. Find the area of the path (π = 22/7).
  1. R = 42, r = 35.Add the width to the radius.
  2. 22/7 × (42 + 35)(42 - 35) = 22/7 × 77 × 7.Ring formula, factorised.
  3. = 22 × 77 = 1694.

Answer: 1694 sq m

hard A regular hexagon has side 4 cm. Find its area in terms of √(3).
  1. 6 equilateral triangles of side 4.A regular hexagon splits into 6 equilateral triangles.
  2. Each: (√(3)/4) × 16 = 4 √(3).Equilateral formula.
  3. 6 × 4 √(3) = 24 √(3) cm2.

Answer: 24 √(3)

hard The radius of a sphere is doubled. How many times does its volume become?
  1. Volume grows with r3.All lengths × 2 means volume × 23.
  2. 23 = 8 times.

Answer: 8

hard Water flows through a pipe of radius 2 cm at 7 m per minute into a cylindrical tank of radius 20 cm. By how much does the water level rise in 10 minutes?
  1. In 10 minutes the water column in the pipe is 7 × 10 = 70 m = 7000 cm long.Speed × time gives the length of water that came out.
  2. Water volume = π × 22 × 7000 = 28000 π cm3.That column is a cylinder of radius 2.
  3. In the tank: π × 202 × h = 28000 π, so 400 h = 28000.The same volume now sits in the wide tank; cancel π.
  4. h = 70 cm.

Answer: 70 cm

Class plan (2 hours, for trainers)

A tested order for teaching this topic in one 2-hour session. Present mode follows the same order.

10 minSpeed maths warm-up: squares 11 to 25 round the class, then two squares of numbers ending in 5.
10 minHook: fence, tiles, water tank. Perimeter vs area vs volume with a grid on the board. Units check habit.
15 minTriangles: draw the six types, let students name them. Angle sum. Area = 1/2 b h and why; Heron with 13-14-15; Pythagorean triples table.
15 minQuadrilaterals: rectangle and square, then parallelogram (cut and slide), rhombus (diagonals), trapezium (two copies make a parallelogram).
10 minCircles: circumference and area, sector as a fraction of 360, semicircle perimeter trap, ring. Wheel-turns example.
5 minPolygons: triangles from one corner, angle sum, exterior angle trick, diagonals.
15 minSolids: cuboid and cube, cylinder (unroll the label), cone (1/3 of a cylinder), sphere and hemisphere. Draw each once.
5 minFrustum, prism, pyramid: one example each.
10 minScaling and melting: k2 and k3, the sphere-to-wire example.
15 minIndividual practice: easy and medium questions on the practice page, timed.
5 minGo through two hard questions on the projector.
5 minRecap and exit ticket: area of a trapezium with parallel sides 8 and 12, height 5 (answer 50).