Chapter 14 of 14

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Bihar (BSEB)Class 8 Mathematics★ 12-15 marks · CBSE Class 8 Maths (Ganita Prakash Part 2, Chapter 7, printed as Chapter 14)

Area

The chapter that reduces every area formula to one idea — count unit squares, then cut everything into triangles. Why perimeter tells you nothing about area, the triangle formula proved even when the altitude falls outside the base, parallelogram, rhombus and trapezium by dissection, the Śulba-Sūtra reshaping problems, and shortest paths by reflection. CBSE Class 8 Maths (Ganita Prakash Part 2, Chapter 7).

⏱ 26 min readMedium difficulty✓ Free · NCERT-aligned

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Measure area by counting unit squares, and explain why perimeter cannot stand in for area
2Prove Area of a triangle = ½ × base × height, including when the foot of the altitude falls outside the base
3Use the fact that a median divides a triangle into two of equal area
4Find the area of any polygon by cutting it into triangles
5Derive the parallelogram, rhombus and trapezium formulas by dissection
6Carry out the Śulba-Sūtra transformations between rectangle, triangle, rhombus and trapezium
7Use the reflection construction to find shortest paths and least-perimeter triangles
8Convert between cm², in², ft², m², km² and acres

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Why this chapter matters

The chapter that turns area from a list of formulas into a single idea: count unit squares, and get every other shape by cutting it into triangles. It settles why perimeter tells you nothing about area, proves the triangle formula for every triangle including the awkward obtuse one, and then derives the parallelogram, rhombus and trapezium rules by dissection rather than assertion. The Śulba-Sūtra reshaping problems running through it are the oldest recorded area problems in Indian mathematics, and the reflection argument for the shortest path is the same one that governs light.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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The Baudhāyana–Pythagoras Theorem

Needed for the dissection with a square hole and for equilateral-triangle heights

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Quadrilaterals

Parallelogram, rhombus and trapezium properties, and diagonals bisecting each other

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Exploring Some Geometric Themes

Congruence criteria (SAS, ASA, RHS), used in every dissection proof here

Area — Class 8 Mathematics (Ganita Prakash Part 2)

Two shapes can have the same boundary and different insides. That single fact is why area has to be measured by counting squares, and why this chapter exists.

1. About the Chapter

This is Chapter 7 of Ganita Prakash Part 2 (pages 148–171), the fourteenth and final chapter of the Class 8 course.

It is not a formula sheet. Every rule in it is derived, and derived the same way: cut the figure into pieces you already understand, or reshape it into one. The chapter opens by asking how many ways a square can be quartered — the answer is infinitely many — and closes by asking you to estimate the area of your own town.

Section	What it establishes
Rectangles and squares	Area = counting unit squares; perimeter is not a measure of area
Triangles	½ × base × height for every triangle; medians halve areas; triangles between parallels
Area of any polygon	Every polygon cuts into triangles, so one formula covers them all
Parallelogram	base × height, by dissection into a rectangle
Rhombus	½ × product of the diagonals, by two different routes
Trapezium	½ × height × (sum of the parallel sides), by three different routes
Areas in real life	cm², in², ft², m², km², acres, and the local units of India

Running right through it are the Śulba-Sūtras — ancient Indian texts on altar construction, where an altar had to have both a prescribed shape and a prescribed area. That forces a very particular kind of problem: reshape this figure into that one without gaining or losing a square inch of ground. Four of the exercises are marked as coming from them.

2. Area Means Counting Unit Squares

Which rangoli takes more powder — a rectangle 7 cm by 4 cm, or one 8 cm by 3 cm?

Pack each with squares of side 1 cm:

7 cm × 4 cm holds 7 × 4 = 28 unit squares
8 cm × 3 cm holds 8 × 3 = 24 unit squares

The first needs more powder. We write these as 28 cm² and 24 cm², or 28 sq. cm and 24 sq. cm.

Area of a rectangle = length × width

Draw a diagonal and the rectangle falls into two congruent triangles, so each has half the area: ½ × 7 × 4 = 14 cm². That is the first sighting of the triangle formula.

Why perimeter can't stand in for area

Look again at those two rangoli rectangles:

Rectangle	Perimeter	Area
7 cm × 4 cm	22 cm	28 cm²
8 cm × 3 cm	22 cm	24 cm²

Same perimeter, different areas. The reverse happens too. A 1 cm × 10 cm strip has perimeter 22 cm and area 10 cm²; a 5 cm × 5 cm square has perimeter 20 cm and area 25 cm². Here the region with the longer boundary encloses less than half the space.

Stretch a rectangle thin and you buy a great deal of boundary for almost no area. A notched comb, a spiral, a star — all do the same. So area cannot be defined by measuring the edge; it has to be defined by counting what is inside.

3. Triangles

The formula, and why it holds for every triangle

Take ∆ABC and draw the line through A parallel to BC. Dropping perpendiculars from B and C to that line boxes the triangle inside a rectangle whose sides are the base and the height. The triangle fills exactly half of it:

Area of a triangle = ½ × base × height

Here height always means the perpendicular distance from the apex to the line of the base — never a slanted side.

The awkward case. If the angle at B is obtuse, the foot of the altitude from A lands outside the segment BC, at a point D on CB extended, and no rectangle on base BC contains the triangle. Write the triangle as a difference instead:

Area(∆ABC) = Area(∆ADC) − Area(∆ADB) = ½ × h × DC − ½ × h × DB = ½ × h × (DC − DB) = ½ × h × BC ✓

So the formula needs no case analysis. It is one rule for all triangles.

Reading the formula backwards

A triangle has three bases and three matching heights, and all three pairs must give the same area. That single observation solves a whole class of problems.

If AX ⊥ BC with AX = 5 and BC = 3, then the area is ½ × 5 × 3 = 15/2. If AC = 4 and BY ⊥ AC, then the area is also ½ × 4 × BY = 2 BY. Equating, BY = 15/4 = 3.75 units. The longer base always carries the shorter height, because the product is fixed.

Medians halve areas

Draw both diagonals of a rectangle and four triangles appear — not congruent, but all equal in area. Take two adjacent ones with bases OD and OB along the same diagonal: the diagonals of a rectangle bisect each other, so OB = OD, and both triangles have the same apex and therefore the same height. Equal bases, equal heights, equal areas.

The general statement is worth committing to memory, because the chapter uses it four more times:

In a triangle, the line joining a vertex to the midpoint of the opposite side divides the triangle into two triangles of equal area.

Triangles between parallel lines

Let l be parallel to BC, and consider every triangle with base BC whose apex lies somewhere on l.

Their areas. All the same. The base is common and the height is the fixed distance between the parallels, so sliding the apex along l changes the shape and nothing else. There is no largest and no smallest.

Their perimeters. These do differ. BC is common, so only AB + AC matters. Treat l as a mirror and reflect C to C′. Then AC = AC′ for every position of A, so

AB + AC = AB + AC′

and that is a path from B to C′ that bends at A. The shortest path from B to C′ is the straight segment BC′, so the best A is where BC′ crosses l. Putting B = (0, 0), C = (c, 0) and l at height h makes C′ = (c, 2h), and the segment BC′ crosses l at x = c/2 — directly above the midpoint of BC.

The triangle of least perimeter is the isosceles one. There is no greatest — push A far along l and the perimeter grows without limit.

The same reflection trick answers Gopal's problem: to carry water from his house to the river and then to his tank by the shortest route, reflect the tank in the river, join the house to that image, and fetch the water where the line crosses. It is also why the angle of incidence equals the angle of reflection for light.

4. The Area of Any Polygon

Quadrilateral. Join a diagonal BD and the quadrilateral becomes two triangles standing on it. Both share that base, so

Area = ½ × BD × h₁ + ½ × BD × h₂ = ½ × d × (h₁ + h₂)

Three measurements — one diagonal and two perpendicular offsets. No sides, no angles.

Pentagon and beyond. Join one vertex to all the others. A pentagon becomes 3 triangles, a hexagon 4, and in general an n-sided polygon becomes n − 2 triangles.

Every straight-sided figure can be cut into triangles. So ½ × base × height is, in principle, the only area formula you need. Everything that follows is a shortcut.

Two consequences worth knowing.

Halving a quadrilateral. Join the midpoints of all four sides. The four corner triangles cut off come to ¼(∆ABC + ∆ACD) + ¼(∆ABD + ∆BCD) = ¼T + ¼T = ½T, so the midpoint quadrilateral has exactly half the original area.
A regular hexagon splits from its centre into six equilateral triangles, so its area is 3ad — or (3√3/2)a² if you know only the side.

5. Parallelogram, Rhombus, Trapezium

Parallelogram — by dissection

Construct AX ⊥ CD. That cuts parallelogram ABCD into ∆AXD and the trapezium ABCX. Extend XC and drop a perpendicular through B, meeting it at Y. Then ∆BYC is precisely the piece missing from ABCX before it becomes the rectangle ABYX, and

BY = AX (ABYX is a rectangle)
∠BYC = ∠AXD = 90°
BC = AD (opposite sides of a parallelogram)

so ∆BYC ≅ ∆AXD by RHS. The cut-off triangle fits the gap exactly. Since DX = CY, adding XC to both gives DC = XY, and therefore

Area of a parallelogram = base × height

Cutting a figure into pieces and reassembling them into a different figure of the same area is called dissection — the technique behind every remaining formula in the chapter.

Either side may serve as the base, provided the height used is the one perpendicular to that side. So base₁ × height₁ = base₂ × height₂ — which is exactly how a second height gets found. If 12 cm goes with 6 cm (area 72 cm²), then the 7.6 cm side carries a height of 72 ÷ 7.6 = 180/19 ≈ 9.47 cm.

A caution. A parallelogram with sides 5 cm and 4 cm does not have area 20 cm². Standing it on the 5 cm base, the height is a leg of a right triangle whose hypotenuse is the 4 cm side, so the height is less than 4 cm. Among all parallelograms with two given sides, the rectangle has the greatest area.

Rhombus — the diagonals do the work

A rhombus is a parallelogram, so base × height still applies. But its diagonals are perpendicular bisectors of each other, which gives something better. Cutting along a diagonal produces two isosceles triangles; turning each into a rectangle and joining them gives a single rectangle with sides AC and BD/2. Hence

Area of a rhombus = ½ × product of the diagonals

The same answer comes from adding two triangles: ½·AO·BD + ½·CO·BD = ½·BD·(AO + CO) = ½·AC·BD. Notice that this argument used only the perpendicularity of the diagonals, not the equal sides — so ½d₁d₂ works for any quadrilateral with perpendicular diagonals, a kite included.

Trapezium — three routes to one formula

Let WXYZ have WX ∥ ZY, with WX = a, ZY = b and height h. Drop WM and XN perpendicular to ZY, writing MZ = x and NY = y. Then WXNM is a rectangle, and

Area = ½xh + ah + ½yh = ½h(x + y + 2a)

Walking along ZY gives b = x + a + y, so x + y = b − a, and substituting:

Area of a trapezium = ½ × height × (sum of the parallel sides)

The bracket (a + b)/2 is the average of the two parallel sides, so a trapezium covers the same ground as a rectangle of that average width and the same height. Setting a = b returns the parallelogram formula.

When the trapezium leans hard — far enough that one perpendicular falls outside the long side — two arguments still work. Either box it in and subtract the overhanging triangle (the terms in the overhang cancel), or draw BG ∥ AD, which splits the figure into a parallelogram of area ah and a triangle of area ½(b − a)h. The second is tidier and works for any trapezium once you name the shorter parallel side a.

The two-copies proof. Take a second copy of the trapezium and rotate it. At the join, x + y = 180° (co-interior angles), so the outline has four corners rather than six; and u + v = 180° makes both pairs of opposite sides parallel. The result is a parallelogram of base a + b and height h, and the trapezium is half of it. This proof never cuts the figure at all, so it needs no separate case for the leaning trapezium.

6. The Śulba-Sūtra Transformations

Altars had to be built to an exact area in an exact shape, so the Śulba-Sūtras — and Euclid's Elements independently — worked out how to reshape figures without changing their areas. Four of these appear as exercises.

Problem	The move	Result
Rectangle → triangle	Extend the base to twice its length, join to a top corner	Triangle of base 2ℓ, height w
Triangle → rectangle	Cut along the midline, half-turn the top piece about a midpoint	Rectangle of base b, height h/2
Isosceles triangle → rectangle	One cut along the altitude, one half turn about its midpoint	Rectangle of width BC/2, height AD
Rectangle → rhombus	Halve it, make each half an isosceles triangle, join base to base	Rhombus of diagonals L and 2w

Each is checkable with card and scissors, which is exactly the point — the Vedic builders had cord and pegs, not algebra.

A related dissection. Cut a square with two perpendicular lines through its centre. The four congruent pieces, re-glued in a pinwheel, form a larger square with a square hole — of side L = √(s² + k²) with a hole of side k, since L² − k² = s². This is Perigal's dissection, and the picture is the Baudhāyana–Pythagoras figure from Chapter 3 read backwards.

7. Composite Regions: Add, Subtract, and Don't Double-Count

A path round a park is the outer rectangle minus the inner one: Area = LW − ℓw. For a uniform width t this becomes 2t(ℓ + w + 2t). Sliding the outer rectangle about does not change the answer at all — as one strip widens, the opposite one narrows by the same amount.

A crosspath through a 14 m × 12 m plot has arms of widths a and b:

Area = 12a + 14b − ab

The subtraction matters. Where the arms cross, that patch belongs to both and has been counted twice.

A bent tube of width 1 obeys the same rule: each corner square is shared by two arms. A spiral with nine arms totalling 120 and eight bends has area 120 − 8 = 112 sq units — equivalently, straightening it out gives a single tube 112 units long.

A pinwheel of rectangles is solved by walking round it: each rectangle hands you a side of the next. In the page-150 figure, the areas 14, 21, 28 and 35 in² all turn out to share a side of 7 in, which is what makes the pinwheel close up.

8. Areas in Real Life

An A4 sheet is 21 cm by 29.7 cm, so its area is 623.7 cm² — about a 25 cm square, and a handy mental yardstick.

Furniture is often measured in inches and feet:

1 in = 2.54 cm 1 ft = 12 in

Area conversions square the length factor. This is the single most common slip in the topic.

Conversion	Value	Why
1 in²	6.4516 cm²	2.54 × 2.54
10 in²	64.516 cm²	ten of them
161.29 cm²	25 in²	161.29 ÷ 6.4516
1 ft²	144 in²	12 × 12
1 km²	1,000,000 m²	1000 × 1000
1 acre	43,560 ft² ≈ 4047 m²	land measure

The same rule explains why doubling a square's side quadruples its area: lengths × k ⟹ areas × k².

Estimating is a skill the chapter takes seriously. Pace out your classroom (a typical one is about 8 m × 6 m ≈ 48 m²), then your school (perhaps 90 m × 60 m = 5400 m², a little over 1.3 acres), then your town from a map. India also uses many local units — bigha, gaj, katha, dhur, cent, ankanam — whose sizes are not standard between states, so any land record has to say which definition it uses. The same caution applies to "the largest city": municipal limits, metropolitan spread and administrative district give quite different figures, so state the boundary you mean and the year.

9. Summary

Area of a triangle = ½ × base × height — for every triangle, with height measured perpendicular to the base's line
The area of any polygon can be found by breaking it into triangles
Area of a parallelogram = base × height
Area of a rhombus = ½ × product of its diagonals
Area of a trapezium = ½ × height × sum of the parallel sides
Perimeter is not a measure of area — in either direction
A median halves a triangle; triangles on the same base between the same parallels have equal areas
Dissection preserves area, which is what every derivation and every Śulba-Sūtra construction relies on
Lengths × k ⟹ areas × k²

Appendix — What This Chapter Does Not Cover

Older Class 8 syllabuses put circles, Heron's formula and the surface area and volume of solids into the "Area" chapter. None of these is in Ganita Prakash Class 8. They are listed here only so you know where they belong, and do not go looking for them in the wrong year:

Topic	Where it actually appears
Area and circumference of a circle, πr² and 2πr	Class 9–10
Heron's formula √(s(s−a)(s−b)(s−c))	Class 9
Surface area and volume of cube, cuboid, cylinder	Class 9–10
Areas related to circles, sectors and segments	Class 10

The reason the chapter stops where it does is structural, not arbitrary: its method is cutting figures into triangles, and a circle has no straight sides to cut along. Class 9 picks up exactly where this leaves off, with Areas of Parallelograms and Triangles, and then adds the cases that need a genuinely new idea.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Rectangle

length × width

The number of unit squares that pack into it

Square

a²

A rectangle with both sides equal

Triangle

½ × base × height

Height is the PERPENDICULAR distance from the apex to the base's line — it may fall outside the base

Any polygon

sum of the triangles it cuts into

An n-sided polygon splits into n − 2 triangles from one vertex

Quadrilateral from a diagonal

½ × d × (h₁ + h₂)

d is the diagonal; h₁ and h₂ the perpendicular offsets to it

Parallelogram

base × height

Either side may be the base, paired with ITS perpendicular height

Rhombus

½ × d₁ × d₂

Half the product of the diagonals; holds for any quadrilateral with perpendicular diagonals

Trapezium

½ × height × (sum of the parallel sides)

Same as a rectangle of the average width and the same height

Regular hexagon

3ad, or (3√3/2)a²

Six equilateral triangles from the centre; d is the centre-to-side distance

Equilateral triangle

(√3/4)a²

Its height (√3/2)a ≈ 0.87a is always less than its side

Area scaling

lengths × k ⟹ areas × k²

Doubling a side quadruples the area; 1 ft = 12 in gives 1 ft² = 144 in²

Length conversions

1 in = 2.54 cm, 1 ft = 12 in, 1 km = 1000 m

Area conversions

1 in² = 6.4516 cm², 1 ft² = 144 in², 1 km² = 10⁶ m², 1 acre = 43,560 ft²

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using a slanted side as the height of a parallelogram or trapezium

✓ The height is the PERPENDICULAR distance between the parallel sides. A parallelogram with sides 5 cm and 4 cm has area LESS than 20 cm², because the height is less than 4 cm.

WATCH OUT

✗ Assuming equal perimeters mean equal areas

✓ 7 cm × 4 cm and 8 cm × 3 cm both have perimeter 22 cm but areas 28 cm² and 24 cm². The two measurements are independent, in both directions.

WATCH OUT

✗ Pairing a base with the wrong height

✓ Every base has its own height. If 12 cm goes with 6 cm, then 7.6 cm goes with 72 ÷ 7.6 ≈ 9.47 cm — never mix a base from one pair with a height from the other.

WATCH OUT

✗ Adding the arms of a crosspath or a bent tube without removing the overlap

✓ Where two arms cross, that patch belongs to both and has been counted twice. Subtract one copy: 12a + 14b − ab for a crosspath, and one 1 × 1 square per bend for a tube.

WATCH OUT

✗ Converting areas with the length factor instead of its square

✓ 1 in = 2.54 cm gives 1 in² = 2.54² = 6.4516 cm², not 2.54 cm². Likewise 1 km² = 1,000,000 m², not 1000 m².

WATCH OUT

✗ Believing the triangle formula needs a rectangle drawn round the triangle

✓ For an obtuse triangle write ∆ABC = ∆ADC − ∆ADB; the difference collapses to ½ × h × BC, so the formula holds for every triangle.

WATCH OUT

✗ Thinking the shape of a path changes its area when the outer rectangle is slid

✓ Area of path = outer − inner, and neither rectangle changes size. One strip widens exactly as much as the opposite one narrows.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

In-text — Rectangles, Squares and Why Perimeter Is Not Area

Section 7.1, pages 148-150 — infinitely many ways to quarter a square, the rangoli rectangles 7 × 4 and 8 × 3 that share a perimeter of 22 cm but not an area, the diagonal halving a rectangle, and regions where a longer boundary encloses less space

Questions

Figure it Out — Rectangles, Paths and Composite Regions

Pages 150-152 — the pinwheel of four rectangles all sharing a side of 7 in, the step figure giving widths 7.25 m and 2.75 m, the path round a park and the crosspath, the spiral tube of 112 sq units, the three regions of a divided square, and the four-piece dissection that opens a square hole

Questions

In-text — Triangles

Pages 153-156 — every triangle on a full side of a rectangle with its apex opposite is half the rectangle, the proof of ½ × base × height for obtuse triangles, finding a second altitude from the area, the four equal triangles cut by a rectangle's diagonals, and triangles between parallel lines

Questions

Figure it Out — Triangles and Sulba-Sutra Transformations

Pages 157-159 — three triangle areas, the altitude BY = 3 units, the isosceles ∆SUB of 48 sq units, the Śulba-Sūtra rectangle-triangle transformations, the three-square figure giving 12.25 and 144 sq units, the midpoint triangle at one quarter, and Gopal's shortest path to the river

Questions

In-text and Figure it Out — Area of any Polygon

Pages 159-160 — cutting a quadrilateral and a pentagon into triangles, the 66 cm² quadrilateral, the 110 cm² shaded region, the measurements a regular hexagon needs, the blue region that is always exactly half the rectangle, and halving a quadrilateral by joining midpoints

Questions

In-text — Parallelogram and Rhombus

Pages 161-166 — converting a parallelogram into a rectangle by dissection with the RHS congruence that makes the pieces fit, XY = DC giving base × height, using either side as the base, and two derivations of ½ × d₁ × d₂ for a rhombus

Questions

Figure it Out — Parallelogram Areas

Pages 162-163 — the seven grid parallelograms that all have area 15 square units but different perimeters, and the four measured parallelograms giving 28, 15, 24 and 8.8 cm²

Questions

Figure it Out — Parallelograms, Triangles and Sulba-Sutra Constructions

Pages 163-164 — QN = 180/19 ≈ 9.47 cm, the rectangle beating every parallelogram on the same two sides, rectangles of the same and of twice a triangle's area, the one-cut isosceles dissection, and the equilateral triangle against the square

Questions

In-text — Trapezium

Pages 166-169 — breaking trapeziums into a rectangle and triangles, the derivation of ½h(a + b) from x + y = b − a, both approaches for a strongly leaning trapezium, and the two-copies argument that builds a parallelogram of base a + b

Questions

Figure it Out — Rhombus and Trapezium

Pages 169-170 — the 150 cm² rhombus, four trapezium areas (136 ft², 420 m², 100 in², 120 ft²), rectangle-to-rhombus and isosceles-trapezium dissections, locating the rectangle EFGH from the midpoints, building a trapezium of area 144 cm², the hexagon ratio 3 : 1 : 2, and the ASA proof that ZYXW equals ∆ZWB

Questions

In-text — Areas in Real Life

Pages 170-171 — the A4 sheet at 623.7 cm², inch and centimetre conversions, 1 in² = 6.4516 cm² and 161.29 cm² = 25 in², 1 ft² = 144 in² and 1 km² = 10⁶ m², estimating a classroom and a school, acres and the local units bigha, gaj, katha, dhur, cent and ankanam

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Area?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~15 marks in Bihar (BSEB) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Area is the number of unit squares that fit; perimeter says nothing about it, in either direction
•Rectangle = length × width; a diagonal halves it
•Triangle = ½ × base × height, for every triangle — for an obtuse one, ∆ABC = ∆ADC − ∆ADB
•A median divides a triangle into two triangles of equal area
•Triangles on the same base between the same parallels all have the same area
•Any polygon = sum of triangles; an n-gon gives n − 2 from one vertex
•Quadrilateral from a diagonal = ½ × d × (h₁ + h₂)
•Parallelogram = base × height, either side serving as the base
•Rhombus = ½ × d₁ × d₂, and this holds whenever the diagonals are perpendicular
•Trapezium = ½ × height × (sum of the parallel sides) = a rectangle of the average width
•Regular hexagon = six equilateral triangles = 3ad = (3√3/2)a²
•Joining the midpoints of a quadrilateral's sides halves its area
•Reflection finds the shortest path and the least-perimeter triangle
•Śulba-Sūtras: triangle ↔ rectangle by the midline cut; isosceles triangle ↔ rectangle by one cut and one half turn
•Lengths × k ⟹ areas × k²: 1 in² = 6.4516 cm², 1 ft² = 144 in², 1 km² = 1,000,000 m²
•1 acre = 43,560 ft² ≈ 4047 m²; A4 = 21 × 29.7 = 623.7 cm²

Bihar (BSEB) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 12-15 marks per chapter — one of the highest weightages in Class 8 Maths

Question type	Marks each	Typical count	What it tests
MCQ / Very Short	1	3	Direct use of a formula; perimeter-versus-area traps; the k² scaling rule
Short Answer	2-3	2-3	Second-altitude problems; trapezium and parallelogram areas; composite regions and paths
Long Answer	4-5	1-2	Dissection proofs, Śulba-Sūtra transformations, and area arguments using medians or midpoints

Prep strategy

Learn the five formulas as one idea — everything is triangles
For every figure, identify the base and ITS perpendicular height before writing anything
Practise the double-counting subtraction: crosspaths, bent tubes, overlapping strips
Be able to prove, not just quote, the parallelogram and trapezium formulas
Rehearse the reflection construction for shortest paths
Square the factor for every area conversion
Always write units, and leave awkward answers as exact fractions

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Rangoli and floor design

Powder or tile needed is proportional to area, not to the length of the outline — the chapter opens with exactly this comparison.

Vedic altar construction

The Śulba-Sūtras prescribe altars of exact shape and exact area, which is why they contain systematic rules for turning a rectangle into a triangle, a triangle into a rectangle, and a rhombus into a rectangle.

Land records

Plots are triangulated from a diagonal and its offsets — one diagonal plus two perpendicular distances gives the area without measuring a single side.

Paths, borders and crossings

A path round a park or a crosspath through a plot is computed as a difference of rectangles, with the overlap subtracted once — the standard estimate for paving and turfing.

Shortest routes

Fetching water from a river and delivering it elsewhere is solved by reflecting the destination in the river — the same construction that makes the angle of incidence equal the angle of reflection for light.

Paper, furniture and land units

A4 at 623.7 cm², tabletops in inches and feet, rooms in ft² or m², land in acres, bigha, gaj, katha, dhur, cent or ankanam — each needing an area conversion, where the factor is squared.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Name the base and its perpendicular height before touching the numbers

For a tilted figure, find the segment carrying the right-angle mark — that is the height

Composite regions: decide whether to add pieces or subtract from a whole, and say which

Subtract every overlap exactly once in paths, crosspaths and bent tubes

Show the dissection or the congruence when a proof is asked for; a formula alone earns little

Leave answers such as 180/19 and 84/29 as exact fractions, then give the rounded value

Square the conversion factor for areas, never copy the length factor

Write units throughout and check the answer is plausible against the drawing

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Varignon's theorem: the midpoint quadrilateral is a parallelogram of half the area

STRETCH

Perigal's dissection and its link to the Baudhāyana–Pythagoras figure

STRETCH

Pick's theorem: the area of a lattice polygon from its boundary and interior points

STRETCH

The isoperimetric problem: among all shapes of a given perimeter, the circle encloses the most

STRETCH

Equidissection: which polygons can be cut into n triangles of equal area

STRETCH

Wallace–Bolyai–Gerwien: any two polygons of equal area are related by a dissection

STRETCH

Śulba-Sūtra circling of the square, and the approximations to √2 it produced

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 8 School ExamVery High

Class 8 Olympiad (IMO/NSTSE)High — dissection and equal-area reasoning

NTSE / NMMSVery High

Class 9 Areas of Parallelograms and TrianglesVery High — direct continuation

Class 9 Heron's FormulaHigh — the case this chapter deliberately leaves out

Class 10 Areas Related to CirclesMedium — builds on the polygon method

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because two regions can share a perimeter and differ in area, and a region with a longer boundary can enclose less. The rangoli rectangles 7 × 4 and 8 × 3 both have perimeter 22 cm but areas 28 cm² and 24 cm²; a 1 cm × 10 cm strip has perimeter 22 cm and area 10 cm² against a 5 cm square's perimeter 20 cm and area 25 cm². Perimeter measures the edge; area measures the inside.

Yes. When the apex leans out past the base you cannot draw a rectangle on that base containing the triangle, so instead write ∆ABC = ∆ADC − ∆ADB, where D is the foot of the altitude. That gives ½h·DC − ½h·DB = ½h(DC − DB) = ½h·BC. No case analysis is needed, provided the height is always taken as a perpendicular distance.

Any of them — but each base has its OWN height, the perpendicular distance to the side opposite it. Both pairings give the same area, so base₁ × height₁ = base₂ × height₂. That is exactly how QN is found on page 163: the area is 12 × 6 = 72 cm², so the height on the 7.6 cm side must be 72 ÷ 7.6 ≈ 9.47 cm.

They are ancient Indian texts on the construction of altars, which had to have an exactly prescribed shape and an exactly prescribed area. That forces problems of the form 'reshape this figure into that one without changing its area' — and the answers are dissections you can carry out with cord and pegs. Euclid's Elements poses the same problems. Every formula in this chapter comes from one.

Because the chapter's method is cutting figures into triangles, and a circle has no straight sides to cut along. Circles, along with Heron's formula and the surface area and volume of solids, come later — in Classes 9 and 10 — and are not part of Class 8 Ganita Prakash.

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