Area Class 8: Rectangles, Triangles, Polygons, Parallelograms, Rhombuses and Trapeziums

What Area is about

Key concepts

Rectangles, and why perimeter is not area

NCERT section 7.1 Rectangle and Squares

A rectangle of length $l$ and width $w$ holds $l\times w$ unit squares, so **area $=l\times w$** (in cm², m², and so on). A diagonal splits it into two congruent triangles, each of area $\frac{1}{2}lw$.

Perimeter cannot measure area. Regions with the same perimeter can have different areas, and a region with a larger perimeter can even have a smaller area:

• a 9 cm by 4 cm rectangle has perimeter 26 cm and area 36 cm²;
• a 12 cm by 2 cm rectangle has perimeter 28 cm (larger) but area only 24 cm² (smaller).

A very thin, wiggly shape can have a huge perimeter and a tiny area. Among rectangles with a fixed perimeter, the square has the greatest area.

Dividing a square into equal areas: there are infinitely many ways; one part can be pushed in on one edge if it is pushed out by the same amount elsewhere.

Worked example

Compare a 9 cm by 4 cm rectangle with a 12 cm by 2 cm rectangle: which has the larger perimeter, and which has the larger area?

Try it first, then show the solutionSolution

1. Perimeters: $2(9+4)=26$ cm and $2(12+2)=28$ cm.
2. Areas: $9\times 4=36$ cm² and $12\times 2=24$ cm².
3. The 12 by 2 rectangle has the larger perimeter but the smaller area.

Answer: 12 × 2 has the larger perimeter (28 cm); 9 × 4 has the larger area (36 cm²).

Paths around and across rectangles

NCERT section 7.1 Rectangle and Squares

Path around a park: area of path $=$ area of outer rectangle $-$ area of inner rectangle. A 40 m by 25 m park with a 3 m path all round it lies inside a $(40+6)$ m by $(25+6)$ m rectangle: path area $=46\times 31-40\times 25=1426-1000=426$ m².

You can also split the path into rectangles: two strips $40\times 3$, two strips $25\times 3$ and four corner squares $3\times 3$: $240+150+36=426$ m².

Cross paths: two paths of width $w$ crossing a plot, parallel to its sides: area $=l\times w+b\times w-w\times w$. The middle square is counted in both strips, so subtract it once.

Moving the outer rectangle around the park (keeping the park inside) does not change the path area, because both rectangles keep their areas.

Worked example

A 14 m by 12 m plot has two paths, each 2 m wide, crossing in the middle parallel to its sides. Find the area of the paths and the cost of paving them at ₹50 per m².

Try it first, then show the solutionSolution

1. Strip along the length: $14\times 2=28$ m²; strip along the width: $12\times 2=24$ m².
2. The 2 m by 2 m crossing square is counted twice: $28+24-4=48$ m².
3. Cost $=48\times 50=₹2,400$.

Answer: 48 m²; ₹2,400.

Area of a triangle

NCERT section Triangles

Draw a line through the top vertex parallel to the base, and drop perpendiculars from the ends of the base: the triangle sits inside a rectangle with the same base and height, and the triangle is exactly half of it (each half of the rectangle is cut in two by a side of the triangle). So

**area of a triangle $=\frac{1}{2}\times$ base $\times$ height**,

where the height is the perpendicular distance from the opposite vertex to the base (line).

For an obtuse triangle where the altitude falls outside, write the triangle as the difference of two right triangles: $\frac{1}{2}h\times DC-\frac{1}{2}h\times DB=\frac{1}{2}h\times BC$. The formula still holds.

Any side can be the base, with its own height. Since the area is the same either way, you can find an unknown altitude: if base BC $=9$ with height 4, the area is 18; with base AC $=12$, the height from B is $\frac{2\times 18}{12}=3$.

Worked example

In $△ABC$, $BC=9$ cm and the altitude from A is 4 cm. If $AC=12$ cm, find the altitude from B to AC.

Try it first, then show the solutionSolution

1. Area $=\frac{1}{2}\times 9\times 4=18$ cm².
2. Using base AC: $\frac{1}{2}\times 12\times BY=18$.
3. $6\times BY=18$, so $BY=3$ cm.

Answer: 3 cm

Equal-area triangles: medians and parallel lines

NCERT section Triangles between Parallel Lines with a Common Base

Because area depends only on base and height:

• A median halves a triangle: the line from a vertex to the midpoint of the opposite side makes two triangles with equal bases and the same height, so equal areas. The two diagonals of a rectangle make four triangles of equal area, though they are not all congruent.
• Same base, between the same parallels: all triangles with base BC and third vertex on a line parallel to BC have the same height, so the same area. Their perimeters, however, differ.
• Least perimeter: among these triangles, the one with the smallest perimeter is found by reflecting C in the parallel line: the path B-A-C has the same length as B-A-C′, which is shortest when A lies on the straight line BC′. This is the isosceles triangle, with A on the perpendicular bisector of BC.

Repeated halving: if D is the midpoint of BC and E the midpoint of AD, then $△ABE$ is half of $△ABD$, which is half of $△ABC$, so it is a quarter.

Worked example

In $△ABC$, D is the midpoint of BC and E is the midpoint of AD. What fraction of the area of $△ABC$ is $△ABE$?

Try it first, then show the solutionSolution

1. AD is a median of $△ABC$, so area($△ABD$) $=\frac{1}{2}$ area($△ABC$).
2. BE is a median of $△ABD$ (E is the midpoint of AD), so area($△ABE$) $=\frac{1}{2}$ area($△ABD$).
3. So area($△ABE$) $=\frac{1}{4}$ area($△ABC$).

Answer: One quarter.

Area of any polygon

NCERT section Area of any Polygon

Every polygon can be divided into triangles, so its area is the sum of the triangles' areas.

Quadrilateral with a diagonal and offsets: join diagonal AC; drop perpendiculars BM and DN to AC (the 'offsets'). Then
$\text{area}=\frac{1}{2}\times AC\times BM+\frac{1}{2}\times AC\times DN=\frac{1}{2}\times AC\times (BM+DN)$.

Composite figures: split into rectangles and triangles, or subtract the unshaded parts from a surrounding rectangle.

Regular hexagon: joining the centre to the six vertices gives six congruent equilateral triangles; you need the side length (and the height of one triangle, from the Baudhayana-Pythagoras theorem).

Worked example

In quadrilateral ABCD, diagonal $AC=18$ m, and the perpendiculars from B and D to AC are 5 m and 7 m. Find the area.

Try it first, then show the solutionSolution

1. Area($△ABC$) $=\frac{1}{2}\times 18\times 5=45$ m².
2. Area($△ADC$) $=\frac{1}{2}\times 18\times 7=63$ m².
3. Total $=108$ m² $=\frac{1}{2}\times 18\times (5+7)$.

Answer: 108 m²

Area of a parallelogram

NCERT section Parallelogram

Dissection: in parallelogram ABCD, drop a perpendicular AX to DC (a height). Cut off $△AXD$ and move it to the other end. It fits exactly as $△BYC$ because $△BYC\cong △AXD$ (RHS: right angles, $BY=AX$, $BC=AD$). The result is a rectangle ABYX with sides AX and $XY=DC$.

**Area of a parallelogram $=$ base $\times$ corresponding height.**

• Any side can be the base, with the perpendicular height to that side: $AB\times h_1=AD\times h_2$.
• Parallelograms on the same base and between the same parallels have equal areas (but different perimeters).
• A rectangle and a (non-rectangular) parallelogram with the same side lengths do not have equal areas: the parallelogram's height is shorter than its slanting side, so the rectangle has more area.

Worked example

In parallelogram ABCD, $AB=12$ cm, $AD=10$ cm and the height to AB is 7.5 cm. Find the area and the height to AD.

Try it first, then show the solutionSolution

1. Area $=12\times 7.5=90$ cm².
2. Using base AD: $10\times h=90$.
3. $h=9$ cm.

Answer: 90 cm²; height to AD is 9 cm.

Area of a rhombus

NCERT section Rhombus

A rhombus is a parallelogram, so base $\times$ height works. The Sulba-Sutras give another dissection: a diagonal BD splits the rhombus into two isosceles triangles, each of which can be cut and rearranged into a rectangle; together they form a rectangle of sides $AC$ and $\frac{BD}{2}$. So

**area of a rhombus $=\frac{1}{2}\times d_1\times d_2$** (half the product of the diagonals).

Check with triangles: the diagonals are perpendicular, so area $=\frac{1}{2}\times AO\times BD+\frac{1}{2}\times CO\times BD=\frac{1}{2}\times AC\times BD$.

With the Baudhayana-Pythagoras theorem you can also find the side: diagonals 16 and 30 give side $\sqrt{8^2+15^2}=17$, and then the height $=\frac{\text{area}}{\text{side}}$.

Worked example

A rhombus has diagonals 16 cm and 30 cm. Find its area, side and height.

Try it first, then show the solutionSolution

1. Area $=\frac{1}{2}\times 16\times 30=240$ cm².
2. Side $=\sqrt{8^2+15^2}=17$ cm.
3. Height $=240÷17\approx 14.1$ cm.

Answer: 240 cm²; side 17 cm; height about 14.1 cm.

Area of a trapezium

NCERT section Trapezium

Let the parallel sides be $a$ and $b$ and the distance between them (the height) $h$.

Method 1 (rectangle and two triangles): drop perpendiculars from the shorter parallel side. With the overhangs $x$ and $y$, $b=a+x+y$ and
area $=\frac{1}{2}hx+ha+\frac{1}{2}hy=\frac{1}{2}h(x+y+2a)=\frac{1}{2}h(a+b)$.

Method 2 (two copies): put two copies together, one turned upside down; they form a parallelogram with base $a+b$ and height $h$. The trapezium is half of it.

**Area of a trapezium $=\frac{1}{2}\times h\times (a+b)$**: half the height times the sum of the parallel sides.

The formula also works for trapeziums whose perpendicular falls outside (use a parallelogram and a triangle, or subtract triangles).

Worked example

A trapezium has parallel sides 14 cm and 22 cm, 9 cm apart. Find its area in two ways.

Try it first, then show the solutionSolution

1. Formula: $\frac{1}{2}\times 9\times (14+22)=\frac{1}{2}\times 9\times 36=162$ cm².
2. Two copies make a parallelogram of base $14+22=36$ and height 9: area 324 cm².
3. Half of that is 162 cm².

Answer: 162 cm²

Maths in real life: Area

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Area of Rectangles, Paths, Triangles and Parallelograms
• Area of Rhombus, Trapezium, Polygons and Area Units

Area formula sheet

Rectangle

$A=l\times w$

Square

$A=s^2$

Triangle

$A=\frac{1}{2}\times b\times h$

Altitude from area

$h=\frac{2A}{b}$

Quadrilateral with offsets

$A=\frac{1}{2}\times d\times (h_1+h_2)$

Parallelogram

$A=b\times h$

Rhombus

$A=\frac{1}{2}{d}_1{d}_2$

Trapezium

$A=\frac{1}{2}h(a+b)$

Path round a rectangle

$A=(l+2w)(b+2w)-lb$

Cross paths

$A=lw+bw-w^2$

How the school exam asks Area

The question patterns to expect, and the method that works for each.

Tricks to remember Area

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Triangle is half a box

  Every triangle is half of the rectangle around it: $\frac{1}{2}\times b\times h$.

  Use it when: Triangle area.

• Slide the slice

  Parallelogram: slice off a triangle, slide it across, get a rectangle: base × height.

  Use it when: Parallelogram area.

• Half the diagonals

  Rhombus area is half of diagonal times diagonal.

  Use it when: Rhombus area.

• Average the parallels

  Trapezium: average of the parallel sides times the height.

  Use it when: Trapezium area.

• Height stands straight

  Height is always perpendicular to the base, never a slanting side.

  Use it when: Avoiding the most costly slip.

Area: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Area

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using the slanting side as the height

  The side is mistaken for the perpendicular distance.

  Wrong working

  Parallelogram with sides 12 cm and 10 cm: area $=12\times 10=120$ cm².

  Right working

  Use the perpendicular height to the base: $12\times 7.5=90$ cm².

  ✓Do this: Look for the right-angle mark; the height is perpendicular to the base.

  Usually costs: Whole answer.

• ✗Mistake: Forgetting the half in triangle and rhombus formulas

  Formulas are mixed with the rectangle.

  Wrong working

  Rhombus with diagonals 16 and 30: area $=480$ cm².

  Right working

  $\frac{1}{2}\times 16\times 30=240$ cm².

  ✓Do this: Triangle, rhombus and trapezium formulas all have a half.

  Usually costs: Answer mark.

• ✗Mistake: Counting the crossing twice in cross paths

  The two strips are simply added.

  Wrong working

  Paths 2 m wide in a 14 m by 12 m plot: $28+24=52$ m².

  Right working

  $28+24-4=48$ m².

  ✓Do this: Subtract the overlap once.

  Usually costs: 1 mark.

• ✗Mistake: Path width added only once to each dimension

  The path on both sides is forgotten.

  Wrong working

  Park 40 by 25 with a 3 m path all round: outer rectangle 43 by 28.

  Right working

  Outer rectangle $46\times 31$ (3 m on each side); path $=1426-1000=426$ m².

  ✓Do this: Add twice the width to each dimension.

  Usually costs: Whole answer.

• ✗Mistake: Judging area by perimeter

  Bigger boundary is assumed to mean bigger area.

  Wrong working

  A 12 by 2 plot has more fencing than a 9 by 4 plot, so it is bigger.

  Right working

  Areas are 24 and 36: the 9 by 4 plot is bigger despite less fencing.

  ✓Do this: Compute areas; perimeter measures the boundary only.

  Usually costs: 1-2 marks.

NCERT exercise guide: Area

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Area

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Which rectangle has the larger area?
   (a) 9 cm by 4 cm
   (b) 10 cm by 3 cm
   (c) they are equal
   (d) cannot be decided

   Hint

   Multiply.

   Answer

   (a) 36 cm² against 30 cm²

2. Q2 · 1 mark · MCQExam level

   The area of a parallelogram with base 9 cm and height 4 cm is
   (a) 13 cm²
   (b) 18 cm²
   (c) 26 cm²
   (d) 36 cm²

   Hint

   Base × height.

   Answer

   (d) 36 cm²

3. Q3 · 1 mark · Very short answerExam level

   A rectangle has area 84 cm² and length 12 cm. Find its width.

   Hint

   Divide.

   Answer

   7 cm

4. Q4 · 2 marks · Very short answerExam level

   Find the area of a trapezium with parallel sides 10 cm and 6 cm, 4 cm apart.

   Hint

   $\frac{1}{2}h(a+b)$.

   Answer

   32 cm²

5. Q5 · 3 marks · Short answerHarder exam level

   A 20 m by 15 m plot has two 3 m wide paths crossing in the middle, parallel to its sides. Find the area of the paths and the cost of paving at ₹40 per m².

   Hint

   Subtract the crossing.

   Answer

   $60+45-9=96$ m²; ₹3,840.

6. Q6 · 3 marks · Short answerHarder exam level

   A rhombus has area 150 cm² and one diagonal 20 cm. Find the other diagonal and the side.

   Hint

   Then use half-diagonals.

   Answer

   Other diagonal 15 cm; side $\sqrt{10^2+7.5^2}=\sqrt{156.25}=12.5$ cm.

7. Q7 · 2 marks · Very short answerHarder exam level

   In $△PQR$, $QR=16$ cm and the altitude from P is 5 cm. Find the altitude from Q to PR if $PR=20$ cm.

   Hint

   Area is 40 cm².

   Answer

   4 cm

8. Q8 · 3 marks · Short answerHarder exam level

   Explain how to turn a triangle into a parallelogram of the same area by one cut, and why the pieces fit.

   Hint

   Use the midpoints of two sides.

   Answer

   In $△ABC$ mark the midpoints M of AB and N of AC and cut along MN. Turn $△AMN$ half a turn about N: since $AN=NC$, A lands on C, and M lands at M′ on MN extended with $N{M}^{\prime }=MN$. Angles and lengths are kept by the turn, so the pieces fit edge to edge, giving parallelogram MBCM′ with base $M{M}^{\prime }=BC$ and height half the triangle's height: area $BC\times \frac{h}{2}=\frac{1}{2}\times BC\times h$.

9. Q9 · 2 marks · Very short answerHarder exam level

   A field ABCD has diagonal AC = 26 m with offsets 7 m and 9 m. Find its area.

   Hint

   $\frac{1}{2}\times 26\times 16$.

   Answer

   208 m²

10. Q10 · 5 marks · Long answerHarder exam level

    An isosceles trapezium has parallel sides 24 m and 36 m, 8 m apart. (a) Find its area. (b) Find the length of each slanting side. (c) Find its perimeter. (d) Fencing costs ₹150 per m; find the fencing cost.

    Hint

    Each overhang is 6 m.

    Answer

    (a) $\frac{1}{2}\times 8\times 60=240$ m². (b) Overhang $\frac{36-24}{2}=6$ m, so slant $=\sqrt{8^2+6^2}=10$ m. (c) $24+36+10+10=80$ m. (d) $80\times 150=₹12,000$.

11. Q11 · 3 marks · Short answerHOTS / competency

    Triangles are drawn on a fixed base BC = 10 cm with the third vertex anywhere on a line parallel to BC and 6 cm away. What is the area of each? Do they all have the same perimeter? Which has the least perimeter?

    Hint

    Height is fixed.

    Answer

    Each has area $\frac{1}{2}\times 10\times 6=30$ cm². Perimeters differ; the least is the isosceles triangle with the third vertex on the perpendicular bisector of BC (found by the reflection argument).

12. Q12 · 2 marks · Very short answerHOTS / competency

    In $△ABC$ of area 64 cm², D is the midpoint of BC and E the midpoint of AD. Find the area of $△AEC$.

    Hint

    Two medians.

    Answer

    $△ADC=32$ cm²; CE is a median of it, so $△AEC=16$ cm².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What does area measure?Show answer

   The number of unit squares (including fractions) needed to cover a region.

2. 2.Why can't perimeter measure area?Show answer

   Regions with the same perimeter can have different areas, and a larger perimeter can go with a smaller area.

3. 3.State the area of a triangle.Show answer

   $\frac{1}{2}\times$ base $\times$ height.

4. 4.What does a median do to the area of a triangle?Show answer

   It splits it into two triangles of equal area.

5. 5.How can you find the area of any polygon?Show answer

   Divide it into triangles and add their areas.

6. 6.State the area of a parallelogram.Show answer

   Base $\times$ corresponding height.

7. 7.What is dissection?Show answer

   Cutting a figure into pieces and rearranging them into another figure of equal area.

8. 8.State the area of a rhombus in terms of its diagonals.Show answer

   $\frac{1}{2}\times d_1\times d_2$.

9. 9.State the area of a trapezium.Show answer

   $\frac{1}{2}\times h\times (a+b)$.

10. 10.Which ancient texts set problems of turning one shape into another of equal area?Show answer

    The Sulba-Sutras (and Euclid's Elements).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you give two rectangles where the larger perimeter has the smaller area?
• ☐Can you find the area of a path around a park in two ways?
• ☐Can you find the area of cross paths without double counting?
• ☐Can you explain why a triangle is half of a rectangle?
• ☐Can you find an altitude from the area and another side?
• ☐Can you show that a median halves a triangle's area?
• ☐Can you find the area of a quadrilateral from a diagonal and offsets?
• ☐Can you convert a parallelogram into a rectangle by dissection?
• ☐Can you find a rhombus's area from its diagonals?
• ☐Can you derive the trapezium formula with two copies?

Area: frequently asked questions