MathematicsClass 8ICSE

The Height Is Never the Slanted Side, However It Looks

Learn the area formulas for a trapezium, parallelogram, triangle and rhombus, use the diagonals of a rhombus, and work backwards to find a missing height, side or diagonal.

Why is a parallelogram's area not just the product of its two sides?

Because the second side is slanted, and slanted length is not height.

Picture a rectangle of card measuring

10

cm by

6

cm, with area

60 cm^{2}

. Now push the top edge sideways so it leans over. Not one side has changed length — it is still

10

cm by

6

cm — but the shape is visibly flatter, so it must enclose less area.

What shrank is the perpendicular distance between the two long sides. Lean it far enough and that distance approaches zero while the sides stay exactly as they were.

So the area of a parallelogram is base

\times

perpendicular height, never base

\times

slant side. Every formula on this page is built on that same perpendicular distance, and nearly every mark lost in this chapter comes from using a slanted measurement instead. This page covers the first part of the ICSE Class 8 Mathematics chapter on the area of a trapezium and a polygon.

Formula

What is the formula for the area of a trapezium?

For a trapezium with parallel sides

a

and

b

and perpendicular height

h

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

Where it comes from. Take two identical trapeziums and turn one upside down against the other. Together they form a parallelogram with base

(a + b)

and height

h

, so its area is

(a + b) h

. One trapezium is half of that.

Another way to see it. Draw a diagonal and the trapezium splits into two triangles, one with base

a

and one with base

b

, both of height

h

\frac{1}{2} ah + \frac{1}{2} bh = \frac{1}{2} (a + b) h

The same formula, which is why you need not memorise both derivations.

Worked example 1. A trapezium has parallel sides

12

cm and

8

cm and height

5

cm.

area = \frac{1}{2} (12 + 8) \times 5 = \frac{1}{2} \times 20 \times 5 = 50 cm^{2}

Worked example 2 — a field. A trapezium-shaped plot has parallel sides

15

m and

25

m, with the perpendicular distance between them

12

area = \frac{1}{2} (15 + 25) \times 12 = \frac{1}{2} \times 40 \times 12 = 240 m^{2}

At a turfing rate of ₹

45

per square metre the cost would be

240 \times 45 = ₹10800

.

Worked example 3 — the sides to add are the parallel ones. A trapezium has parallel sides

9

cm and

13

cm, slanted sides

6

cm and

7

cm, and height

5

cm.

area = \frac{1}{2} (9 + 13) \times 5 = \frac{1}{2} \times 22 \times 5 = 55 cm^{2}

The slanted sides of

6

cm and

7

cm play no part in the calculation. A question that lists them is testing whether you know which two sides the formula wants — and adding

9 + 13 + 6 + 7

or using

6

as the height are the two traps it is set for.

Check the units and the size. An area is always in square units. And a useful sanity check: the trapezium's area must lie between the areas of two rectangles, one of width

a

and one of width

b

, both of height

h

. For example

1

, those are

8 \times 5 = 40

and

12 \times 5 = 60

, and the answer

50

sits between them — as it must, since

\frac{1}{2} (a + b)

is the average of the two parallel sides.

How do you find the area of a parallelogram, triangle and rhombus?

Each is a version of base times perpendicular height, and the rhombus has a second formula using its diagonals.

parallelogram = b \times h, triangle = \frac{1}{2} bh

rhombus = b \times h = \frac{1}{2} d_{1} d_{2}

Worked example 1 — parallelogram. Base

14

cm, height

9

cm:

area = 14 \times 9 = 126 cm^{2}

Worked example 2 — triangle. Base

18

cm, height

7

cm:

area = \frac{1}{2} \times 18 \times 7 = 63 cm^{2}

Worked example 3 — rhombus from its diagonals. Diagonals

24

cm and

10

cm:

area = \frac{1}{2} \times 24 \times 10 = 120 cm^{2}

The same rhombus can be described by its side and height. Since the diagonals bisect each other at right angles, the half-diagonals

12

cm and

5

cm give

side = 1 2^{2} + 5^{2} = 144 + 25 = 169 = 13 cm

and then the height follows from the area:

h = \frac{area}{base} = \frac{120}{13} \approx 9.23 cm

Both formulas describe the same rhombus, so they must agree — and using one to find what the other needs is a standard question.

Worked example 4 — a cleaner rhombus. Diagonals

16

cm and

12

cm:

area = \frac{1}{2} \times 16 \times 12 = 96 cm^{2}, side = 8^{2} + 6^{2} = 10 cm

h = \frac{96}{10} = 9.6 cm

Why the half-diagonal formula works. The two diagonals cut the rhombus into four identical right-angled triangles, each with legs

\frac{d _{1}}{2}

and

\frac{d _{2}}{2}

. Four of them give

4 \times \frac{1}{2} \times \frac{d _{1}}{2} \times \frac{d _{2}}{2} = \frac{1}{2} d_{1} d_{2}

A parallelogram has two base-height pairs, and both give the same area. Take a parallelogram with sides

12

cm and

16

cm, where the height on the

12

cm base is

8

cm:

area = 12 \times 8 = 96 cm^{2}

Using the other side as the base, the height must satisfy

16 \times h = 96

, so

h = 6

cm. The longer base carries the shorter height, which makes sense — the same area is being spread differently.

Do not pair a base with the wrong height. In that parallelogram,

16 \times 8 = 128 cm^{2}

is simply wrong: the

8

cm height belongs to the

12

cm base. Each base has its own perpendicular height, and matching them correctly is the substance of the question.

The half-diagonal formula is for a rhombus, not any quadrilateral. It works because the diagonals meet at right angles. A parallelogram with diagonals

16

cm and

12

cm does not have area

96 cm^{2}

unless those diagonals happen to be perpendicular — which is to say, unless it is a rhombus.

How do you find a missing height, side or diagonal from the area?

Substitute what you know into the formula and solve the equation. Every formula on this page rearranges.

h = \frac{2 A}{a + b}, a + b = \frac{2 A}{h}, b = \frac{A}{h}, d_{2} = \frac{2 A}{d _{1}}

Worked example 1 — a missing height. A trapezium of area

84 cm^{2}

has parallel sides

10

cm and

18

cm. Find its height.

84 = \frac{1}{2} (10 + 18) h = 14 h \Rightarrow h = \frac{84}{14} = 6 cm

Check by substituting back:

\frac{1}{2} (28) (6) = 84 cm^{2}

. Correct.

Worked example 2 — a missing parallel side. A trapezium of area

180 cm^{2}

has height

10

cm and one parallel side

14

cm. Find the other.

180 = \frac{1}{2} (14 + b) \times 10 = 5 (14 + b) \Rightarrow 14 + b = 36 \Rightarrow b = 22 cm

Check:

\frac{1}{2} (14 + 22) (10) = \frac{1}{2} (36) (10) = 180 cm^{2}

. Correct.

Worked example 3 — a missing base. A parallelogram has area

150 cm^{2}

and height

12

cm.

b = \frac{150}{12} = 12.5 cm

Worked example 4 — a missing diagonal. A rhombus has area

120 cm^{2}

and one diagonal

16

cm.

120 = \frac{1}{2} \times 16 \times d_{2} = 8 d_{2} \Rightarrow d_{2} = 15 cm

Worked example 5 — a missing triangle height. A triangle of area

63 cm^{2}

has base

14

cm.

63 = \frac{1}{2} \times 14 \times h = 7 h \Rightarrow h = 9 cm

Worked example 6 — two steps, then a consistency check. A rhombus has area

336 cm^{2}

and one diagonal

48

cm. Find the other diagonal, the side and the height.

The other diagonal, from area

= \frac{1}{2} d_{1} d_{2}

d_{2} = \frac{2 \times 336}{48} = \frac{672}{48} = 14 cm

The side, from the half-diagonals

24

cm and

7

cm:

side = 2 4^{2} + 7^{2} = 576 + 49 = 625 = 25 cm

The height, from area

= b \times h

h = \frac{336}{25} = 13.44 cm

The two formulas must agree, and here they do:

\frac{1}{2} \times 48 \times 14 = 336

and

25 \times 13.44 = 336

. If a question's figures make them disagree, the figures are inconsistent — and noticing that is worth more than forcing out an answer.

Keep the factor of one half in view. The commonest rearrangement error is dividing by

(a + b)

without first doubling the area. Writing the formula down in full before substituting, and then checking by substituting your answer back, costs two lines and removes the risk entirely.

Exam tip

Exam tip: use the perpendicular height, and check by substituting back

The height is always the perpendicular distance, never a slanted side. A parallelogram with sides

10

cm and

6

cm and height

5

cm has area

50 cm^{2}

, not

60 cm^{2}

.

For a trapezium, add only the two parallel sides:

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

. Slanted sides listed in the question are there to distract.

Each base has its own height. In a parallelogram of area

96 cm^{2}

, the

12

cm base carries a height of

8

cm and the

16

cm base a height of

6

cm. Never pair one base with the other's height.

For a rhombus, use

\frac{1}{2} d_{1} d_{2}

— and remember it works because the diagonals are perpendicular, so it does not apply to a general parallelogram.

The half-diagonals give the side by Pythagoras:

16

and

12

give

64 + 36 = 10

cm.

When working backwards, double the area first:

h = \frac{2 A}{a + b}

, so area

84 cm^{2}

with sides

10

and

18

gives

h = 6

cm.

Substitute your answer back into the original formula — two lines, complete certainty.

Write square units on every area, and keep all lengths in the same unit before calculating.

And sanity-check a trapezium: its area must lie between

a \times h

and

b \times h

, because

\frac{1}{2} (a + b)

is their average.

Did you know

Why sliding the top of a shape sideways changes nothing

Take a stack of identical playing cards squared into a neat block. Now push the stack sideways so it leans into a sloping parallelogram. Not one card has moved up or down, and no card has changed size — so the total amount of card is exactly what it was.

That is why a parallelogram and a rectangle on the same base with the same height have the same area. The parallelogram is the rectangle with its layers slid sideways, and sliding never adds or removes material.

The same reasoning explains the triangle. Two identical triangles fit together into a parallelogram on the same base and height, so a triangle is exactly half of one — and it means every triangle with base

18

cm and height

7

cm has area

63 cm^{2}

, no matter how lopsided it looks. A very slanted triangle and a neat upright one are equal in area provided their base and perpendicular height agree.

This also settles the misconception at the top of the page from the other direction. Leaning the stack further does not change the area because the height has not changed. What reduces the area is squashing the stack down — reducing the perpendicular height — and that is a genuinely different action from leaning it.

So the formulas are fewer than they appear. A rectangle, a parallelogram, a triangle and a trapezium are four views of one measurement: how much base, multiplied by how much perpendicular height.

Key takeaways

Areas of trapeziums, parallelograms and rhombuses: quick revision

- Height means perpendicular distance, never a slanted side. Leaning a shape over reduces its area without changing any side length.
- Trapezium:

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

, where

a

and

b

are the parallel sides. Two copies make a parallelogram of base

(a + b)

.
- Parallel sides

12

cm and

8

cm with

h = 5

cm give

50 cm^{2}

;

15

m and

25

m with

h = 12

m give

240 m^{2}

, costing ₹

10800

at ₹

45

per square metre.
- Slanted sides are not used: sides

9

cm and

13

cm with

h = 5

cm give

55 cm^{2}

whatever the slants are.
- The answer must lie between

a \times h

and

b \times h

, since

\frac{1}{2} (a + b)

is their average.
- Parallelogram:

b \times h

. Base

14

cm, height

9

cm gives

126 cm^{2}

.
- Triangle:

\frac{1}{2} bh

. Base

18

cm, height

7

cm gives

63 cm^{2}

.
- Rhombus:

\frac{1}{2} d_{1} d_{2}

, because the diagonals cut it into four right-angled triangles. Diagonals

24

cm and

10

cm give

120 cm^{2}

;

16

cm and

12

cm give

96 cm^{2}

.
- Half-diagonals give the side by Pythagoras:

12

and

5

give

13

cm;

8

and

6

give

10

cm. Then

h = \frac{A}{b}

, so

\frac{120}{13} \approx 9.23

cm and

\frac{96}{10} = 9.6

cm.
- Each base has its own height: area

96 cm^{2}

means

12 \times 8

16 \times 6

, and never

16 \times 8

.
- The

\frac{1}{2} d_{1} d_{2}

formula needs perpendicular diagonals, so it is for a rhombus, not any parallelogram.
- Working backwards:

h = \frac{2 A}{a + b}

a + b = \frac{2 A}{h}

b = \frac{A}{h}

d_{2} = \frac{2 A}{d _{1}}

.
- Area

84 cm^{2}

with sides

10

and

18

gives

h = 6

cm; area

180 cm^{2}

with

h = 10

and one side

14

gives the other as

22

cm.
- Area

150 cm^{2}

with

h = 12

gives base

12.5

cm; a rhombus of area

120 cm^{2}

with one diagonal

16

cm has the other

15

cm; a triangle of area

63 cm^{2}

on base

14

cm has height

9

cm.
- Double the area before dividing, and always substitute your answer back to check.

Work one trapezium question forwards and then hide the answer and work it backwards from the area — being able to travel both ways through a formula is what the harder questions in this chapter actually test.

Why is a parallelogram's area not just the product of its two sides?

What is the formula for the area of a trapezium?

How do you find the area of a parallelogram, triangle and rhombus?

How do you find a missing height, side or diagonal from the area?

Exam tip: use the perpendicular height, and check by substituting back

Why sliding the top of a shape sideways changes nothing

Areas of trapeziums, parallelograms and rhombuses: quick revision

Related study notes

Ready to put this into practice?