An Area from Three Sides Without Measuring Any Height
Find a triangle's area from its base and height or from its three sides, handle equilateral, isosceles and right-angled cases, compute areas of quadrilaterals, and split a composite figure into parts.
How do you find a triangle's area when you cannot measure its height?
A triangular plot of land has sides of m, m and m. You can walk the boundary with a tape, so those three numbers are easy. The height is not: it runs from one corner to a point somewhere on the opposite side, and there may be a wall, a tree or a neighbour's land in the way.
The formula is useless here — you have no height. But the three sides are enough:
and once you know that, the height on any side falls out. On the m side,
So the height was never independent information. Three sides determine a triangle completely — SSS from the congruence chapter — so they must determine its area as well, and Heron's formula is what performs the calculation.
That is the theme of this chapter: every figure's area can be reached from whatever measurements you actually have, provided you know which formula fits which data. The perimeter goes round the boundary and is measured in centimetres or metres; the area covers the inside and is measured in square centimetres or square metres. Confusing the two units is the single most penalised error in this chapter.
This page covers the first part of the ICSE Class 9 Mathematics chapter on area and perimeter: the triangle by base-height and by Heron's formula, special triangles, the quadrilaterals, and composite figures.
The formula is useless here — you have no height. But the three sides are enough:
and once you know that, the height on any side falls out. On the m side,
So the height was never independent information. Three sides determine a triangle completely — SSS from the congruence chapter — so they must determine its area as well, and Heron's formula is what performs the calculation.
That is the theme of this chapter: every figure's area can be reached from whatever measurements you actually have, provided you know which formula fits which data. The perimeter goes round the boundary and is measured in centimetres or metres; the area covers the inside and is measured in square centimetres or square metres. Confusing the two units is the single most penalised error in this chapter.
This page covers the first part of the ICSE Class 9 Mathematics chapter on area and perimeter: the triangle by base-height and by Heron's formula, special triangles, the quadrilaterals, and composite figures.
Formula
What is Heron's formula and how do you use it?
Half the perimeter, then subtract each side from it, then multiply the four numbers and take the square root.
That is the semi-perimeter — half the way round, not the whole way. Using the full perimeter is the commonest mistake with this formula, and it gives an answer far too large.
Worked example 1. Find the area of a triangle with sides m, m and m.
The perimeter is m and the area is m — different quantities in different units, and a question may ask for either.
Multiply in a helpful order. Pairing the numbers as keeps the arithmetic small, and spotting square factors is even faster: and , so the product is , whose root is .
Worked example 2 — the heights follow. For that same triangle, find the height on each side.
The longest side carries the shortest height, since base times height is fixed at m — the same inverse relation you met in the area theorems chapter.
Worked example 3 — a check on a right-angled triangle. Find the area of a triangle with sides cm, cm and cm, both ways.
By Heron: , so
By base and height: , so the triangle is right-angled and the two legs are base and height:
The two routes agree, which is the best possible check that you have used Heron's formula correctly. Whenever a triangle turns out to be right-angled, use the legs — it is quicker and less error-prone than Heron.
That is the semi-perimeter — half the way round, not the whole way. Using the full perimeter is the commonest mistake with this formula, and it gives an answer far too large.
Worked example 1. Find the area of a triangle with sides m, m and m.
The perimeter is m and the area is m — different quantities in different units, and a question may ask for either.
Multiply in a helpful order. Pairing the numbers as keeps the arithmetic small, and spotting square factors is even faster: and , so the product is , whose root is .
Worked example 2 — the heights follow. For that same triangle, find the height on each side.
The longest side carries the shortest height, since base times height is fixed at m — the same inverse relation you met in the area theorems chapter.
Worked example 3 — a check on a right-angled triangle. Find the area of a triangle with sides cm, cm and cm, both ways.
By Heron: , so
By base and height: , so the triangle is right-angled and the two legs are base and height:
The two routes agree, which is the best possible check that you have used Heron's formula correctly. Whenever a triangle turns out to be right-angled, use the legs — it is quicker and less error-prone than Heron.
How do you find the area of an equilateral, isosceles or right-angled triangle?
Each special triangle has a shortcut, and each shortcut comes from dropping one perpendicular.
The equilateral triangle. Drop the perpendicular from the apex; it bisects the base, so it forms a right-angled triangle with hypotenuse and base :
Worked example 1. Find the height and area of an equilateral triangle of side cm.
Check with Heron: , so , as required.
Worked example 2 — backwards. The area of an equilateral triangle is cm. Find its side.
The isosceles triangle. The perpendicular from the apex bisects the base — proved in the isosceles chapter — so with equal sides and base ,
Worked example 3. Find the area of an isosceles triangle with equal sides cm and base cm.
Check with Heron: , and cm, as required.
The right-angled triangle needs no shortcut at all: the two legs are perpendicular, so one is the base and the other the height. With legs cm and cm the area is cm, the hypotenuse is cm and the perimeter is cm.
One boundary condition on the isosceles formula. The expression under the root needs , that is — the base must be shorter than the two equal sides put together. That is the triangle inequality again, appearing here as the point where the formula stops producing a real answer.
The equilateral triangle. Drop the perpendicular from the apex; it bisects the base, so it forms a right-angled triangle with hypotenuse and base :
Worked example 1. Find the height and area of an equilateral triangle of side cm.
Check with Heron: , so , as required.
Worked example 2 — backwards. The area of an equilateral triangle is cm. Find its side.
The isosceles triangle. The perpendicular from the apex bisects the base — proved in the isosceles chapter — so with equal sides and base ,
Worked example 3. Find the area of an isosceles triangle with equal sides cm and base cm.
Check with Heron: , and cm, as required.
The right-angled triangle needs no shortcut at all: the two legs are perpendicular, so one is the base and the other the height. With legs cm and cm the area is cm, the hypotenuse is cm and the perimeter is cm.
One boundary condition on the isosceles formula. The expression under the root needs , that is — the base must be shorter than the two equal sides put together. That is the triangle inequality again, appearing here as the point where the formula stops producing a real answer.
What are the area and perimeter formulas for the quadrilaterals?
Every one of them is either base times height or a sum of triangles.
- Rectangle: area , perimeter
- Square: area , perimeter , diagonal
- Parallelogram: area , perimeter
- Rhombus: area or base height, perimeter
- Trapezium: area , where and are the parallel sides
Worked example 1 — a trapezium. The parallel sides of a trapezium are cm and cm and the distance between them is cm. Find the area.
The trapezium formula is the mid-point theorem in disguise. From the quadrilaterals chapter, the segment joining the mid-points of the slant sides is — so the area is that mid-segment times the height, exactly as if the trapezium had been straightened into a rectangle.
Worked example 2 — a rhombus, three ways. A rhombus has diagonals of cm and cm. Find its side, perimeter, area and height.
The diagonals bisect at right angles, so the side is
Since a rhombus is also a parallelogram, area base height, so
Two formulas for one figure, and they agree — which is how you check a rhombus answer. Note that the height cm is less than the side cm, as a perpendicular distance must be.
Worked example 3 — equal perimeters, different areas. A rectangle measures cm by cm. Compare its area with that of a square of the same perimeter.
A square of perimeter cm has side cm, so its area is
The square encloses more with the same boundary, and the difference is cm. Perimeter does not determine area — a fact worth holding on to, because questions are built on the assumption that students think it does.
- Rectangle: area , perimeter
- Square: area , perimeter , diagonal
- Parallelogram: area , perimeter
- Rhombus: area or base height, perimeter
- Trapezium: area , where and are the parallel sides
Worked example 1 — a trapezium. The parallel sides of a trapezium are cm and cm and the distance between them is cm. Find the area.
The trapezium formula is the mid-point theorem in disguise. From the quadrilaterals chapter, the segment joining the mid-points of the slant sides is — so the area is that mid-segment times the height, exactly as if the trapezium had been straightened into a rectangle.
Worked example 2 — a rhombus, three ways. A rhombus has diagonals of cm and cm. Find its side, perimeter, area and height.
The diagonals bisect at right angles, so the side is
Since a rhombus is also a parallelogram, area base height, so
Two formulas for one figure, and they agree — which is how you check a rhombus answer. Note that the height cm is less than the side cm, as a perpendicular distance must be.
Worked example 3 — equal perimeters, different areas. A rectangle measures cm by cm. Compare its area with that of a square of the same perimeter.
A square of perimeter cm has side cm, so its area is
The square encloses more with the same boundary, and the difference is cm. Perimeter does not determine area — a fact worth holding on to, because questions are built on the assumption that students think it does.
How do you find the area of an irregular composite figure?
Cut it into triangles and quadrilaterals whose formulas you know, find each area, and add — or subtract, when a piece has been removed.
Worked example 1 — a rectangle with a triangular end. A plot consists of a rectangle m by m with a triangular strip on one short side, of base m and height m. Find the total area.
Worked example 2 — subtraction. An L-shaped room is formed by removing a m by m rectangle from one corner of an m by m rectangle. Find the floor area.
Check by splitting instead of subtracting: the L divides into m and m, and m, as required. Two independent routes to the same number is the best check available on a composite figure.
Worked example 3 — the surveyor's method. A four-sided field is measured along one diagonal m. From and , perpendicular offsets to that diagonal measure m and m. Find the area.
The diagonal splits the field into and , both standing on the base :
or in one step,
This is how land is actually measured. A field book records one long diagonal and a set of perpendicular offsets to it, because a tape can measure those without entering the standing crop. The area then needs only additions and one halving, and it works for any number of sides.
The two rules for splitting.
- Every piece must have a formula you know. If a cut leaves a shape with no formula, cut differently
- Every length you use must be given or derivable. A triangle's height has to be a stated offset or come from the Pythagoras theorem, not from the look of the figure
And watch the units when they are mixed. A question giving one length in metres and another in centimetres needs converting before any multiplication — and since area involves two lengths, an unconverted centimetre becomes an error of ten thousand in square metres.
Worked example 1 — a rectangle with a triangular end. A plot consists of a rectangle m by m with a triangular strip on one short side, of base m and height m. Find the total area.
Worked example 2 — subtraction. An L-shaped room is formed by removing a m by m rectangle from one corner of an m by m rectangle. Find the floor area.
Check by splitting instead of subtracting: the L divides into m and m, and m, as required. Two independent routes to the same number is the best check available on a composite figure.
Worked example 3 — the surveyor's method. A four-sided field is measured along one diagonal m. From and , perpendicular offsets to that diagonal measure m and m. Find the area.
The diagonal splits the field into and , both standing on the base :
or in one step,
This is how land is actually measured. A field book records one long diagonal and a set of perpendicular offsets to it, because a tape can measure those without entering the standing crop. The area then needs only additions and one halving, and it works for any number of sides.
The two rules for splitting.
- Every piece must have a formula you know. If a cut leaves a shape with no formula, cut differently
- Every length you use must be given or derivable. A triangle's height has to be a stated offset or come from the Pythagoras theorem, not from the look of the figure
And watch the units when they are mixed. A question giving one length in metres and another in centimetres needs converting before any multiplication — and since area involves two lengths, an unconverted centimetre becomes an error of ten thousand in square metres.
Exam tip
What layout keeps area marks safe in the exam?
Write the formula, substitute, then evaluate — and put the unit on every line. Area questions are marked for the formula and the unit as much as for the number.
- **State as its own step** before using Heron's formula, and write it as . Forgetting the halving is the classic error
- Test for a right angle first. If , use leg leg and skip Heron entirely
- Keep surds exact. cm is the answer; give the decimal only if asked, and take when you do
- Never mix perimeter and area units. Perimeter in cm, area in cm — an answer in the wrong unit usually loses the mark even when the number is right
- Convert units before multiplying, not after. Two lengths multiply, so an error in one length squares itself in the area
- For a composite figure, draw and label the split, number the pieces, and show a small table of areas as a list before adding
- Check a rhombus two ways — half the product of the diagonals, and base times height
- Check a Heron answer against base-height if any height is known
The misconception to name. Doubling every side does not double the area — it quadruples it, because area involves two lengths. A square of side cm has area cm and one of side cm has area cm. Scale factors square when they act on areas, and that idea returns as the area ratio of similar figures in Class 10.
- **State as its own step** before using Heron's formula, and write it as . Forgetting the halving is the classic error
- Test for a right angle first. If , use leg leg and skip Heron entirely
- Keep surds exact. cm is the answer; give the decimal only if asked, and take when you do
- Never mix perimeter and area units. Perimeter in cm, area in cm — an answer in the wrong unit usually loses the mark even when the number is right
- Convert units before multiplying, not after. Two lengths multiply, so an error in one length squares itself in the area
- For a composite figure, draw and label the split, number the pieces, and show a small table of areas as a list before adding
- Check a rhombus two ways — half the product of the diagonals, and base times height
- Check a Heron answer against base-height if any height is known
The misconception to name. Doubling every side does not double the area — it quadruples it, because area involves two lengths. A square of side cm has area cm and one of side cm has area cm. Scale factors square when they act on areas, and that idea returns as the area ratio of similar figures in Class 10.
Did you know
Which triangle encloses the most land for a given length of fencing?
You have m of fencing and want to enclose a triangular plot. Which triangle gives the most land?
Try three with that exact perimeter. All of them have , so Heron's formula is quick:
- **Right-angled, sides , , **: m
- **Isosceles, sides , , **: m
- **Equilateral, sides , , **: m
The equilateral triangle wins, and it is not a close-run thing — it beats the right-angled one by more than m from identical fencing. Among all triangles of a given perimeter, the most symmetric one encloses the most area, and Heron's formula shows why: the product is largest when the three factors are equal.
The same principle keeps going. With that m of fencing:
- a square gives side m and area m
- a circle gives radius m and area m
More sides means more area, and the circle is the limit. That is why a farmer fencing a field prefers a shape close to a square, why a water tank is round rather than square, and why a soap bubble is spherical — the surface pulls itself into the shape that holds the most for the least boundary.
And it answers the rectangle question from the previous section properly. The by rectangle lost to the square of the same perimeter for the same reason: of all rectangles with a fixed perimeter, the square has the greatest area — and you can see it by writing the sides as and , whose product is . That is largest when , which is the square.
Try three with that exact perimeter. All of them have , so Heron's formula is quick:
- **Right-angled, sides , , **: m
- **Isosceles, sides , , **: m
- **Equilateral, sides , , **: m
The equilateral triangle wins, and it is not a close-run thing — it beats the right-angled one by more than m from identical fencing. Among all triangles of a given perimeter, the most symmetric one encloses the most area, and Heron's formula shows why: the product is largest when the three factors are equal.
The same principle keeps going. With that m of fencing:
- a square gives side m and area m
- a circle gives radius m and area m
More sides means more area, and the circle is the limit. That is why a farmer fencing a field prefers a shape close to a square, why a water tank is round rather than square, and why a soap bubble is spherical — the surface pulls itself into the shape that holds the most for the least boundary.
And it answers the rectangle question from the previous section properly. The by rectangle lost to the square of the same perimeter for the same reason: of all rectangles with a fixed perimeter, the square has the greatest area — and you can see it by writing the sides as and , whose product is . That is largest when , which is the square.
Exam relevance
How are area formulas used in JEE and NEET questions?
This is foundation work, and its formulas are used as tools inside later questions rather than being the question themselves.
Where it leads. In Class 10 these plane areas become the bases and cross-sections of surface areas and volumes, and the area ratio of similar figures introduces the squaring of scale factors you met above. In Class 11 the triangle's area is computed two new ways: as a determinant from three coordinates, and as in trigonometry — which is the same with the height written as . Both are standard JEE Main items.
Where Heron's formula itself survives. It appears in JEE Properties of Triangles, where the area is written as and linked to the circumradius and inradius by and . That second relation uses the **same you compute here, so the semi-perimeter is a quantity you will be writing for years.
Worth trying now**: for the -- triangle with and , the inradius is and the circumradius is . Both come out of numbers you already have, which is why this triangle appears so often in textbooks.
Where it appears outside Mathematics. Physics uses areas constantly — the area under a graph, the cross-section of a wire for resistance, the area for pressure — and the unit discipline is exactly what is tested. In NEET Biology, surface-area-to-volume arguments about cells depend on the squaring-versus-cubing idea from the tip above.
Question types to expect. At this level: direct areas, perimeters and composite figures. In competitive papers: area from coordinates, , maximum-area problems with a fixed perimeter, and the , , , relations.
The single trap that costs marks. Using the perimeter instead of the semi-perimeter, and mixing units. In JEE the equivalent slip is forgetting that area scales as the square of a length, which appears in assertion-reason items about doubling dimensions.
Board versus competitive emphasis. ICSE marks the formula, the substitution and the unit; a competitive paper marks the number. The transferable habit is checking an area two ways — Heron against base-height, or diagonals against base-height — because a second route costs one line and catches almost every slip.
Where it leads. In Class 10 these plane areas become the bases and cross-sections of surface areas and volumes, and the area ratio of similar figures introduces the squaring of scale factors you met above. In Class 11 the triangle's area is computed two new ways: as a determinant from three coordinates, and as in trigonometry — which is the same with the height written as . Both are standard JEE Main items.
Where Heron's formula itself survives. It appears in JEE Properties of Triangles, where the area is written as and linked to the circumradius and inradius by and . That second relation uses the **same you compute here, so the semi-perimeter is a quantity you will be writing for years.
Worth trying now**: for the -- triangle with and , the inradius is and the circumradius is . Both come out of numbers you already have, which is why this triangle appears so often in textbooks.
Where it appears outside Mathematics. Physics uses areas constantly — the area under a graph, the cross-section of a wire for resistance, the area for pressure — and the unit discipline is exactly what is tested. In NEET Biology, surface-area-to-volume arguments about cells depend on the squaring-versus-cubing idea from the tip above.
Question types to expect. At this level: direct areas, perimeters and composite figures. In competitive papers: area from coordinates, , maximum-area problems with a fixed perimeter, and the , , , relations.
The single trap that costs marks. Using the perimeter instead of the semi-perimeter, and mixing units. In JEE the equivalent slip is forgetting that area scales as the square of a length, which appears in assertion-reason items about doubling dimensions.
Board versus competitive emphasis. ICSE marks the formula, the substitution and the unit; a competitive paper marks the number. The transferable habit is checking an area two ways — Heron against base-height, or diagonals against base-height — because a second route costs one line and catches almost every slip.
Key takeaways
What should you be able to calculate before moving on to circles?
Every figure here reduces to a base and a height, or to a sum of triangles.
- Triangle: , or Heron's formula with — the semi-perimeter
- Test for a right angle first; if the two legs are base and height
- Equilateral: area , height
- Isosceles: drop the perpendicular to bisect the base, then
- Rectangle , square with diagonal , parallelogram base height
- Rhombus: or base height — use both as a check
- Trapezium: , which is the mid-segment times the height
- Composite figures: split into shapes you have formulas for, or subtract a missing piece, and check by a second split
- The surveyor's rule: diagonal sum of the offsets
- Perimeter and area are different quantities in different units, and equal perimeters do not mean equal areas
The quickest self-test is the -- triangle. Find its area by Heron, then the height on each of the three sides, and check that base times height comes to m every time.
- Triangle: , or Heron's formula with — the semi-perimeter
- Test for a right angle first; if the two legs are base and height
- Equilateral: area , height
- Isosceles: drop the perpendicular to bisect the base, then
- Rectangle , square with diagonal , parallelogram base height
- Rhombus: or base height — use both as a check
- Trapezium: , which is the mid-segment times the height
- Composite figures: split into shapes you have formulas for, or subtract a missing piece, and check by a second split
- The surveyor's rule: diagonal sum of the offsets
- Perimeter and area are different quantities in different units, and equal perimeters do not mean equal areas
The quickest self-test is the -- triangle. Find its area by Heron, then the height on each of the three sides, and check that base times height comes to m every time.