Two Triangles Can Have the Same Area and No Common Shape
Prove that parallelograms on the same base between the same parallels are equal in area, that a triangle is half of one, and use equal areas to prove riders and to deduce equal altitudes.
How can two completely different triangles have exactly the same area?
Draw a line segment of length cm, and a second line parallel to it at a perpendicular distance of cm. Now pick any point on that parallel line and join it to and .
Every triangle you get has area
One of them may be tall and narrow, another wide and slanting, another so stretched that its apex is far off to the side. **They are not congruent, not similar, and share no angle — and all of them measure cm.
That is the whole idea of this chapter. Area depends on only two things: the base and the perpendicular distance to the opposite vertex or side. Slide the apex along a parallel line and the distance never changes, so neither does the area.
The phrase to learn precisely is on the same base and between the same parallels.** It means two figures share one side, and their opposite sides or vertices lie on one line parallel to it. Whenever a question contains that arrangement, an equal-area statement is available for free.
This page covers the ICSE Class 9 Mathematics chapter on area theorems: parallelograms on the same base, the triangle as half a parallelogram, equal-area riders, and what equal areas tell you about altitudes.
Every triangle you get has area
One of them may be tall and narrow, another wide and slanting, another so stretched that its apex is far off to the side. **They are not congruent, not similar, and share no angle — and all of them measure cm.
That is the whole idea of this chapter. Area depends on only two things: the base and the perpendicular distance to the opposite vertex or side. Slide the apex along a parallel line and the distance never changes, so neither does the area.
The phrase to learn precisely is on the same base and between the same parallels.** It means two figures share one side, and their opposite sides or vertices lie on one line parallel to it. Whenever a question contains that arrangement, an equal-area statement is available for free.
This page covers the ICSE Class 9 Mathematics chapter on area theorems: parallelograms on the same base, the triangle as half a parallelogram, equal-area riders, and what equal areas tell you about altitudes.
Why are two parallelograms on the same base between the same parallels equal in area?
Because the two triangles left over at the ends are congruent, so the same region is being measured twice.
The proof. Let and be parallelograms on the same base , with , , and all on one line parallel to .
In and :
- (corresponding angles, )
- (corresponding angles, )
- (opposite sides of parallelogram )
So by AAS, and their areas are equal.
Now look at how each parallelogram is built from pieces:
The two triangles are equal in area and the middle region is shared, so
Both parallelograms have the same base and the same height, which is the reason behind the appearance: the perpendicular distance between the two parallel lines is one number, and both figures sit across it.
Worked example. A parallelogram on base cm has height cm. A second parallelogram stands on the same base, between the same parallels, with a slant so extreme that one of its sides measures cm. Find its area.
for both of them. The cm side is irrelevant — the slant side of a parallelogram never enters its area, only the base and the perpendicular height.
The condition that is easy to lose. Between the same parallels means the opposite sides lie on one line. Two parallelograms on the same base whose far sides are on two different parallel lines have different heights and different areas. Name the two parallel lines explicitly in your proof, and this cannot go wrong.
The proof. Let and be parallelograms on the same base , with , , and all on one line parallel to .
In and :
- (corresponding angles, )
- (corresponding angles, )
- (opposite sides of parallelogram )
So by AAS, and their areas are equal.
Now look at how each parallelogram is built from pieces:
The two triangles are equal in area and the middle region is shared, so
Both parallelograms have the same base and the same height, which is the reason behind the appearance: the perpendicular distance between the two parallel lines is one number, and both figures sit across it.
Worked example. A parallelogram on base cm has height cm. A second parallelogram stands on the same base, between the same parallels, with a slant so extreme that one of its sides measures cm. Find its area.
for both of them. The cm side is irrelevant — the slant side of a parallelogram never enters its area, only the base and the perpendicular height.
The condition that is easy to lose. Between the same parallels means the opposite sides lie on one line. Two parallelograms on the same base whose far sides are on two different parallel lines have different heights and different areas. Name the two parallel lines explicitly in your proof, and this cannot go wrong.
Formula
Why is a triangle exactly half of the parallelogram on its base?
Because a diagonal cuts a parallelogram into two congruent triangles, and one of them stands on the same base between the same parallels.
The proof. Let and parallelogram stand on the same base , with , and on one parallel line.
Complete the parallelogram on the base with on the same parallel. By the theorem above, , since both are on base between the same parallels.
The diagonal of parallelogram divides it into and , which are congruent by SSS — so each is half the parallelogram. Hence
And that is where the familiar formula comes from. Since a parallelogram is base times height, the triangle is half of it — so is not an independent fact to memorise but a consequence of this theorem.
Worked example 1. A parallelogram and a triangle stand on the same base of cm between the same parallels, cm apart. Find both areas.
Worked example 2 — reading the theorem backwards. A triangle of area cm stands on a base of cm. Find its height, and the area of a parallelogram on the same base between the same parallels.
The parallelogram is twice the triangle, so cm.
The misconception this theorem invites. Equal in area does not mean congruent. The cm triangle above can be drawn in infinitely many shapes, and none of them is congruent to the others. Congruence is a statement about every side and angle; equal area is a statement about one number. Every rider in this chapter depends on keeping those two apart.
The proof. Let and parallelogram stand on the same base , with , and on one parallel line.
Complete the parallelogram on the base with on the same parallel. By the theorem above, , since both are on base between the same parallels.
The diagonal of parallelogram divides it into and , which are congruent by SSS — so each is half the parallelogram. Hence
And that is where the familiar formula comes from. Since a parallelogram is base times height, the triangle is half of it — so is not an independent fact to memorise but a consequence of this theorem.
Worked example 1. A parallelogram and a triangle stand on the same base of cm between the same parallels, cm apart. Find both areas.
Worked example 2 — reading the theorem backwards. A triangle of area cm stands on a base of cm. Find its height, and the area of a parallelogram on the same base between the same parallels.
The parallelogram is twice the triangle, so cm.
The misconception this theorem invites. Equal in area does not mean congruent. The cm triangle above can be drawn in infinitely many shapes, and none of them is congruent to the others. Congruence is a statement about every side and angle; equal area is a statement about one number. Every rider in this chapter depends on keeping those two apart.
How do you use equal areas to prove a rider?
Find two triangles on the same base between the same parallels, write down that their areas are equal, and then add or subtract a shared piece. That subtraction step is what turns the theorem into a proof.
Rider 1 — a median halves the area. In , is the median to . Prove that .
Since is the mid-point, , so the two triangles have equal bases. They also share the same apex , so the perpendicular from to the line is the same height for both. Therefore
Worked numbers: if cm, each half is cm. Drawing all three medians cuts the triangle into six triangles of equal area, cm each.
Rider 2 — a trapezium's diagonals. In trapezium with , the diagonals meet at . Prove that .
Triangles and stand on the same base and between the same parallels and , so
Now subtract the shared piece from both:
Worked numbers. Take cm, cm and height cm. The trapezium's area is cm, and
- cm
- cm, and cm, as required
The diagonals divide each other in the ratio , so splitting each of those triangles in that ratio gives
- cm and cm
- cm and cm
Check: the two middle values are both cm, exactly as the proof promised, and the four parts add to cm.
Rider 3 — a parallelogram's diagonals. Both diagonals of a parallelogram divide it into four triangles of equal area, since each diagonal halves it and each half is halved again by the other. A parallelogram of area cm therefore gives four triangles of cm.
The pattern across all three riders is the same two-step move: an equal-area statement from the theorem, then a shared region added or taken away. Look for the piece the two figures have in common — it is almost always the triangle at the centre.
Rider 1 — a median halves the area. In , is the median to . Prove that .
Since is the mid-point, , so the two triangles have equal bases. They also share the same apex , so the perpendicular from to the line is the same height for both. Therefore
Worked numbers: if cm, each half is cm. Drawing all three medians cuts the triangle into six triangles of equal area, cm each.
Rider 2 — a trapezium's diagonals. In trapezium with , the diagonals meet at . Prove that .
Triangles and stand on the same base and between the same parallels and , so
Now subtract the shared piece from both:
Worked numbers. Take cm, cm and height cm. The trapezium's area is cm, and
- cm
- cm, and cm, as required
The diagonals divide each other in the ratio , so splitting each of those triangles in that ratio gives
- cm and cm
- cm and cm
Check: the two middle values are both cm, exactly as the proof promised, and the four parts add to cm.
Rider 3 — a parallelogram's diagonals. Both diagonals of a parallelogram divide it into four triangles of equal area, since each diagonal halves it and each half is halved again by the other. A parallelogram of area cm therefore gives four triangles of cm.
The pattern across all three riders is the same two-step move: an equal-area statement from the theorem, then a shared region added or taken away. Look for the piece the two figures have in common — it is almost always the triangle at the centre.
What can you deduce if two triangles on the same base have equal areas?
Their altitudes to that base are equal — and so their apexes lie on a line parallel to the base. This is the converse of the theorem the chapter began with, and it is the version used in numerical problems.
If on the common base , then
Since both apexes are the same perpendicular distance from the line and on the same side of it, .
Worked example 1. Two triangles on the same base of cm have equal areas of cm. Find each altitude.
Both altitudes are cm, so both apexes lie on a line parallel to the base and cm from it.
Worked example 2 — different bases. has area cm. Find the altitude on each side if cm, cm and cm.
The area is one number, so each altitude is found from its own base:
Notice that the longest side carries the shortest altitude. Base times height is fixed at cm, so the two are inversely related — and that is a useful check on any answer of this kind.
Worked example 3 — a ratio of areas. In , lies on with . If cm, find the areas of and .
Both triangles share the apex , so they have the same height, and their areas are in the ratio of their bases:
Check: cm, as required.
The general statement is worth keeping. Triangles with the same height have areas in the ratio of their bases, and triangles on the same base have areas in the ratio of their heights. Almost every area rider in this chapter is one of those two sentences applied once.
If on the common base , then
Since both apexes are the same perpendicular distance from the line and on the same side of it, .
Worked example 1. Two triangles on the same base of cm have equal areas of cm. Find each altitude.
Both altitudes are cm, so both apexes lie on a line parallel to the base and cm from it.
Worked example 2 — different bases. has area cm. Find the altitude on each side if cm, cm and cm.
The area is one number, so each altitude is found from its own base:
Notice that the longest side carries the shortest altitude. Base times height is fixed at cm, so the two are inversely related — and that is a useful check on any answer of this kind.
Worked example 3 — a ratio of areas. In , lies on with . If cm, find the areas of and .
Both triangles share the apex , so they have the same height, and their areas are in the ratio of their bases:
Check: cm, as required.
The general statement is worth keeping. Triangles with the same height have areas in the ratio of their bases, and triangles on the same base have areas in the ratio of their heights. Almost every area rider in this chapter is one of those two sentences applied once.
Exam tip
How should an area-theorem proof be written out?
**Quote the phrase on the same base and between the same parallels, and name the base and both parallel lines. That sentence is the theorem, and an examiner is checking that you can identify the arrangement rather than recite the conclusion.
- Write the base and the two parallels explicitly**: * and are on the same base and between the same parallels and .* Naming both lines is worth the mark
- **Use the notation** rather than the word area inside equations. It keeps a subtraction step readable
- State the shared piece before you subtract it: *subtracting from both sides*. That line is the proof's engine
- For a median, say that the apex is common so the height is the same. The equal bases alone do not finish the argument
- Never conclude congruence from equal areas. If a question asks to prove two triangles equal in area, proving them congruent is more than needed and often impossible
- Check numerical answers by adding the parts back to the whole figure, as in the trapezium example
- Remember the inverse relation: on a fixed area, a longer base means a shorter altitude
The trap that costs the most marks. Applying the theorem when the two figures are not between the same parallels. Two triangles on the same base with apexes on different parallel lines have different areas, and a figure that merely looks that way is not evidence. If you cannot name the single line both apexes lie on, the theorem does not apply.
- Write the base and the two parallels explicitly**: * and are on the same base and between the same parallels and .* Naming both lines is worth the mark
- **Use the notation** rather than the word area inside equations. It keeps a subtraction step readable
- State the shared piece before you subtract it: *subtracting from both sides*. That line is the proof's engine
- For a median, say that the apex is common so the height is the same. The equal bases alone do not finish the argument
- Never conclude congruence from equal areas. If a question asks to prove two triangles equal in area, proving them congruent is more than needed and often impossible
- Check numerical answers by adding the parts back to the whole figure, as in the trapezium example
- Remember the inverse relation: on a fixed area, a longer base means a shorter altitude
The trap that costs the most marks. Applying the theorem when the two figures are not between the same parallels. Two triangles on the same base with apexes on different parallel lines have different areas, and a figure that merely looks that way is not evidence. If you cannot name the single line both apexes lie on, the theorem does not apply.
Did you know
How can a crooked field boundary be straightened without either owner losing land?
Two neighbouring fields meet along a boundary that bends at a corner. Both owners would prefer a straight fence, and neither is willing to give up any area. The area theorems settle it exactly, with a ruler and a pair of compasses.
Suppose the boundary runs from to and then turns to . Join to . Now draw a line through **parallel to **, and let it meet the line extended at . Replace the two-part boundary -- with the single straight line -.
Why nothing is lost: and stand on the same base and between the same parallels and , so
The piece of land one owner gives up is exactly the piece the other gives back. The boundary has become straight and both areas are unchanged — and the proof is one application of the theorem in this chapter.
The same construction does something else useful. It converts any polygon into a triangle of equal area, one corner at a time: each step replaces two sides by one and leaves the area alone. A five-sided field can be reduced to a triangle in two steps, and a triangle's area needs only a base and a height.
That is why this theorem appears in land measurement rather than only in textbooks. A surveyor who can turn an irregular plot into an equal-area triangle can measure it with two numbers instead of many — and the farmer with the crooked fence gets a straight one for free.
The habit worth taking from it is noticing that parallel is the instruction that preserves area. Whenever you need to move a vertex without changing an area, move it along a line parallel to the opposite side.
Suppose the boundary runs from to and then turns to . Join to . Now draw a line through **parallel to **, and let it meet the line extended at . Replace the two-part boundary -- with the single straight line -.
Why nothing is lost: and stand on the same base and between the same parallels and , so
The piece of land one owner gives up is exactly the piece the other gives back. The boundary has become straight and both areas are unchanged — and the proof is one application of the theorem in this chapter.
The same construction does something else useful. It converts any polygon into a triangle of equal area, one corner at a time: each step replaces two sides by one and leaves the area alone. A five-sided field can be reduced to a triangle in two steps, and a triangle's area needs only a base and a height.
That is why this theorem appears in land measurement rather than only in textbooks. A surveyor who can turn an irregular plot into an equal-area triangle can measure it with two numbers instead of many — and the farmer with the crooked fence gets a straight one for free.
The habit worth taking from it is noticing that parallel is the instruction that preserves area. Whenever you need to move a vertex without changing an area, move it along a line parallel to the opposite side.
Exam relevance
Where do the area theorems show up in JEE preparation?
This is foundation work that becomes a computation and then a locus in later classes.
Where it leads. In Class 10 Similar Triangles, the ratio of areas becomes the square of the ratio of corresponding sides, and the first step of that proof is triangles with the same height have areas in the ratio of their bases — the sentence you met above. In Class 11 Coordinate Geometry the area of a triangle from three vertices is a determinant, and JEE Main regularly asks for the locus of a point that makes a triangle of fixed area with two given points. The answer is a pair of lines parallel to the base, which is exactly the converse theorem in this chapter, written in coordinates.
That locus is worth seeing now. Fixing the base and the area fixes the height, so the third vertex can lie anywhere on either of the two parallel lines at that distance — a pair of parallel lines, one on each side. Candidates who give only one line lose the question, and they lose it because they never saw the geometric version.
Where the subtraction technique leads. Adding and subtracting shared regions is the whole method of the definite integral for the area between two curves in Class 12, and it appears earlier in Physics whenever an area under a velocity-time graph is split into a rectangle and a triangle.
Question types to expect. At this level: proofs and riders, and numerical work with altitudes and ratios. In competitive papers: area from coordinates, the fixed-area locus, ratios of areas created by a point on a side, and centroid questions using the six-equal-triangles result.
The single trap that costs marks. Treating equal area as congruence, or forgetting the second parallel line in the locus answer. Both come from the same oversight — that area fixes only one number, and one number cannot fix a shape or a side.
Board versus competitive emphasis. ICSE marks the named arrangement and the subtraction step; a competitive paper marks a determinant or a locus. **The sentence that carries across is same height means areas in the ratio of bases** — learn that one and most ratio questions in both papers become a single line.
Where it leads. In Class 10 Similar Triangles, the ratio of areas becomes the square of the ratio of corresponding sides, and the first step of that proof is triangles with the same height have areas in the ratio of their bases — the sentence you met above. In Class 11 Coordinate Geometry the area of a triangle from three vertices is a determinant, and JEE Main regularly asks for the locus of a point that makes a triangle of fixed area with two given points. The answer is a pair of lines parallel to the base, which is exactly the converse theorem in this chapter, written in coordinates.
That locus is worth seeing now. Fixing the base and the area fixes the height, so the third vertex can lie anywhere on either of the two parallel lines at that distance — a pair of parallel lines, one on each side. Candidates who give only one line lose the question, and they lose it because they never saw the geometric version.
Where the subtraction technique leads. Adding and subtracting shared regions is the whole method of the definite integral for the area between two curves in Class 12, and it appears earlier in Physics whenever an area under a velocity-time graph is split into a rectangle and a triangle.
Question types to expect. At this level: proofs and riders, and numerical work with altitudes and ratios. In competitive papers: area from coordinates, the fixed-area locus, ratios of areas created by a point on a side, and centroid questions using the six-equal-triangles result.
The single trap that costs marks. Treating equal area as congruence, or forgetting the second parallel line in the locus answer. Both come from the same oversight — that area fixes only one number, and one number cannot fix a shape or a side.
Board versus competitive emphasis. ICSE marks the named arrangement and the subtraction step; a competitive paper marks a determinant or a locus. **The sentence that carries across is same height means areas in the ratio of bases** — learn that one and most ratio questions in both papers become a single line.
Key takeaways
What should you be able to prove about areas before moving on?
The whole chapter rests on one observation: area needs a base and a perpendicular height, and nothing else.
- Parallelograms on the same base and between the same parallels are equal in area, proved by an AAS congruence of the two end triangles
- A triangle on the same base between the same parallels is half the parallelogram, which is where comes from
- Name the base and both parallel lines whenever you quote either theorem
- A median divides a triangle into two equal areas; all three medians give six equal parts
- In a trapezium, the two side triangles made by the diagonals are equal in area, found by subtracting the shared central triangle
- Both diagonals of a parallelogram give four triangles of equal area
- Equal areas on the same base mean equal altitudes, so the apexes lie on a line parallel to the base
- Same height means areas in the ratio of bases; same base means areas in the ratio of heights
- Equal area is never congruence — the slant side of a parallelogram never enters its area
The test of this chapter is the trapezium rider. Take cm, cm and a height of cm, work out all four triangles made by the diagonals, and see whether the two middle ones come out equal without you having to look up why.
- Parallelograms on the same base and between the same parallels are equal in area, proved by an AAS congruence of the two end triangles
- A triangle on the same base between the same parallels is half the parallelogram, which is where comes from
- Name the base and both parallel lines whenever you quote either theorem
- A median divides a triangle into two equal areas; all three medians give six equal parts
- In a trapezium, the two side triangles made by the diagonals are equal in area, found by subtracting the shared central triangle
- Both diagonals of a parallelogram give four triangles of equal area
- Equal areas on the same base mean equal altitudes, so the apexes lie on a line parallel to the base
- Same height means areas in the ratio of bases; same base means areas in the ratio of heights
- Equal area is never congruence — the slant side of a parallelogram never enters its area
The test of this chapter is the trapezium rider. Take cm, cm and a height of cm, work out all four triangles made by the diagonals, and see whether the two middle ones come out equal without you having to look up why.