MathsICSE Class 9 · Area Theorems

Parallelograms and Triangles on the Same Base and Between the Same Parallels (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Parallelograms on the same base and between the same parallels are equal in area. A triangle on that base and between those parallels has half the area of the parallelogram. Two triangles on the same base and between the same parallels are equal in area. Every one of these follows from the fact that they share the same height.

All three shapes share base AB and height h, so the parallelograms are equal and the triangle is half of each.

Illustration drawn for QuestionBuddy.

Understand Parallelograms and Triangles on the Same Base and Between the Same Parallels

Two figures are on the same base and between the same parallels when they share a side lying on one line and their opposite vertices (or sides) lie on a second line parallel to it. The perpendicular distance between the two lines is the common height $h$ .

Three results.

1. Parallelograms on the same base, between the same parallels: equal areas, each $= bh$ .
2. A triangle and a parallelogram on the same base, between the same parallels: $ar(triangle) = \frac{1}{2} ar(parallelogram)$ .
3. Triangles on the same base, between the same parallels: equal areas, each $= \frac{1}{2} bh$ (used without proof in ICSE Class IX).

Why theorem 1 holds (outline for a proof question). Let $A B C D$ and $A B E F$ stand on $A B$ with $D, C, F, E$ on one line parallel to $A B$ . In $△ A D F$ and $△ B C E$ : $A D = B C$ , $A F = B E$ (opposite sides) and $∠ D A F = ∠ C B E$ (arms parallel in the same sense). So the triangles are congruent (SAS). Subtracting equal triangles from the trapezium $A B E D$ leaves equal parallelograms.

Why theorem 2 holds. Draw $B Q ∥ A P$ to complete a parallelogram $A B QP$ on the same base and between the same parallels. A diagonal bisects a parallelogram, so $ar (△ A B P) = \frac{1}{2} ar (A B QP) = \frac{1}{2} ar (A B C D)$ by theorem 1.

Typical consequences you must be able to prove and use.

If $P$ is any point on side $D C$ of parallelogram $A B C D$ , then $ar (△ A P B) = \frac{1}{2} ar (A B C D)$ , so the other two triangles $A P D$ and $B P C$ together also make half.
If $P$ is any point inside the parallelogram, then $ar (△ A P B) + ar (△ C P D) = \frac{1}{2} ar (A B C D) = ar (△ A P D) + ar (△ B P C)$ . (Draw a line through $P$ parallel to $A B$ and apply theorem 2 twice.)
In a trapezium $A B C D$ with $A B ∥ D C$ and diagonals meeting at $O$ : $ar (△ A B D) = ar (△ A B C)$ , so removing the common $△ A O B$ gives $ar (△ A O D) = ar (△ B O C)$ .

Proof language. Always name the base, name the two parallel lines and write "same base and between the same parallels" before you equate areas. Examiners give the method marks for that sentence.

Parallelograms and Triangles on the Same Base and Between the Same Parallels: formulas to remember

Parallelogram: $ar = base \times height$; Height is perpendicular to the base.
Triangle: $ar = \frac{1}{2} \times base \times height$; Same height as the parallelogram on the same base.
Triangle and parallelogram: $ar (△ A B P) = \frac{1}{2} ar (A B C D)$; Same base AB, P on line DC.
Point inside a parallelogram: $ar (△ A P B) + ar (△ C P D) = \frac{1}{2} ar (A B C D)$; Opposite triangles at P make half.
Trapezium diagonals: $ar (△ A O D) = ar (△ B O C)$; AB parallel to DC, diagonals meet at O.

How to solve Parallelograms and Triangles on the Same Base and Between the Same Parallels questions

Identify base and parallels first, then convert each area to a half or whole of one parallelogram.

1Name the base and the two parallel lines in the figure.
2State "same base and between the same parallels" for each pair you compare.
3Use theorem 1 for parallelogram = parallelogram, theorem 2 for triangle = half of parallelogram, theorem 3 for triangle = triangle.
4For a point inside a parallelogram, pair opposite triangles: each pair totals half of the parallelogram.
5Add or subtract areas of parts to reach the required region; check that the parts add up to the whole.
6Write the numerical value with units (cm² or m²).

Parallelograms and Triangles on the Same Base and Between the Same Parallels in real life

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.

Equal areas on a shifting boundary

Suppose a farmer owns a field shaped like a parallelogram $A B C D$ of area 360 m², with the boundary $D C$ lying along a straight canal parallel to $A B$ . A new corner post $P$ is placed anywhere on $D C$ .

Find the area of the triangular patch $A P B$ no matter where $P$ is placed on $D C$ .

Try it first, then show the solutionSolution

Triangle $A P B$ and parallelogram $A B C D$ share base $A B$ and lie between the parallels $A B$ and $D C$ .
So $ar (△ A P B) = \frac{1}{2} \times 360 = 180$ m².

Answer: 180 m², for every position of $P$ on $D C$ .

Memory hook: Slide the top vertex along the parallel line: the height does not change, so the area does not change.

Triangle on a roof truss

Imagine a steel beam of length 24 cm drawn in a design as the base of a triangular plate whose apex moves along a rail parallel to the beam, 7.5 cm away.

Find the area of the plate and say how it changes as the apex slides along the rail.

Try it first, then show the solutionSolution

$ar = \frac{1}{2} \times 24 \times 7.5 = 90$ cm².
The base and the distance between the parallels stay fixed, so the area stays 90 cm² at every position.

Answer: 90 cm²; unchanged as the apex slides.

Memory hook: Same base + same parallels = same area.

Tricks to remember Parallelograms and Triangles on the Same Base and Between the Same Parallels

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

Slide the vertex
If the top vertex moves along a line parallel to the base, the triangle changes shape but not area.
Use it when: Recognising theorem 3 situations in a figure.
Half the parallelogram
Triangle and parallelogram on one base between the same parallels: triangle = half.
Use it when: Theorem 2 in one line.
Opposite pairs at P
For P inside a parallelogram, APB with CPD and APD with BPC each make half.
Use it when: Quick answers in the point-inside questions.

Parallelograms and Triangles on the Same Base and Between the Same Parallels: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Triangle from a parallelogram · 1 mark · MCQBoard level

$A B C D$ is a parallelogram of area $96$ cm² and $P$ is any point on side $D C$ . The area of $△ A P B$ is:
(a) 24 cm²
(b) 48 cm²
(c) 96 cm²
(d) 192 cm²

Try it first, then show the solutionSolution

$△ A P B$ and $A B C D$ are on the same base $A B$ and between the parallels $A B$ and $D C$ .
$ar (△ A P B) = \frac{1}{2} \times 96 = 48$ cm².

Answer: (b) 48 cm²

Example 2: Two parallelograms on one base · 3 marks · Short answerBoard level

Parallelograms $P QR S$ and $P QT U$ lie on the same base $P Q = 16$ cm and between the same parallels. If $ar (P QR S) = 136$ cm², find $ar (P QT U)$ and the distance between the parallels.

Try it first, then show the solutionSolution

Same base and same parallels, so $ar (P QT U) = ar (P QR S) = 136$ cm².
Height $= \frac{136}{16} = 8.5$ cm.

Answer: $ar (P QT U) = 136$ cm²; distance between the parallels $= 8.5$ cm.

Example 3: A point inside the parallelogram · 4 marks · Short answerHarder board level

$P$ is a point inside parallelogram $A B C D$ of area $100$ cm². If $ar (△ A P B) = 18$ cm² and $ar (△ A P D) = 21$ cm², find $ar (△ C P D)$ , $ar (△ B P C)$ and check that the four triangles add up to the parallelogram.

Try it first, then show the solutionSolution

Opposite triangles at $P$ : $ar (△ A P B) + ar (△ C P D) = \frac{1}{2} \times 100 = 50$ .
$ar (△ C P D) = 50 - 18 = 32$ cm².
Also $ar (△ A P D) + ar (△ B P C) = 50$ , so $ar (△ B P C) = 50 - 21 = 29$ cm².
Check: $18 + 32 + 21 + 29 = 100$ cm².

Answer: $ar (△ C P D) = 32$ cm², $ar (△ B P C) = 29$ cm²; total $100$ cm².

Example 4: Trapezium with known areas · 4 marks · Short answerHarder board level

In trapezium $A B C D$ , $A B ∥ D C$ and the diagonals meet at $O$ . If $ar (△ A O B) = 36$ cm² and $ar (△ A O D) = 24$ cm², find $ar (△ B O C)$ , $ar (△ C O D)$ and the area of the trapezium.

Try it first, then show the solutionSolution

$△ A B D$ and $△ A B C$ are on base $A B$ between the parallels $A B$ , $D C$ , so they are equal in area. Removing common $△ A O B$ : $ar (△ B O C) = ar (△ A O D) = 24$ cm².
Triangles $A O B$ and $B O C$ share the vertex $B$ and have bases on the same line $A C$ , so $\frac{A O}{O C} = \frac{36}{24} = \frac{3}{2}$ .
Triangles $A O D$ and $C O D$ have the same ratio of bases: $\frac{24}{ar ( △ C O D )} = \frac{3}{2}$ , so $ar (△ C O D) = 16$ cm².
Trapezium $= 36 + 24 + 24 + 16 = 100$ cm².

Answer: $ar (△ B O C) = 24$ cm², $ar (△ C O D) = 16$ cm², trapezium $= 100$ cm².

Example 5: A proof with a number · 4 marks · Long answerHOTS / competency

$A B C D$ is a parallelogram and $E$ is any point on side $B C$ . Prove that $ar (△ A B E) + ar (△ C D E) = ar (△ A D E)$ . If $ar (A B C D) = 84$ cm², find $ar (△ A B E) + ar (△ C D E)$ .

Try it first, then show the solutionSolution

$△ A D E$ and parallelogram $A B C D$ are on the same base $A D$ and between the same parallels $A D$ and $B C$ , so $ar (△ A D E) = \frac{1}{2} ar (A B C D)$ .
The parallelogram is made up of $△ A B E$ , $△ C D E$ and $△ A D E$ , so $ar (△ A B E) + ar (△ C D E) = ar (A B C D) - ar (△ A D E)$ .
$= 2 ar (△ A D E) - ar (△ A D E) = ar (△ A D E)$ . Hence proved.
Numerically $ar (△ A D E) = \frac{84}{2} = 42$ cm², so the sum is $42$ cm².

Answer: Proved; $ar (△ A B E) + ar (△ C D E) = 42$ cm².

Where students lose marks in Parallelograms and Triangles on the Same Base and Between the Same Parallels

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

Mistake: Skipping the same-parallels statement
Area equalities are only valid for the same base and the same parallels; the reason is what earns the marks.
Wrong working
$ar (△ A P B) = ar (△ C P B)$ . (No reason.)
Right working
$△ A P B$ and $△ C P B$ would be equal only if $A$ and $C$ lie on a line parallel to $P B$ ; state the base and the parallels first.
Do this: Name the base and both parallel lines every time.
Usually costs: Method marks of a proof.
Mistake: Taking the slant side as the height
The height is the perpendicular distance, not the slanting side.
Wrong working
Parallelogram with base 10 and slant side 8: area $= 80$ .
Right working
Area $= 10 \times (perpendicular height)$ , for example $10 \times 6 = 60$ .
Do this: Draw the perpendicular and label it.
Usually costs: Final answer.
Mistake: Forgetting the half
A triangle on the parallelogram base has half the area, not equal area.
Wrong working
$ar (△ A P B) = ar (A B C D) = 96$ .
Right working
$ar (△ A P B) = \frac{1}{2} \times 96 = 48$ .
Do this: Triangle = half whenever a parallelogram is involved.
Usually costs: Final answer.

Exam tips

Class IX is examined in school papers; this page is set at the hardest school-exam and foundation-olympiad level for the chapter and labelled honestly.
Draw the figure large and mark the two parallel lines with arrows before writing anything.
A proof question usually asks for the statement of a theorem and its application: write "same base and between the same parallels" in words.
After splitting a figure into parts, add the parts back to check against the whole.

Practice questions on Parallelograms and Triangles on the Same Base and Between the Same Parallels

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

Q1 · 1 mark · MCQBoard level
A parallelogram has base $15$ cm and height $8$ cm. A triangle stands on the same base and between the same parallels. Its area is:
(a) 120 cm²
(b) 60 cm²
(c) 30 cm²
(d) 23 cm²
Hint
Triangle = half of parallelogram.
Answer
(b). Parallelogram $= 15 \times 8 = 120$ ; triangle $= 60$ cm².
Q2 · 3 marks · Short answerHarder board level
$A B C D$ is a parallelogram of area $180$ cm² and $E$ is the mid-point of $B C$ . Find $ar (△ A B E)$ and $ar (A E C D)$ .
Hint
Diagonal $A C$ halves the parallelogram; $A E$ is a median of $△ A B C$ .
Answer
$ar (△ A B C) = \frac{180}{2} = 90$ cm². $E$ is the mid-point of $B C$ , so $A E$ is a median and $ar (△ A B E) = 45$ cm². $ar (A E C D) = 180 - 45 = 135$ cm².
Q3 · 4 marks · Short answerHarder board level
In trapezium $A B C D$ , $A B ∥ D C$ and the diagonals meet at $O$ . If $ar (△ A O D) = 18$ cm² and $ar (△ C O D) = 12$ cm², find $ar (△ A O B)$ and the area of the trapezium.
Hint
$ar (△ B O C) = 18$ ; the ratio $A O : O C = 18 : 12$ .
Answer
$ar (△ B O C) = ar (△ A O D) = 18$ . $\frac{A O}{O C} = \frac{18}{12} = \frac{3}{2}$ , so $ar (△ A O B) = \frac{3}{2} \times 18 = 27$ cm². Trapezium $= 27 + 18 + 18 + 12 = 75$ cm².
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Triangles $A B C$ and $P B C$ have equal areas when $A$ and $P$ lie on a line parallel to $B C$ .
Reason (R): Both triangles have the same base $B C$ and the same height.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Height is the distance between the parallels.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A plot is in the shape of parallelogram $A B C D$ of area $240$ m². A water tank is placed at an interior point $P$ . Wheat is grown in $△ A P B$ and $△ C P D$ ; vegetables in $△ A P D$ and $△ B P C$ .
(a) Find the total area of the wheat region.
(b) If $ar (△ A P B) = 75$ m², find $ar (△ C P D)$ .
(c) If $ar (△ A P D) = 70$ m², find $ar (△ B P C)$ .
Hint
Each opposite pair of triangles makes half of the plot.
Answer
(a) $120$ m² (b) $120 - 75 = 45$ m² (c) $120 - 70 = 50$ m².

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.What do "same base and between the same parallels" guarantee?Show answer
The same height, so area depends only on the base and the shape type.
2.How does a triangle compare with a parallelogram on the same base and between the same parallels?Show answer
It has half the area.
3.In a trapezium ABCD with AB parallel to DC, which two triangles at O are equal in area?Show answer
Triangles AOD and BOC.
4.P is inside parallelogram ABCD. What is ar(APB) + ar(CPD)?Show answer
Half of ar(ABCD).

Parallelograms and Triangles on the Same Base and Between the Same Parallels: frequently asked questions

Do I have to prove the theorem about triangles on the same base?

The ICSE Class IX syllabus asks for it to be used without proof; the parallelogram theorems are the ones to prove.

What does "between the same parallels" mean?

The base lies on one line and the opposite side or vertex lies on another line parallel to it.

Why do the equal triangles at the trapezium diagonals help?

Removing the common triangle AOB from the equal triangles ABD and ABC leaves AOD and BOC.