Area Theorems ICSE Class 9: Parallelograms and Triangles Between the Same Parallels

What Area Theorems is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Area Theorems is in the 4. Geometry (ii) Rectilinear Figures (c) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Same base, same parallels, same height

Area of a parallelogram $=\text{base}\times \text{height}$; area of a triangle $=\frac{1}{2}\times \text{base}\times \text{height}$. The height is the perpendicular distance between the base line and the parallel line through the opposite side or vertex.

Figures on the same base and between the same parallels all have the same height (the distance between the parallels), so their areas depend only on their shape type:

• every parallelogram there has area $bh$;
• every triangle there has area $\frac{1}{2}bh$.

Worked example

A parallelogram and a rectangle stand on the same base of 10 cm and between the same parallels. The rectangle's other side is 6 cm. Find the area of the parallelogram. If its slanting side is 8 cm, find the height on that side.

Try it first, then show the solutionSolution

1. Area of parallelogram $=$ area of rectangle $=10\times 6=60$ cm².
2. Taking the 8 cm side as base: $8\times h=60$, so $h=7.5$ cm.

Answer: 60 cm²; 7.5 cm

Theorem 1: parallelograms on the same base

Theorem: parallelograms that share a base and lie between the same two parallel lines have equal areas.

Proof: let parallelograms $ABCD$ and $ABEF$ stand on base $AB$, with $D$, $C$, $F$, $E$ on one line parallel to $AB$ ($D$ and $E$ the outermost points). In $△ADF$ and $△BCE$:

• $AD=BC$ (opposite sides of $ABCD$);
• $AF=BE$ (opposite sides of $ABEF$);
• $\angle DAF=\angle CBE$ (their arms are parallel: $AD\parallel BC$ and $AF\parallel BE$).

So $△ADF\cong △BCE$ (SAS) and they have equal areas. Now take the trapezium $ABED$:

• $\text{ar}(ABCD)=\text{ar}(ABED)-\text{ar}(△BCE)$;
• $\text{ar}(ABEF)=\text{ar}(ABED)-\text{ar}(△ADF)$.

The right-hand sides are equal, so the parallelograms are equal in area.

Worked example

Parallelograms $ABCD$ and $ABEF$ are on the same base $AB=9$ cm and between the same parallels. If $\text{ar}(ABCD)=54$ cm², find $\text{ar}(ABEF)$ and the distance between the parallels.

Try it first, then show the solutionSolution

1. By Theorem 1, $\text{ar}(ABEF)=54$ cm².
2. Height $=\frac{54}{9}=6$ cm.

Answer: 54 cm²; 6 cm

Theorem 2: triangle and parallelogram

Theorem: if a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half the area of the parallelogram.

Proof: let $△ABP$ and parallelogram $ABCD$ be on base $AB$, with $P$ on line $DC$. Through $B$ draw $BQ\parallel AP$ meeting line $DC$ at $Q$. Then $ABQP$ is a parallelogram, and its diagonal $BP$ divides it into two congruent triangles, so $\text{ar}(△ABP)=\frac{1}{2}\text{ar}(ABQP)$. By Theorem 1, $\text{ar}(ABQP)=\text{ar}(ABCD)$. Hence $\text{ar}(△ABP)=\frac{1}{2}\text{ar}(ABCD)$.

Worked example

$ABCD$ is a parallelogram of area 80 cm² and $P$ is any point on $DC$. Find $\text{ar}(△APB)$ and $\text{ar}(△APD)+\text{ar}(△BPC)$.

Try it first, then show the solutionSolution

1. $△APB$ and $ABCD$ share base $AB$ and lie between $AB$ and $DC$: $\text{ar}(△APB)=40$ cm².
2. The rest of the parallelogram: $80-40=40$ cm².

Answer: 40 cm² and 40 cm²

Theorem 3: triangles on the same base

Result (use without proof): triangles on the same base and between the same parallels are equal in area. (Each is half of the same parallelogram, by Theorem 2.)

In particular, if $AB\parallel DC$, then $\text{ar}(△ABC)=\text{ar}(△ABD)$ and $\text{ar}(△ADC)=\text{ar}(△BDC)$.

A median divides a triangle into two triangles of equal area, because they have equal bases ($BD=DC$) and the same height from $A$.

Worked example

In trapezium $ABCD$ with $AB\parallel DC$, the diagonals meet at $O$. Prove that $\text{ar}(△AOD)=\text{ar}(△BOC)$.

Try it first, then show the solutionSolution

1. $△ABD$ and $△ABC$ are on base $AB$ between $AB\parallel DC$: equal areas.
2. Subtract the common $△AOB$ from each.
3. $\text{ar}(△AOD)=\text{ar}(△BOC)$.

Answer: Proved.

Theorem 4: equal areas, equal altitudes

Theorem: triangles with equal areas on the same base (or equal bases) have equal corresponding altitudes.

Proof: if $\text{ar}(△ABC)=\text{ar}(△ABD)$ on base $AB$, with altitudes $h_1$ and $h_2$, then $\frac{1}{2}AB\cdot h_1=\frac{1}{2}AB\cdot h_2$, so $h_1=h_2$.

Use: if two triangles of equal area stand on the same base on the same side of it, their vertices are at the same height, so the line joining the vertices is parallel to the base: $CD\parallel AB$.

Worked example

$△ABC$ and $△ABD$ lie on the same side of $AB$ and have equal areas. Show that $CD\parallel AB$.

Try it first, then show the solutionSolution

1. Equal areas on the same base give equal altitudes: $C$ and $D$ are the same distance from $AB$.
2. Two points on the same side of a line at equal distances from it lie on a line parallel to it.
3. So $CD\parallel AB$.

Answer: CD ∥ AB

Splitting areas: medians and mid-points

Combine the theorems with mid-points:

• A median halves the area of a triangle.
• If $D$ is the mid-point of $BC$ and $E$ the mid-point of $AD$, then $BE$ is a median of $△ABD$, so $\text{ar}(△BED)=\frac{1}{2}\text{ar}(△ABD)=\frac{1}{4}\text{ar}(△ABC)$.
• The diagonal of a parallelogram halves it; both diagonals cut it into four triangles of equal area.

Work step by step, writing which triangle each median belongs to.

Worked example

In $△ABC$ of area 48 cm², $D$ is the mid-point of $BC$ and $E$ is the mid-point of $AD$. Find $\text{ar}(△BEC)$.

Try it first, then show the solutionSolution

1. $BE$ is a median of $△ABD$: $\text{ar}(△BED)=\frac{1}{2}\times 24=12$ cm².
2. Similarly $\text{ar}(△CED)=12$ cm².
3. $\text{ar}(△BEC)=24$ cm².

Answer: 24 cm²

Maths in real life: Area Theorems

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

• Medians, Equal Areas and Equal Altitudes of Triangles
• Parallelograms and Triangles on the Same Base and Between the Same Parallels

Area Theorems formula sheet

Parallelogram

$\text{ar}=b\times h$

h: distance between the parallels.

Triangle

$\text{ar}=\frac{1}{2}b\times h$

Any side can be the base.

Theorem 1

$\text{ar}(ABCD)=\text{ar}(ABEF)$

Same base AB, same parallels.

Theorem 2

$\text{ar}(△ABP)=\frac{1}{2}\text{ar}(ABCD)$

P on the opposite parallel.

Theorem 3

$\text{ar}(△ABC)=\text{ar}(△ABD)$

Same base, same parallels (no proof).

Median

$\text{ar}(△ABD)=\text{ar}(△ADC)$

D mid-point of BC.

Area Theorems: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Area Theorems

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

Tricks to remember Area Theorems

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

• Slide the top

  Slide the top side of a parallelogram along its parallel line: the area stays the same.

  Use it when: Theorem 1.

• Triangle is half

  Same base, same parallels: triangle = half the parallelogram.

  Use it when: Theorem 2.

• Median makes twins

  A median cuts a triangle into two equal-area twins.

  Use it when: Fraction-of-area problems.

• Equal area, equal height

  Same base and equal area: the vertices sit at the same height, so they lie on a parallel line.

  Use it when: Theorem 4 and its use.

Area Theorems: 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 B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 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 Theorems

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

• ✗Mistake: Not stating 'same base and between the same parallels'

  The theorems apply only when both conditions hold; quoting them without naming the base and the parallels loses method marks.

  ✓Do this: Write: 'on the same base AB and between the same parallels AB and DC'.

  Usually costs: 1 mark.

• ✗Mistake: Using a slanting side as the height

  Height is the perpendicular distance between the parallels, not the side length.

  Wrong working

  Area $=16\times 12$ for sides 16 and 12

  Right working

  Area $=16\times 9$ with perpendicular height 9

  ✓Do this: Draw the perpendicular before calculating.

  Usually costs: The answer.

• ✗Mistake: Proving Theorem 3 at length

  The syllabus states it without proof.

  ✓Do this: Quote it and use it.

  Usually costs: Time.

• ✗Mistake: Forgetting the common region

  In Theorem 1 and trapezium proofs, the key step is adding or subtracting the shared part; students jump to the conclusion.

  ✓Do this: Name the common region explicitly.

  Usually costs: 1 mark.

• ✗Mistake: Mixing up which triangle a median belongs to

  In repeated halving, students halve the wrong triangle.

  ✓Do this: Write 'BE is a median of △ABD' before halving.

  Usually costs: 1 to 2 marks.

Practice questions on Area Theorems

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

   A triangle and a parallelogram are on the same base and between the same parallels. If the parallelogram has area 50 cm², the triangle has area:
   (a) 100 cm²
   (b) 50 cm²
   (c) 25 cm²
   (d) 12.5 cm²

   Hint

   Theorem 2.

   Answer

   (c)

2. Q2 · 2 marks · Short answerExam level

   In $△PQR$, $S$ is the mid-point of $QR$. If $\text{ar}(△PQS)=18$ cm², find $\text{ar}(△PQR)$.

   Hint

   PS is a median.

   Answer

   36 cm².

3. Q3 · 2 marks · Short answerExam level

   A parallelogram has base 15 cm and area 120 cm². Its other side is 10 cm. Find both heights.

   Hint

   Area = base × corresponding height.

   Answer

   $\frac{120}{15}=8$ cm and $\frac{120}{10}=12$ cm.

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): In trapezium $ABCD$ with $AB\parallel DC$, $\text{ar}(△ADC)=\text{ar}(△BDC)$.
   Reason (R): Triangles on the same base and between the same parallels are equal in area.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Base DC, parallels DC and AB.

   Answer

   (c)

5. Q5 · 3 marks · Short answerHarder exam level

   $ABCD$ is a parallelogram and $O$ is any point inside it. Prove that $\text{ar}(△OAB)+\text{ar}(△OCD)=\frac{1}{2}\text{ar}(ABCD)$.

   Hint

   Draw a line through O parallel to AB.

   Answer

   Through $O$ draw $PQ\parallel AB$ with $P$ on $AD$, $Q$ on $BC$. $△OAB$ is half of parallelogram $ABQP$ and $△OCD$ is half of $PQCD$ (Theorem 2). Adding: half of $ABCD$.

6. Q6 · 3 marks · Short answerHarder exam level

   In $△ABC$, $E$ is any point on the median $AD$. Prove that $\text{ar}(△ABE)=\text{ar}(△ACE)$.

   Hint

   ED is a median of △EBC.

   Answer

   $\text{ar}(△ABD)=\text{ar}(△ACD)$ ($AD$ median) and $\text{ar}(△EBD)=\text{ar}(△ECD)$ ($ED$ median of $△EBC$). Subtract: $\text{ar}(△ABE)=\text{ar}(△ACE)$.

7. Q7 · 3 marks · Short answerHarder exam level

   $ABCD$ is a parallelogram. $X$ and $Y$ are the mid-points of $AB$ and $DC$. Prove that $\text{ar}(AXYD)=\frac{1}{2}\text{ar}(ABCD)$.

   Hint

   AXYD is a parallelogram with half the base.

   Answer

   $AX=DY$ and $AX\parallel DY$, so $AXYD$ is a parallelogram with base $AX=\frac{1}{2}AB$ and the same height. Its area is $\frac{1}{2}AB\times h=\frac{1}{2}\text{ar}(ABCD)$.

8. Q8 · 4 marks · Long answerHarder exam level

   $ABCDE$ is a pentagon. A line through $B$ parallel to $AC$ meets $DC$ produced at $F$. Prove that $\text{ar}(△ACB)=\text{ar}(△ACF)$ and $\text{ar}(AEDF)=\text{ar}(ABCDE)$.

   Hint

   Same base AC, parallels AC and BF.

   Answer

   $△ACB$ and $△ACF$ are on base $AC$ between $AC\parallel BF$: equal. Add $\text{ar}(ACDE)$ to both: $\text{ar}(ABCDE)=\text{ar}(AEDF)$.

9. Q9 · 4 marks · Long answerHarder exam level

   The area of parallelogram $ABCD$ is 96 cm². $P$ is a point on $BC$ with $BP:PC=1:3$. Find $\text{ar}(△APD)$ and $\text{ar}(△ABP)$.

   Hint

   △APD is half the parallelogram; △ABP has base BP = ¼BC.

   Answer

   $\text{ar}(△APD)=48$ cm² (base $AD$, parallels $AD$, $BC$). $△ABC=48$; $△ABP$ has $\frac{1}{4}$ of base $BC$ with the same height: $12$ cm².

10. Q10 · 3 marks · Short answerHarder exam level

    $D$ is a point on side $BC$ of $△ABC$ such that $BD:DC=2:3$. If $\text{ar}(△ABC)=40$ cm², find $\text{ar}(△ABD)$ and $\text{ar}(△ADC)$.

    Hint

    Same height, bases in ratio 2 : 3.

    Answer

    16 cm² and 24 cm².

11. Q11 · 4 marks · Long answerHOTS / competency

    $△ABC$ and $△DBC$ are on the same base $BC$, on opposite sides of it, with equal areas. Prove that $BC$ bisects $AD$.

    Hint

    Drop perpendiculars AL and DM to BC; let AD meet BC at O.

    Answer

    Equal areas on base $BC$ give $AL=DM$. In $△ALO$ and $△DMO$: $\angle ALO=\angle DMO=90^{\circ }$, $\angle AOL=\angle DOM$ (vertically opposite), $AL=DM$, so they are congruent (AAS) and $AO=DO$.

12. Q12 · 4 marks · Long answerHOTS / competency

    $P$, $Q$, $R$, $S$ are the mid-points of the sides of parallelogram $ABCD$. Prove that $\text{ar}(PQRS)=\frac{1}{2}\text{ar}(ABCD)$.

    Hint

    Join PR (parallel to AD) and use Theorem 2 twice.

    Answer

    With $P$, $R$ mid-points of $AB$, $CD$: $PR\parallel AD$ and $APRD$, $PBCR$ are parallelograms. $△PSR$ is half of $APRD$ (same base $PR$, between $PR$ and $AD$) and $△PQR$ is half of $PBCR$. Adding: $\text{ar}(PQRS)=\frac{1}{2}\text{ar}(ABCD)$.

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.Parallelograms on the same base and between the same parallels are …Show answer

   Equal in area.

2. 2.A triangle and a parallelogram on the same base and between the same parallels: relation?Show answer

   Triangle area = ½ parallelogram area.

3. 3.Which area result is used without proof?Show answer

   Triangles on the same base and between the same parallels are equal in area.

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

   Divides it into two equal areas.

5. 5.Two triangles of equal area on the same base: what about their altitudes?Show answer

   They are equal.

6. 6.In trapezium ABCD (AB ∥ DC) with diagonals meeting at O, which triangles are equal in area?Show answer

   △AOD and △BOC.

7. 7.Into how many equal-area triangles do the diagonals divide a parallelogram?Show answer

   Four.

8. 8.Area of a parallelogram with base 8 cm and height 5 cm?Show answer

   40 cm².

Quick revision checklist

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

• ☐Can you prove that parallelograms on the same base are equal in area?
• ☐Can you prove the triangle is half the parallelogram?
• ☐Do you state the base and the parallels each time?
• ☐Do you use perpendicular heights, not slant sides?
• ☐Can you prove ar(AOD) = ar(BOC) in a trapezium?
• ☐Can you find ¼ or ⅛ of a triangle using medians?
• ☐Can you deduce parallel lines from equal areas?

Area Theorems: frequently asked questions