Rectilinear Figures ICSE Class 9: Parallelogram Theorems and Constructions

What Rectilinear Figures 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. Rectilinear Figures is in the 4. Geometry (ii) Rectilinear Figures (a)-(b) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Angles of polygons (tools for this chapter)

For a polygon with $n$ sides:

• sum of interior angles $=(n-2)\times 180^{\circ }$;
• sum of exterior angles (one at each vertex) $=360^{\circ }$.

For a regular polygon (all sides and all angles equal): each exterior angle $=\frac{360^{\circ }}{n}$ and each interior angle $=180^{\circ }-\frac{360^{\circ }}{n}$.

So a regular hexagon has interior angles of $120^{\circ }$ and a quadrilateral's angles add to $360^{\circ }$.

Worked example

Each interior angle of a regular polygon is $156^{\circ }$. How many sides does it have?

Try it first, then show the solutionSolution

1. Exterior angle $=180^{\circ }-156^{\circ }=24^{\circ }$.
2. $n=\frac{360^{\circ }}{24^{\circ }}=15$.

Answer: 15 sides

Parallelogram: sides and angles

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

• Opposite sides are equal (use without proof).
• Opposite angles are equal (prove): in parallelogram $ABCD$, join $AC$. $\angle BAC=\angle DCA$ and $\angle BCA=\angle DAC$ (alternate angles), so adding, $\angle A=\angle C$. Similarly, with diagonal $BD$, $\angle B=\angle D$.
• Adjacent angles are supplementary: $\angle A+\angle B=180^{\circ }$ (co-interior angles between parallels).

Worked example

In parallelogram $ABCD$, $\angle A=(3x+10{)}^{\circ }$ and $\angle B=(2x+20{)}^{\circ }$. Find all the angles.

Try it first, then show the solutionSolution

1. Adjacent angles: $(3x+10)+(2x+20)=180$, so $5x=150$ and $x=30$.
2. $\angle A=100^{\circ }$, $\angle B=80^{\circ }$.
3. $\angle C=\angle A=100^{\circ }$, $\angle D=\angle B=80^{\circ }$.

Answer: 100°, 80°, 100°, 80°

Parallelogram: diagonals

The diagonals of a parallelogram bisect each other (prove): let diagonals $AC$ and $BD$ meet at $O$. In $△AOB$ and $△COD$: $AB=CD$ (opposite sides), $\angle OAB=\angle OCD$ and $\angle OBA=\angle ODC$ (alternate angles). So $△AOB\cong △COD$ (AAS) and $AO=OC$, $BO=OD$.

Each diagonal bisects the parallelogram (prove): $△ABC\cong △CDA$ (SSS: $AB=CD$, $BC=DA$, $AC$ common), so the diagonal divides it into two triangles of equal area.

Test for a parallelogram (use without proof): if one pair of opposite sides is equal and parallel, the quadrilateral is a parallelogram.

Worked example

The diagonals of parallelogram $PQRS$ meet at $O$. If $PO=(2x+1)$ cm, $OR=(x+4)$ cm and $QS=16$ cm, find $PR$ and $OQ$.

Try it first, then show the solutionSolution

1. $PO=OR$: $2x+1=x+4$, so $x=3$.
2. $PR=2\times 7=14$ cm.
3. $OQ=\frac{1}{2}QS=8$ cm.

Answer: PR = 14 cm; OQ = 8 cm

Rhombus, rectangle and square

All three are parallelograms, so they have every parallelogram property, plus:

• Rhombus (all sides equal): the diagonals meet at right angles and bisect the angles. Proof: in $△AOB$ and $△AOD$, $AB=AD$, $BO=DO$, $AO$ common, so they are congruent (SSS) and $\angle AOB=\angle AOD$; as they form a linear pair, each is $90^{\circ }$.
• Rectangle (all angles $90^{\circ }$): the diagonals are equal. Proof: $△ABC\cong △BAD$ (SAS: $AB$ common, $BC=AD$, $\angle B=\angle A=90^{\circ }$), so $AC=BD$.
• Square (rectangle and rhombus): diagonals are equal and meet at right angles, and each makes $45^{\circ }$ with the sides.

Worked example

The diagonals of rectangle $ABCD$ meet at $O$. If $\angle AOB=70^{\circ }$, find $\angle OAB$ and $\angle OBC$.

Try it first, then show the solutionSolution

1. Diagonals of a rectangle are equal and bisect each other, so $OA=OB$.
2. $\angle OAB=\angle OBA=\frac{180^{\circ }-70^{\circ }}{2}=55^{\circ }$.
3. $\angle OBC=90^{\circ }-55^{\circ }=35^{\circ }$.

Answer: ∠OAB = 55°; ∠OBC = 35°

Constructing quadrilaterals

With ruler and compasses only, a quadrilateral can be drawn when five independent measurements are given (for example four sides and a diagonal, or three sides and two included angles). For special quadrilaterals fewer are needed, because the properties supply the rest:

• Parallelogram: two adjacent sides and the included angle, or two adjacent sides and a diagonal, or the two diagonals and the angle between them.
• Rhombus: one side and one angle, or the two diagonals (draw one diagonal, construct its perpendicular bisector, mark half the other diagonal each side).

Angles of $60^{\circ }$, $90^{\circ }$, $120^{\circ }$, $45^{\circ }$, $30^{\circ }$ and $75^{\circ }$ can be constructed with compasses. Always write the steps of construction and leave arcs visible.

Worked example

Describe how to construct a rhombus with diagonals 8 cm and 6 cm, and find its side.

Try it first, then show the solutionSolution

1. Draw $AC=8$ cm and construct its perpendicular bisector, meeting $AC$ at $O$.
2. On the bisector mark $OB=OD=3$ cm on opposite sides of $AC$.
3. Join $AB$, $BC$, $CD$, $DA$. Side $=\sqrt{4^2+3^2}=5$ cm.

Answer: Rhombus ABCD; side 5 cm

Constructing a regular hexagon

In a regular hexagon each side equals the radius of its circumscribed circle, because the hexagon splits into six equilateral triangles at the centre ($\frac{360^{\circ }}{6}=60^{\circ }$ each).

Method 1 (circle): draw a circle with radius equal to the side; starting at any point on it, step round with the compasses set to the same radius, marking six points; join them in order.

Method 2 (angles): draw one side $AB$; at $A$ and $B$ construct $120^{\circ }$ angles and mark the next sides equal to $AB$; continue round.

Worked example

A regular hexagon is constructed with side 4 cm. Find the radius of its circumscribed circle and its perimeter.

Try it first, then show the solutionSolution

1. Six equilateral triangles of side 4 cm meet at the centre, so the radius is 4 cm.
2. Perimeter $=6\times 4=24$ cm.

Answer: Radius 4 cm; perimeter 24 cm

Maths in real life: Rectilinear Figures

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

• Angles of Polygons and Properties of a Parallelogram
• Rhombus, Rectangle and Square: Properties and Diagonals

Rectilinear Figures formula sheet

Interior angle sum

$(n-2)\times 180^{\circ }$

Quadrilateral: 360°.

Regular polygon exterior angle

$\frac{360^{\circ }}{n}$

Interior = 180° − exterior.

Parallelogram angles

$\angle A=\angle C,\angle A+\angle B=180^{\circ }$

Opposite equal; adjacent supplementary.

Diagonals bisect

$AO=OC,BO=OD$

Any parallelogram.

Rhombus side

$s^2={\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2$

Diagonals at 90°.

Square diagonal

$d=a\sqrt{2}$

Equal diagonals at 90°.

Rectilinear Figures: diagrams to practise

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

How the school exam asks Rectilinear Figures

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

Tricks to remember Rectilinear Figures

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

• Rhombus: right; Rectangle: equal

  Rhombus diagonals meet at Right angles; Rectangle diagonals are Equal. A square does both.

  Use it when: Diagonal properties.

• Neighbours add to 180

  Next-door angles of a parallelogram are supplementary; opposite ones are twins.

  Use it when: Angle problems.

• Hexagon = six triangles

  A regular hexagon is six equilateral triangles round a point, so side = radius.

  Use it when: Constructing the hexagon.

• 360 shared out

  Exterior angles always total 360°; divide by n for a regular polygon.

  Use it when: Finding the number of sides.

Rectilinear Figures: 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 Rectilinear Figures

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

• ✗Mistake: Assuming diagonals of every parallelogram are equal or perpendicular

  Only rectangles have equal diagonals and only rhombuses perpendicular ones.

  ✓Do this: Check which special figure you have before using a diagonal property.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Proving a result the syllabus says to use without proof

  Time is wasted proving 'opposite sides are equal'.

  ✓Do this: Quote it as a property; prove only the ones the syllabus lists for proof.

  Usually costs: Time.

• ✗Mistake: Using a protractor in constructions

  The syllabus requires ruler and compasses only; angles like 60° and 120° must be constructed.

  ✓Do this: Learn the compass constructions for 60°, 90°, 120°, 45°, 30°.

  Usually costs: Construction marks.

• ✗Mistake: No steps of construction

  A figure without written steps may not get full marks.

  ✓Do this: Number the steps briefly and leave all arcs visible.

  Usually costs: 1 mark.

• ✗Mistake: Wrong angle bisected by a diagonal

  In a rhombus each diagonal bisects the angles at its own ends, not the others.

  ✓Do this: Mark which vertices the diagonal joins before halving any angle.

  Usually costs: 1 mark.

Practice questions on Rectilinear Figures

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

   Each interior angle of a regular polygon is $140^{\circ }$. The number of sides is:
   (a) 7
   (b) 8
   (c) 9
   (d) 10

   Hint

   Exterior angle 40°.

   Answer

   (c)

2. Q2 · 2 marks · Short answerExam level

   Find the sum of the interior angles of an octagon, and each interior angle if it is regular.

   Hint

   (n − 2) × 180.

   Answer

   $1080^{\circ }$; $135^{\circ }$.

3. Q3 · 2 marks · Short answerExam level

   In parallelogram $ABCD$, $\angle A:\angle B=2:3$. Find all its angles.

   Hint

   Adjacent angles add to 180°.

   Answer

   $\angle A=72^{\circ }$, $\angle B=108^{\circ }$, $\angle C=72^{\circ }$, $\angle D=108^{\circ }$.

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

   Assertion (A): Every rhombus is a square.
   Reason (R): All sides of a rhombus are equal.
   (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

   Does a rhombus need right angles?

   Answer

   (b)

5. Q5 · 3 marks · Short answerHarder exam level

   The diagonals of square $ABCD$ meet at $O$. If $AC=10$ cm, find $OB$, the side of the square and $\angle OAB$.

   Hint

   Diagonals equal, bisect at 90°.

   Answer

   $OB=5$ cm; side $=\frac{10}{\sqrt{2}}=5\sqrt{2}$ cm; $\angle OAB=45^{\circ }$.

6. Q6 · 3 marks · Short answerHarder exam level

   In parallelogram $ABCD$, the bisector of $\angle A$ meets $DC$ at $E$. If $AD=6$ cm and $AB=10$ cm, find $DE$ and $EC$.

   Hint

   ∠DAE = ∠EAB = ∠DEA.

   Answer

   $\angle DEA=\angle EAB$ (alternate) $=\angle DAE$, so $DE=AD=6$ cm and $EC=10-6=4$ cm.

7. Q7 · 3 marks · Short answerHarder exam level

   $ABCD$ is a rectangle with diagonals meeting at $O$. If $\angle OBC=32^{\circ }$, find $\angle AOB$ and $\angle ODC$.

   Hint

   OB = OC.

   Answer

   $\angle OCB=32^{\circ }$, so $\angle BOC=116^{\circ }$ and $\angle AOB=64^{\circ }$. $\angle ODC=\angle OBA=90^{\circ }-32^{\circ }=58^{\circ }$.

8. Q8 · 4 marks · Long answerHarder exam level

   Prove that a quadrilateral whose diagonals bisect each other at right angles is a rhombus.

   Hint

   First a parallelogram, then equal adjacent sides.

   Answer

   Diagonals bisecting each other make it a parallelogram (opposite triangles congruent by SAS give equal, parallel opposite sides). In $△AOB$ and $△AOD$: $BO=DO$, $AO$ common, $\angle AOB=\angle AOD=90^{\circ }$, so $AB=AD$ (SAS). A parallelogram with two adjacent sides equal has all sides equal: a rhombus.

9. Q9 · 3 marks · Short answerHarder exam level

   Write the steps to construct a rhombus $ABCD$ with side 4 cm and $\angle A=60^{\circ }$, and state the length of diagonal $BD$.

   Hint

   Triangle ABD is equilateral.

   Answer

   Draw $AB=4$ cm; construct $60^{\circ }$ at $A$ and mark $AD=4$ cm; arcs of 4 cm from $B$ and $D$ meet at $C$; join. $△ABD$ has $AB=AD$ and $\angle A=60^{\circ }$, so it is equilateral: $BD=4$ cm.

10. Q10 · 3 marks · Short answerHarder exam level

    The angles of a pentagon are $x^{\circ }$, $(x+20{)}^{\circ }$, $(x+40{)}^{\circ }$, $(x+60{)}^{\circ }$ and $(x+80{)}^{\circ }$. Find the largest angle.

    Hint

    Sum is 540°.

    Answer

    $5x+200=540$, $x=68$. Largest $=148^{\circ }$.

11. Q11 · 4 marks · Long answerHOTS / competency

    In parallelogram $ABCD$, $P$ and $Q$ are points on diagonal $BD$ with $BP=DQ$. Prove that $APCQ$ is a parallelogram.

    Hint

    Use the intersection O of the diagonals.

    Answer

    Let $AC$, $BD$ meet at $O$; $OA=OC$, $OB=OD$. Then $OP=OB-BP=OD-DQ=OQ$. The diagonals $AC$ and $PQ$ of $APCQ$ bisect each other, so $APCQ$ is a parallelogram.

12. Q12 · 4 marks · Long answerHOTS / competency

    How many sides does a regular polygon have if each interior angle exceeds each exterior angle by $108^{\circ }$? Can a regular polygon have each exterior angle $50^{\circ }$?

    Hint

    Interior + exterior = 180°.

    Answer

    Exterior $=\frac{180-108}{2}=36^{\circ }$, so $n=10$. For $50^{\circ }$, $n=\frac{360}{50}=7.2$, not a whole number, so no.

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.Sum of interior angles of a hexagon?Show answer

   720°.

2. 2.Each exterior angle of a regular octagon?Show answer

   45°.

3. 3.Which parallelogram has perpendicular diagonals?Show answer

   A rhombus (and a square).

4. 4.Which parallelogram has equal diagonals?Show answer

   A rectangle (and a square).

5. 5.Adjacent angles of a parallelogram add to?Show answer

   180°.

6. 6.Which parallelogram results are used without proof?Show answer

   Opposite sides are equal; one pair of opposite sides equal and parallel gives a parallelogram.

7. 7.Why is the side of a regular hexagon equal to its circumradius?Show answer

   It is made of six equilateral triangles meeting at the centre.

8. 8.What instruments are allowed in constructions?Show answer

   Ruler and compasses only.

Quick revision checklist

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

• ☐Can you find the number of sides from an interior angle?
• ☐Can you prove opposite angles of a parallelogram are equal?
• ☐Can you prove the diagonals bisect each other?
• ☐Can you prove a rhombus has perpendicular diagonals and a rectangle equal ones?
• ☐Can you show a figure is a parallelogram using equal and parallel sides?
• ☐Can you construct 60°, 90° and 120° with compasses?
• ☐Can you construct a regular hexagon and explain why it works?

Rectilinear Figures: frequently asked questions