Quadrilaterals

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CLASS VIII Mathematics ~4–5 marks Ch 4 of 14
Quadrilaterals

Class 8 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • A quadrilateral is a four-sided closed figure (Latin quadri = four, latus = side). Its four angles always add up to $360^\circ$.
  • The chapter studies five special quadrilaterals — rectangle, square, parallelogram, rhombus, trapezium (and the kite) — each defined by its sides, angles and diagonals.
  • Key diagonal facts: rectangle — equal & bisect; square — equal, bisect at $90^\circ$, bisect the angles; parallelogram — bisect only; rhombus — bisect at $90^\circ$ & bisect angles.
  • The shapes are nested: every square is a rectangle and a rhombus; every rectangle and every rhombus is a parallelogram; every parallelogram is a trapezium.
  • Board weightage: ~4–5 marks — usually a "find the remaining angle" calculation, a property/true-false question, and a Venn-relationship MCQ.
Detailed notes

1. What is a quadrilateral?

A quadrilateral is a closed figure made of four straight sides. The four angles are the angles between consecutive sides. A figure that is open, has curved sides, or crosses itself is not a quadrilateral.

Familiar examples are the rectangle and the square — the chapter starts there and then relaxes the rules step by step to discover the parallelogram, rhombus, kite and trapezium. Throughout, properties are not just measured — they are deduced (proved) using congruent triangles, vertically-opposite angles and the transversal rules from earlier classes.

Big idea: every special quadrilateral is fixed by what we demand of its sides, its angles, or its diagonals. Knowing one set of facts lets us prove the rest.

2. Rectangle — definition and properties

Definition (sides & angles): A rectangle is a quadrilateral in which (i) all angles are right angles ($90^\circ$), and (ii) opposite sides are equal.

The chapter shows this can be tightened to a shorter definition:

Rectangle: a quadrilateral in which all the angles are $90^\circ$. (Equal opposite sides then follow automatically — see §3.)

The four properties of a rectangle:

  • P1: All angles are $90^\circ$.
  • P2: Opposite sides are equal.
  • P3: Opposite sides are parallel ($AB\parallel DC,\ AD\parallel BC$).
  • P4: The diagonals are equal in length and bisect each other (cross at their midpoints).

Why opposite sides are parallel (P3): $AB$ is a transversal cutting $AD$ and $BC$. Since $\angle A+\angle B=90^\circ+90^\circ=180^\circ$ (co-interior angles), the lines $AD\parallel BC$. Same logic gives $AB\parallel DC$.

3. The Carpenter's Problem — diagonals of a rectangle

A carpenter joins two thin strips so that a thread through their ends forms a rectangle. One strip is $8$ cm. The chapter answers three questions by deduction:

  • Deduction 1 — length of the other diagonal? In rectangle $ABCD$, $AB=CD$, $\angle BAD=\angle CDA=90^\circ$, and $AD$ is common, so $\triangle ADC\cong\triangle DAB$ (SAS). Hence $AC=BD$: the diagonals are equal, so the second strip is also $8$ cm.
  • Deduction 2 — where do they intersect? Using vertically-opposite angles and $\angle 3+\angle 2=90^\circ$, one shows $\triangle AOB\cong\triangle COD$ (AAS). So $OA=OC$ and $OB=OD$: the diagonals bisect each other at their midpoints.
  • Deduction 3 — angle between diagonals? It can be any value. The chapter checks that even when the angle at $O$ is $60^\circ$, the four corner angles still each work out to $90^\circ$.

So the strips must be equal and joined at their midpoints — a trick really used by carpenters in Europe and house-builders in Mozambique. This gives the alternative definition:

Rectangle (diagonal definition): a quadrilateral whose diagonals are equal and bisect each other.
NCERT Deduction 3 (general) — equal diagonals bisecting at angle $x$

Let the diagonals be equal and bisect each other, crossing at angle $x$. The four central angles are $x,\,x,\,180^\circ-x,\,180^\circ-x$. Each base angle of the isosceles triangles is $a=90^\circ-\tfrac{x}{2}$ or $b=\tfrac{x}{2}$, and each corner angle $=a+b=\left(90^\circ-\tfrac{x}{2}\right)+\tfrac{x}{2}=90^\circ$. So whatever the angle between the diagonals, the figure is a rectangle.

4. Square — the special rectangle

Definition: A square is a quadrilateral in which all angles are $90^\circ$ and all sides are equal.

So every square is a rectangle (with the extra condition of equal sides), but not every rectangle is a square. The five properties:

  • P1: All sides are equal.
  • P2: Opposite sides are parallel.
  • P3: All angles are $90^\circ$.
  • P4: Diagonals are equal, bisect each other, and meet at $90^\circ$.
  • P5: Each diagonal bisects the corner angles (splits each $90^\circ$ into two $45^\circ$).

Why the diagonals meet at $90^\circ$ (Deduction 5): for a square, $\triangle BOA\cong\triangle BOC$ (SSS, since the half-diagonals and a side are equal). So $\angle BOA=\angle BOC$, and as they form a straight line $\angle BOA+\angle BOC=180^\circ\Rightarrow$ each $=90^\circ$.

NCERT — diagonal bisects a corner of a square (P5)

In a square $ABCD$ the diagonal $AC$ makes $\angle 1$ and $\angle 3$ with the sides at $A$ and $C$. In $\triangle ADC$: $\angle 1+\angle 3+90^\circ=180^\circ$, and since $AD=DC$ the triangle is isosceles so $\angle 1=\angle 3$. Hence $\angle 1=\angle 3=45^\circ$ — each diagonal cuts the right angle in half.

5. Angle sum of a quadrilateral $=360^\circ$

Is a quadrilateral with three $90^\circ$ angles and a fourth $\neq90^\circ$ possible? No — and here is why.

Take quadrilateral $SOME$ and draw the diagonal $SM$. It splits into two triangles. Each triangle's angles add to $180^\circ$:

$$(\angle 1+\angle 2+\angle 3)+(\angle 4+\angle 5+\angle 6)=180^\circ+180^\circ=360^\circ$$

Regrouping, the four angles of the quadrilateral sum to:

$$\boxed{\text{Sum of all angles of a quadrilateral}=360^\circ}$$

So three right angles already use $270^\circ$, forcing the fourth to be exactly $90^\circ$ — three-right-angles-and-one-different is impossible. This single fact drives most exam "find the missing angle" sums.

6. Parallelogram — relaxing to parallel sides

Drop the right-angle demand but keep both pairs of opposite sides parallel:

Parallelogram: a quadrilateral in which both pairs of opposite sides are parallel.

Every rectangle is a parallelogram (a special one with $90^\circ$ angles). The three (plus diagonal) properties, all proved with transversals and congruence:

  • P1: Opposite sides are equal.
  • P2: Opposite sides are parallel.
  • P3: Adjacent (co-interior) angles add to $180^\circ$; opposite angles are equal.
  • P4: The diagonals bisect each other (but are not generally equal, and need not meet at $90^\circ$).

Deduction 6 (angles): since $AB\parallel CD$ with transversal $AD$, $\angle A+\angle D=180^\circ$. So opposite angles are equal and any two adjacent angles are supplementary.

Deduction 7 (sides): drawing diagonal $BD$, alternate angles give $\triangle ABD\cong\triangle CDB$ (ASA/AAS), so $AD=CB$ and $AB=CD$ — opposite sides equal.

Deduction 8 (diagonals): in parallelogram $EASY$, alternate angles make $\triangle AOE\cong\triangle SOY$ (ASA), so $OA=OY$ and $OE=OS$ — diagonals bisect each other.

NCERT Example — parallelogram with $\angle A=30^\circ$

Adjacent sides $4$ cm and $5$ cm, $\angle A=30^\circ$. Then $\angle D=180^\circ-30^\circ=150^\circ$ (co-interior). Opposite angles equal: $\angle C=\angle A=30^\circ$ and $\angle B=\angle D=150^\circ$. Opposite sides equal: $DC=AB=4$ cm and $BC=AD=5$ cm.

7. Rhombus — all four sides equal

Now demand equal sides but not right angles:

Rhombus: a quadrilateral in which all four sides are equal.

Because its opposite sides come out parallel, a rhombus is also a parallelogram — so it inherits all parallelogram properties, plus two of its own. Full list:

  • P1: All sides equal.   P2: Opposite sides parallel.
  • P3: Adjacent angles add to $180^\circ$; opposite angles equal.
  • P4: Diagonals bisect each other at $90^\circ$ (right angles).
  • P5: Each diagonal bisects the angles of the rhombus.

Deduction 9 (angles): in rhombus $GAME$, the isosceles triangles formed by a diagonal force the four angles a diagonal makes to be equal ($a=b=c=d$).

Deduction 10 (diagonals at $90^\circ$): $\triangle GEO\cong\triangle MEO$, so $\angle GOE=\angle MOE$; since they form a straight line they are $90^\circ$ each.

NCERT — rhombus with one angle $50^\circ$

Construct a rhombus with $\angle A=50^\circ$. The half-angle at $A$ is $25^\circ$; in $\triangle ADB$, $a+a+50^\circ=180^\circ\Rightarrow a=65^\circ$. So the angles of the rhombus are $50^\circ,130^\circ,50^\circ,130^\circ$: i.e. $\angle A=\angle C=50^\circ$ and $\angle B=\angle D=180^\circ-50^\circ=130^\circ$.

8. Kite — two pairs of equal adjacent sides

Kite: a quadrilateral $ABCD$ with two non-overlapping pairs of equal adjacent sides, e.g. $AB=BC$ and $CD=DA$ (or labelled $AB=AD,\ CB=CD$). Made by joining two triangles of sides $6,9,12$ cm back-to-back along the $12$ cm edge.

Property 1: the diagonal joining the vertices between the unequal sides (here $BD$):

  • (i) bisects the angles at those two vertices, and
  • (ii) bisects the other diagonal at $90^\circ$ — i.e. $AO=OC$ and $BD\perp AC$ (since $\triangle AOB\cong\triangle COB$).

So in a kite the diagonals are perpendicular, and the "axis" diagonal cuts the other in half (but they do not bisect each other in general). A rhombus is a special kite in which all four sides are equal.

9. Trapezium — only one pair of parallel sides

Relax even further — ask for just one pair of parallel sides:

Trapezium: a quadrilateral with at least one pair of parallel opposite sides.

Since a parallelogram has two such pairs, every parallelogram is also a trapezium. In trapezium $PQRS$ with $PQ\parallel SR$:

  • P1: Co-interior angles on each non-parallel side are supplementary: $\angle S+\angle P=180^\circ$ and $\angle R+\angle Q=180^\circ$.

Isosceles trapezium: when the two non-parallel sides are equal. Then the base angles on each parallel side are equal — P2: the angles opposite the equal sides are equal ($\angle U=\angle V$). (Proved by dropping perpendiculars and showing $\triangle UXY\cong\triangle VWZ$.)

NCERT "Figure it Out" — trapezium angles

A trapezium has parallel sides, and two of its base angles are $135^\circ$ and $105^\circ$. The angle co-interior to $135^\circ$ is $180^\circ-135^\circ=45^\circ$; the one co-interior to $105^\circ$ is $180^\circ-105^\circ=75^\circ$. (Check: $135+105+45+75=360^\circ$.)

10. How the shapes are nested (Venn picture)

The whole chapter is summed up by which shape sits inside which. From the outside in:

$$\text{Trapezium}\supset\text{Parallelogram}\supset\{\text{Rectangle},\ \text{Rhombus}\}\supset\text{Square}$$
  • Every square is both a rectangle and a rhombus (it is exactly their overlap).
  • Every rectangle and every rhombus is a parallelogram.
  • Every parallelogram is a trapezium (two parallel pairs $\Rightarrow$ at least one).
  • A kite overlaps the rhombus only at the square/rhombus region: a kite with all sides equal is a rhombus.

This is why a square truthfully answers "I am a rectangle and a rhombus" — just as a person can be both Indian and Malayali.

11. NCERT "Figure it Out" — solved highlights

Rectangle angles (page 94, Q1).

  • (i) $\angle ABD=30^\circ$ given. Diagonals equal & bisect, so $\triangle AOB$ is isosceles $\Rightarrow\angle BAC=30^\circ$ region: $\angle CAD=60^\circ,\ \angle ADB=60^\circ,\ \angle ACB=60^\circ,\ \angle BDC=30^\circ,\ \angle ACD=30^\circ$.
  • (ii) angle at centre $\angle POS=110^\circ\Rightarrow\angle QOP=70^\circ$ (linear pair). Isosceles triangles give $\angle OQR=\angle ORQ=35^\circ$ and $\angle OPQ=\angle OQP=55^\circ$, etc.

Q2. Draw a quadrilateral whose diagonals are equal ($8$ cm) and bisect each other at $30^\circ,40^\circ,90^\circ,140^\circ$ — every one gives a rectangle (at $90^\circ$ it is a square).

Q3. Two perpendicular diameters $PL,AM$ of a circle (equal radii) give equal diagonals bisecting at $90^\circ$ $\Rightarrow APML$ is a square.

Q4. Two equal sticks joined at their midpoints have equal diagonals that bisect each other $\Rightarrow$ a rectangle, so the corner angle is exactly $90^\circ$ — a stick-and-thread way to build a right angle.

Parallelogram angles (page 102, Q1). (i) $\angle P=40^\circ\Rightarrow\angle R=40^\circ$ (opposite), $\angle A=\angle E=140^\circ$. (ii) given $110^\circ$: opposite $=110^\circ$, the other pair $=70^\circ$ each.

True/False (page 108, Q11).

  • (ii) "Three right angles $\Rightarrow$ rectangle." True — the fourth must also be $90^\circ$ since angles sum to $360^\circ$.
  • (iii) "Diagonals bisect each other $\Rightarrow$ parallelogram." True.
  • (iv) "Perpendicular diagonals $\Rightarrow$ rhombus." False — they must also bisect each other (kites have $\perp$ diagonals too).
  • (vii) "Isosceles trapeziums are parallelograms." False — only one pair of sides is parallel.

12. Common mistakes to avoid

  • Thinking a parallelogram's diagonals are equal — only the rectangle/square have that; parallelogram diagonals merely bisect.
  • Saying a rhombus has $90^\circ$ corner angles — no; only its diagonals cross at $90^\circ$ (corners are $90^\circ$ only if it is a square).
  • Forgetting that "$\perp$ diagonals" alone does not make a rhombus (a kite also has them).
  • Using "sum of angles $=180^\circ$" — that is a triangle; a quadrilateral is $360^\circ$.
  • Calling every parallelogram a square/rectangle — the nesting goes one way only.

13. Quick revision checklist

  • Angles of any quadrilateral sum to $360^\circ$.
  • Rectangle: angles $90^\circ$; diagonals equal & bisect.
  • Square: rectangle + equal sides; diagonals equal, bisect at $90^\circ$, bisect corners.
  • Parallelogram: opposite sides parallel & equal; opposite angles equal; diagonals bisect.
  • Rhombus: parallelogram + all sides equal; diagonals bisect at $90^\circ$ and bisect angles.
  • Kite: two adjacent equal pairs; diagonals perpendicular, one bisects the other.
  • Trapezium: at least one parallel pair; co-interior angles $=180^\circ$; isosceles $\Rightarrow$ base angles equal.
  • Nesting: Square $\subset$ {Rectangle, Rhombus} $\subset$ Parallelogram $\subset$ Trapezium.
Practice MCQs
1. The sum of all four angles of a quadrilateral is:
  1. $180^\circ$
  2. $270^\circ$
  3. $360^\circ$
  4. $540^\circ$
Answer: (C) $360^\circ$ — split into two triangles, $180^\circ+180^\circ$.
2. Three angles of a quadrilateral are $90^\circ,80^\circ$ and $110^\circ$. The fourth angle is:
  1. $70^\circ$
  2. $80^\circ$
  3. $90^\circ$
  4. $100^\circ$
Answer: (B) $360^\circ-(90^\circ+80^\circ+110^\circ)=80^\circ$.
3. The diagonals of a rectangle are always:
  1. perpendicular
  2. equal and bisect each other
  3. unequal
  4. bisect the angles
Answer: (B) equal in length and bisecting each other (Property 4).
4. In which quadrilateral do the diagonals bisect each other at $90^\circ$ but the angles are not all $90^\circ$?
  1. Rectangle
  2. Square
  3. Rhombus
  4. Trapezium
Answer: (C) rhombus — diagonals meet at $90^\circ$, corners are not.
5. Every square is a:
  1. rectangle only
  2. rhombus only
  3. both a rectangle and a rhombus
  4. kite only
Answer: (C) a square lies in the overlap of rectangles and rhombuses.
6. One angle of a parallelogram is $65^\circ$. Its adjacent angle is:
  1. $65^\circ$
  2. $115^\circ$
  3. $125^\circ$
  4. $295^\circ$
Answer: (B) adjacent angles are supplementary: $180^\circ-65^\circ=115^\circ$.
7. The diagonal of a square divides each corner angle into:
  1. $30^\circ$ each
  2. $45^\circ$ each
  3. $60^\circ$ each
  4. $90^\circ$ each
Answer: (B) it bisects the $90^\circ$ corner, giving $45^\circ+45^\circ$ (Property 5).
8. A quadrilateral with at least one pair of parallel sides is a:
  1. kite
  2. rhombus
  3. trapezium
  4. rectangle
Answer: (C) trapezium — "at least one" pair of parallel sides.
9. In a kite the diagonals are:
  1. equal
  2. perpendicular to each other
  3. both bisected at the centre
  4. parallel
Answer: (B) perpendicular; the axis diagonal bisects the other but they need not bisect each other.
10. A quadrilateral whose diagonals are equal and bisect each other must be a:
  1. rhombus
  2. kite
  3. rectangle
  4. trapezium
Answer: (C) rectangle — exactly the carpenter's diagonal definition.
11. In a rhombus with one angle $70^\circ$, the angle opposite to it is:
  1. $70^\circ$
  2. $110^\circ$
  3. $140^\circ$
  4. $20^\circ$
Answer: (A) opposite angles of a rhombus (a parallelogram) are equal $=70^\circ$.
12. Which statement is FALSE?
  1. Every rectangle is a parallelogram
  2. Every rhombus is a parallelogram
  3. Every parallelogram is a rectangle
  4. Every square is a rhombus
Answer: (C) the nesting is one-way — a parallelogram need not have $90^\circ$ angles.
Assertion–Reason
A: The diagonals of a rectangle are equal.   R: In a rectangle $\triangle ADC\cong\triangle DAB$ by SAS, making $AC=BD$.
Answer: Both A and R are true, and R is the correct explanation of A — exactly Deduction 1.
A: Perpendicular diagonals make a quadrilateral a rhombus.   R: A kite also has perpendicular diagonals.
Answer: A is false, R is true — perpendicular diagonals alone are not enough (a kite has them too); they must also bisect each other.
Exam-style questions
Q1. In a rectangle the diagonal makes a $30^\circ$ angle with one side. Find all four angles the diagonals make at the centre, and prove the diagonals are equal. (4 marks)
Outline: isosceles $\triangle AOB$ with base angles $30^\circ$ gives central angle $120^\circ$; the other central angles are $60^\circ,120^\circ,60^\circ$. Equality of diagonals: $\triangle ADC\cong\triangle DAB$ (SAS) $\Rightarrow AC=BD$.
Q2. The angles of a quadrilateral are in the ratio $1:2:3:4$. Find each angle and name the most specific special quadrilateral it could be. (3 marks)
Answer: $1x+2x+3x+4x=360^\circ\Rightarrow x=36^\circ$, so angles are $36^\circ,72^\circ,108^\circ,144^\circ$. No pair of opposite angles is equal, so it is a general quadrilateral (a trapezium at most, since $108^\circ+72^\circ=180^\circ$ gives one parallel pair).
Q3. Prove that the diagonals of a parallelogram bisect each other. (3 marks)
Outline (Deduction 8): in parallelogram $EASY$, $AE=YS$ and alternate angles equal $\Rightarrow\triangle AOE\cong\triangle YOS$ (ASA). Hence $OA=OY$ and $OE=OS$, so $O$ is the midpoint of both diagonals.
Q4. A trapezium has $PQ\parallel SR$. If $\angle P=75^\circ$ and $\angle Q=80^\circ$, find $\angle S$ and $\angle R$. Then state when it would be isosceles. (4 marks)
Answer: co-interior: $\angle S=180^\circ-75^\circ=105^\circ$ and $\angle R=180^\circ-80^\circ=100^\circ$ (check $75+80+105+100=360^\circ$). It is isosceles only if the non-parallel sides are equal, which would force $\angle P=\angle Q$ — not the case here.
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