NCERT Solutions for Class 9 Maths Chapter 8: Quadrilaterals — Free PDF

Quadrilateral: A closed figure with four sides, four vertices, and four angles. The sum of its interior angles is always $360^\circ$.

Parallelogram: A quadrilateral in which both pairs of opposite sides are parallel. Notation: $\parallel$gm $ABCD$ means $AB \parallel DC$ and $AD \parallel BC$.

Special Parallelograms:
- Rectangle: A parallelogram with one right angle (then all angles are $90^\circ$). Diagonals are equal.
- Rhombus: A parallelogram with two adjacent sides equal (then all sides are equal). Diagonals are perpendicular.
- Square: A parallelogram that is both a rectangle and a rhombus. All sides equal, all angles $90^\circ$, diagonals equal and perpendicular.

• Theorem 8.1: A diagonal of a parallelogram divides it into two congruent triangles.
  - Theorem 8.2: In a parallelogram, opposite sides are equal.
  - Theorem 8.3: In a parallelogram, opposite angles are equal.
  - Theorem 8.4: The diagonals of a parallelogram bisect each other.
  - Theorem 8.5: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
  - Theorem 8.6: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
  - Theorem 8.7: A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.

Midpoint Theorem (Theorem 8.9): The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

If $D$ and $E$ are midpoints of $AB$ and $AC$ in $\triangle ABC$:

Converse of Midpoint Theorem (Theorem 8.10): A line through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

A quadrilateral is a parallelogram if any ONE of the following holds:
1. Both pairs of opposite sides are parallel (definition).
2. Both pairs of opposite sides are equal.
3. Both pairs of opposite angles are equal.
4. Diagonals bisect each other.
5. One pair of opposite sides is both parallel and equal.

Any single condition is sufficient — you do not need to prove multiple conditions.

Problem: The angles of a quadrilateral are in the ratio $3:5:9:13$. Find all angles.

Solution:
Let the angles be $3x, 5x, 9x, 13x$.

Angle sum of a quadrilateral $= 360^\circ$:

Angles: $36^\circ, 60^\circ, 108^\circ, 156^\circ$.

Problem: If the diagonals of a parallelogram are equal, show that it is a rectangle.

Solution:
Let $ABCD$ be a parallelogram with $AC = BD$.

In $\triangle ABC$ and $\triangle DCB$:
- $AB = DC$ (opposite sides of parallelogram)
- $BC = CB$ (common)
- $AC = DB$ (given)

By SSS: $\triangle ABC \cong \triangle DCB$

By CPCT: $\angle ABC = \angle DCB$

But $\angle ABC + \angle DCB = 180^\circ$ (co-interior angles, $AB \parallel DC$)

$\therefore 2\angle ABC = 180^\circ \implies \angle ABC = 90^\circ$

A parallelogram with one right angle is a rectangle. $\blacksquare$

Problem: Show that the diagonals of a rhombus are perpendicular to each other.

Solution:
Let $ABCD$ be a rhombus with diagonals meeting at $O$.

In $\triangle AOB$ and $\triangle COB$:
- $OA = OC$ (diagonals of a parallelogram bisect each other)
- $AB = CB$ (all sides of a rhombus are equal)
- $OB = OB$ (common)

By SSS: $\triangle AOB \cong \triangle COB$

By CPCT: $\angle AOB = \angle COB$

$\angle AOB + \angle COB = 180^\circ$ (linear pair)
$\therefore \angle AOB = \angle COB = 90^\circ$

Hence diagonals are perpendicular. $\blacksquare$

Problem: Show that the diagonals of a square are equal and perpendicular.

Solution:
A square is both a rectangle and a rhombus.
- From Problem 2: Since a square is a rectangle, diagonals are equal.
- From Problem 3: Since a square is a rhombus, diagonals are perpendicular.

Hence the diagonals of a square are equal and bisect each other at right angles. $\blacksquare$

Problem: Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:
Let diagonals $AC$ and $BD$ of quadrilateral $ABCD$ bisect each other at right angles at $O$.
So $OA = OC$, $OB = OD$, and $\angle AOB = 90^\circ$.

In $\triangle AOB$ and $\triangle COB$:
- $OA = OC$ (given)
- $OB = OB$ (common)
- $\angle AOB = \angle COB = 90^\circ$

By SAS: $\triangle AOB \cong \triangle COB$ $\implies AB = CB$ ...(i)

Similarly, $\triangle BOC \cong \triangle DOC$ $\implies BC = DC$ ...(ii)

And $\triangle COD \cong \triangle AOD$ $\implies CD = DA$ ...(iii)

From (i), (ii), (iii): $AB = BC = CD = DA$

Since all sides are equal, $ABCD$ is a rhombus. $\blacksquare$

Problem: $ABCD$ is a quadrilateral in which $P$, $Q$, $R$, $S$ are the midpoints of sides $AB$, $BC$, $CD$, $DA$ respectively. Show that $PQRS$ is a parallelogram.

Solution:
Join diagonal $AC$.

In $\triangle ABC$: $P$ is the midpoint of $AB$, $Q$ is the midpoint of $BC$.
By the Midpoint Theorem: $PQ \parallel AC$ and $PQ = \dfrac{1}{2}AC$ ...(i)

In $\triangle ACD$: $S$ is the midpoint of $AD$, $R$ is the midpoint of $CD$.
By the Midpoint Theorem: $SR \parallel AC$ and $SR = \dfrac{1}{2}AC$ ...(ii)

From (i) and (ii): $PQ \parallel SR$ and $PQ = SR$

Since one pair of opposite sides is both parallel and equal, $PQRS$ is a parallelogram. $\blacksquare$

Problem: $ABCD$ is a rhombus. Show that the quadrilateral formed by joining the midpoints of the sides is a rectangle.

Solution:
From Problem 1, we know that the quadrilateral $PQRS$ formed by joining the midpoints is a parallelogram.

Since $ABCD$ is a rhombus, diagonals $AC \perp BD$.

From the midpoint theorem: $PQ \parallel AC$ and $QR \parallel BD$.

Since $AC \perp BD$, we get $PQ \perp QR$, i.e., $\angle PQR = 90^\circ$.

A parallelogram with one right angle is a rectangle. Hence $PQRS$ is a rectangle. $\blacksquare$

Problem: Show that the line segment joining the midpoints of the parallel sides of a trapezium is parallel to both sides and equal to half their sum.

Solution:
Let $ABCD$ be a trapezium with $AB \parallel DC$. Let $E$ and $F$ be midpoints of $AD$ and $BC$.

Join $DF$ and extend it to meet $AB$ produced at $G$.

In $\triangle DCF$ and $\triangle BGF$:
- $CF = BF$ ($F$ is midpoint of $BC$)
- $\angle DCF = \angle GBF$ (alternate interior angles, $DC \parallel AG$)
- $\angle DFC = \angle GFB$ (vertically opposite angles)

By ASA: $\triangle DCF \cong \triangle BGF$

By CPCT: $DF = GF$ and $DC = BG$.

In $\triangle DAG$: $E$ is the midpoint of $DA$ and $F$ is the midpoint of $DG$.
By the Midpoint Theorem: $EF \parallel AG$ (i.e., $EF \parallel AB$) and:

Problem: In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$. Show that $APCQ$ is a parallelogram.

Solution:
In $\triangle APD$ and $\triangle CQB$:
- $AD = CB$ (opposite sides of $\parallel$gm $ABCD$)
- $DP = BQ$ (given)
- $\angle ADP = \angle CBQ$ (alternate interior angles, $AD \parallel BC$ with transversal $BD$)

By SAS: $\triangle APD \cong \triangle CQB$
By CPCT: $AP = CQ$ ...(i)

Similarly, in $\triangle AQB$ and $\triangle CPD$:
- $AB = CD$ (opposite sides of $\parallel$gm)
- $BQ = DP$ (given)
- $\angle ABQ = \angle CDP$ (alternate interior angles, $AB \parallel DC$)

By SAS: $\triangle AQB \cong \triangle CPD$
By CPCT: $AQ = CP$ ...(ii)

From (i) and (ii), both pairs of opposite sides of $APCQ$ are equal, so $APCQ$ is a parallelogram. $\blacksquare$

Problem: $ABC$ is a triangle right-angled at $C$. A line through the midpoint $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ at $D$. Show that $D$ is the midpoint of $AC$ and $MD \perp AC$.

Solution:
$M$ is the midpoint of $AB$, and $MD \parallel BC$.

By the converse of the midpoint theorem, $D$ is the midpoint of $AC$. ...(i)

Since $MD \parallel BC$ and $\angle ACB = 90^\circ$, the corresponding angle $\angle ADM = 90^\circ$.

Hence $MD \perp AC$. ...(ii)

Problem: Prove that the quadrilateral formed by joining the midpoints of a rectangle is a rhombus.

Solution:
Let $ABCD$ be a rectangle with $P, Q, R, S$ midpoints of $AB, BC, CD, DA$.

From the general result (Exercise 8.2, Problem 1), $PQRS$ is a parallelogram.

Now, in a rectangle, diagonals are equal: $AC = BD$.

From the midpoint theorem:

Since $AC = BD$, we get $PQ = QR$.

A parallelogram with adjacent sides equal is a rhombus. Hence $PQRS$ is a rhombus. $\blacksquare$

Question: In a parallelogram $ABCD$, $\angle A = 60^\circ$. Find all four angles.

Answer: In a parallelogram, consecutive angles are supplementary.
$\angle A = 60^\circ$, $\angle B = 180^\circ - 60^\circ = 120^\circ$.
Opposite angles are equal: $\angle C = 60^\circ$, $\angle D = 120^\circ$.

Question: The diagonals of a quadrilateral $ABCD$ bisect each other. If $\angle A = 35^\circ$, determine $\angle B$.

Answer: Since diagonals bisect each other, $ABCD$ is a parallelogram.
Consecutive angles are supplementary: $\angle B = 180^\circ - 35^\circ = 145^\circ$.

Question: $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$. Show that $AP = CQ$.

Answer: In $\triangle APB$ and $\triangle CQD$:
- $AB = CD$ (opposite sides)
- $\angle ABP = \angle CDQ$ (alternate interior angles, $AB \parallel CD$)
- $\angle APB = \angle CQD = 90^\circ$ (given)
By AAS: $\triangle APB \cong \triangle CQD$, so $AP = CQ$ by CPCT.

Question: $D$, $E$, $F$ are midpoints of sides $BC$, $CA$, $AB$ of $\triangle ABC$. Show that $BDEF$ is a parallelogram.

Answer: By the midpoint theorem in $\triangle ABC$: $EF \parallel BC$ and $EF = \dfrac{1}{2}BC = BD$.
Since $EF \parallel BD$ and $EF = BD$, one pair of opposite sides is parallel and equal. Hence $BDEF$ is a parallelogram.