Quadrilaterals (Mathematics) Class 9 - NCERT Questions

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

SOLUTION:Let the angles of a quadrilateral be 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360°

[Angle sum property of a quadrilateral]

∴ 3x = 3 × 12° = 36°

5x = 5 × 12° = 60°

9x = 9 × 12° = 108°

13x = 13 × 12° = 156°

∴ The required angles of a quadrilateral are 36°, 60°, 108° and 156°.

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

SOLUTION:ABCD is a parallelogram such that AC = BD.

In ∆ABC and ∆DCB,

AC = DB [Given] AB = DC

[Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∥ABC = ∥DCB [By C.P.C.T.] …(1)

Now, AB ∥ DC and BC is a transversal.

[∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° …(2)

[Co-interior angles]

From (1) and (2), we have ∠ABC = ∠DCB = 90°

i.e. ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

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

SOLUTION:We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ∆AOB and ∆AOD, we have

AO = OA [Common]

OB = OD [O is the midpoint of BD]

∠AOB = ∠AOD [Each 90°]

∴ ∆AOB ≅ ∆AOD [By SAS congruency]

∴ AB = AD [By C.P.C.T.] …(1)

Similarly, AB = BC …(2)

BC = CD …(3)

CD = DA …(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, a quadrilateral ABCD is a rhombus.

Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will results in rhombus

Show that the diagonals of a square are equal and bisect each other at right angles.

SOLUTION: We have a square ABCD such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, i.e., AC = BD. In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Sides of a square ABCD]

∠ABC = ∠BAD [∵ each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

⇒ AC = BD [By C.P.C.T.] ...(1)

(ii) ∵ AD ∥ BC and AC is a transversal.

[∵ A square is a parallelogram]

∴ ∠1 = ∠3

[Alternate interior angles are equal]

Similarly, ∠2 = ∠4

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides of a square ABCD]

∠1 = ∠3 [Proved]

∠2 = ∠4 [Proved]

∴ ∆OAD ≅ ∆OCB [By ASA congruency]

⇒ OA = OC and OD = OB [By C.P.C.T.]

i.e., the diagonals AC and BD bisect each other at O. …(2)

(iii) In ∆OBA and ∆ODA, we have

OB = OD [Proved]

BA = DA [Sides of a square]

OA = AO [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency]

⇒ ∠AOB = ∠AOD …(3)

[By C.P.C.T.]

∵ ∠AOB and ∠AOD form a linear pair.

∴ ∠AOB + ∠AOD = 180°

∴ ∠AOB = ∠AOD = 90° [By(3)]

⇒ AC ⊥ BD …(4)

From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

SOLUTION: We have a quadrilateral ABCD such that O is the midpoint of AC and BD.

Also, AC ⊥ BD.

Now, in ∆AOD and ∆AOB, we have

∠AOD = ∠AOB [Each 90°]

AO = OA [Common]

OD = OB [∵ O is the midpoint of BD]

∴ ∆AOD ≅ ∆AOB [By SAS congruency]

⇒ AD = AB [By C.P.C.T.] …(1)

Similarly, we have AB = BC …(2)

BC = CD …(3)

CD = DA …(4)

From (1), (2), (3) and (4), we have

AB = BC = CD = DA

∴ Quadrilateral ABCD have all sides equal.

In ∆AOD and ∆COB, we have

AO = CO [Given]

OD = OB [Given]

∠AOD = ∠COB

[Vertically opposite angles]

So, ∆AOD ≅ ∆COB [By SAS congruency]

∴ ∠1 = ∠2 [By C.P.C.T.]

But, they form a pair of alternate interior angles.

∴ AD ∥ BC

Similarly, AB ∥ DC

∴ ABCD is a parallelogram

∴ Parallelogram having all its sides equal is a rhombus.

∴ ABCD is a rhombus.

Now, in ∆ABC and ∆BAD, we have

AC = BD [Given]

BC = AD [Proved]

AB = BA [Common]

∴ ∆ABC ≅ ∆BAD [By SSS congruency]

∴ ∠ABC = ∠BAD [By C.P.C.T.] …(5)

Since, AD ∥ BC and AB is a transversal.

∴ ∠ABC + ∠BAD = 180° …(6)

[Adjencent angles are supplementary]

⇒ ∠ABC = ∠BAD = 90° [By (5) & (6)]

So, rhombus ABCD is having one angle equal to 90°.

Thus, ABCD is a square.

Diagonal AC of a parallelogram ABCD bisects ∠A(see figure). Show that

(i) it bisects ∠C also.

(ii)ABCD is a rhombus.

We have a parallelogram ABCD in which diagonal AC bisects ∠A ⇒ ∠DAC = ∠BAC

(i) Since, ABCD is a parallelogram.

∴ AB ∥ DC and AC is a transversal.

∴ ∠1 = ∠3 [∵ alternate interior angles are equal] …(1)

Also, BC ∥ AD and AC is a transversal.

∴ ∠2 = ∠4 [Alternate interior angles]
…(2)

Also, ∠1 = ∠2 [∵ AC bisects ∠A] …(3)

From (1), (2) and (3), we have

∠3 = ∠4 ⇒ AC bisects ∠C.

(ii) In ΔABC, we have

∠1 = ∠4 [From (2) and (3)]

⇒ BC = AB …(4)

[∵ sides opposite to equal angles of a ∆ are equal]

Similarly, AD = DC …(5)

But, ABCD is a parallelogram [Given]

∴ AB = DC …(6)

From (4), (5) and (6), we have

AB = BC = CD = DA

Thus, ABCD is a rhombus.

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

SOLUTION:Since ABCD is a rhombus

⇒ AB = BC = CD = DA

Also, AB ∥ CD and AD ∥ BC

Now, AD = CD ⇒ ∠1 = ∠2 …(1)

[∵ angles opposite to equal sides of a triangle are equal]

Also, AD BC and AC is the transversal.

[∵ every rhombus is a parallelogram]

∠1 = ∠3 …(2)

[∵ Alternate interior angles are equal]

From (1) and (2), we have

∠2 = ∠3 …(3)

Since, AB DC and AC is transversal.

∴ ∠2 = ∠4 …(4)

[∵ Alternate interior angles are equal]

From (1) and (4), we have

∠1 = ∠4

⇒ AC bisects ∠C as well as ∠A.

Similarly, we can prove that BD bisects ∠B as well as ∠D.

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:

(i) ABCD is a square.

(ii) diagonal BD bisects ∠B as well as ∠D.

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.

i.e. ∠1 = ∠4 and ∠2 = ∠3 …(1)

(i) Since, every rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB ∥ CD and AC is a transversal.

∴ ∠2 = ∠4 [Alternate interior angles]
…(2)

From (1) and (2), we have

∠3 = ∠4

In ΔABC, ∠3 = ∠4

⇒ AB = BC

[∵ sides opposite to equal angles of a ∆ are equal]

⇒ ABCD is a rectangle having adjacent sides equal.

⇒ ABCD is a square.

(ii) Since, ABCD is a square and diagonals of a square bisect the opposite angles.

So, BD bisects ∠B as well as ∠D.

In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:

(i) ∆APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD (iv) AQ = CP

(v) APCQ is a parallelogram

We have parallelogram ABCD, BD is the diagonal and points P and Q are such that

PD = QB [Given]

(i) Since, AD ∥ BC and BD is a transversal.

∴ ∠ADB = ∠CBD

[∵ Alternate interior angles are equal]

⇒ ∠ADP = ∠CBQ

Now, in ∆APD and ∆CQB, we have

AD = CB [Opposite sides of a parallelogram

ABCD are equal]

PD = QB [Given]

∠ADP = ∠CBQ [Proved]

∴ ∆APD ≅ ∆CQB [By SAS congruency]

(ii) Since, ∆APD ≅ ∆CQB [Proved]

⇒ AP = CQ [By C.P.C.T.]

(iii) Since, AB ∥ CD and BD is a transversal.

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

AB = CD [∵ Opposite sides of a parallelogram ABCD are equal]

∴ ∆AQB ≅ ∆CPD [By SAS congruency]

(iv) Since, ∆AQB ≅ ∆CPD [Proved]

⇒ AQ = CP [By C.P.C.T.]

(v) ∵ In a quadrilateral APCQ

Opposite sides are equal. [Proved]

∴ APCQ is a parallelogram.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that :

(i) ∆APB≅ ∆CQD

(ii) AP = CQ

(i) In ∆APB and ∆CQD, we have

∠APB = ∠CQD [Each 90°]

AB = CD [∵ Opposite sides of a parallelogram ABCD are equal]

∠ABP = ∠CDQ [∵ Alternate angles are euqal as AB∥ CD and BD is a transversal]

∴ ∆APB ≅ ∆CQD [By AAS congruency]

(ii) Since, ∆APB ≅ ∆CQD [Proved]

⇒ AP = CQ [By C.P.C.T.]

In ∆ABC and ∆DEF, AB = DE, AB ∥DE, BC = EF and BC ∥ EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that

(i) quadrilateral ABED is a parallelogram.

(ii) quadrilateral BEFC is a parallelogram.

(iii) AD ∥ CF and AD = CF.

(iv) quadrilateral ACFD is a parallelogram.

(v) AC = DF

(i) We have

AB = DE [Given]

AB DE [Given]

i.e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.

∴ ABED is a parallelogram.

(ii) BC = EF [Given]

and BC ∥ EF [Given]

i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.

∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram [Proved]

∴ AD ∥ BE and AD = BE [∵ Opposite sides of a parallelogram are equal and parallel] …(1)

Also, BEFC is a parallelogram. [Proved]

∴ BE ∥ CF and BE = CF [∵ Opposite sides of a parallelogram are equal and parallel] …(2)

From (1) and (2), we have

AD ∥CF and AD = CF

(iv) Since, AD ∥ CF and AD = CF [Proved]

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and equal in length.

∴ Quadrilateral ACFD is a parallelogram.

(v) Since, ACFD is a parallelogram.

[Proved]

So, AC = DF [∵ opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DEF, we have

AB = DE [∵ opposite sides of a parallelogram ABED are equal]

BC = EF [∵ opposite sides of a parallelogram BEFC are equal]

AC = DF [Proved in (v)]

∴ ∆ABC ≅ ∆DEF [By SSS congruency]

ABCD is a trapezium in which AB ∥ CD and AD = BC (see figure). Show that

(i) ∠A = ∠ B

(ii) ∠C = ∠D

(iii) ∆ABC ≅ ∆BAD

(iv) Diagonal AC = Diagonal BD

Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.

(i) Produce AB to E and draw CE ∥ AD.

∵ AB ∥ DC ⇒ AE DC

Also AD ∥ CE [By construction]

∴ AECD is a parallelogram.

⇒ AD = CE …(1) [∵opposite sides of the parallelogram AECD are equal]

But AD = BC …(2) [Given]

By (1) and (2)

BC = CE

Now, in ∆BCE, we have BC = CE

⇒ ∠CEB = ∠CBE ...(1)

[∵ angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° …(2)

[Linear pair]

and ∠A + ∠CEB = 180° …(3)

[co-interior angles of a parallelogram ADCE

]
From (2) and (3), we get

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From (1)]

⇒ ∠B = ∠A …(4)

(ii) AB ∥ CD and AD is a transversal.

∴ ∠A + ∠D = 180° …(5)

[Co-interior angles of a paralelogram]

Similarly, ∠B + ∠C = 180° …(6)

From (5) and (6), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (4)]

(iii) In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

(iv) Since, ∆ABC ≅ ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]

ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that:

(ii) PQ = SR

(iii) PQRS is a parallelogram.

(i) In ∆ACD, we have

S is the midpoint of AD and R is the midpoint of CD.

ABCD is a rhombus and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

SOLUTION:Join AC. In ∆ABC, P and Q are the midpoints of AB and BC respectively.

[By mid-point theorem]

In ∆ADC, R and S are the midpoints of CD and DA respectively.

CE = EC [Common]

∴ ∆ERC ≅ ∆EQC [By SAS congruency]

⇒ ∠3 = ∠4 …(4) [By C.P.C.T.]

But ∠3 + ∠4 = 180° …(5) [Linear pair]

From (4) and (5), we get

⇒ ∠3 = ∠4 = 90°

Now, ∠RQP = 180° – ∠5

[∵ Co-interior angles for PQ ∥ AC and EQ is transversal]

But ∠5 = ∠3 [∵ Vertically opposite angles are equal]

∴ ∠5 = 90°

So, ∠RQP = 180° – ∠5 = 90°

∴ One angle of parallelogram PQRS is 90°.

Thus, PQRS is a rectangle.

ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

SOLUTION:

∴ ∆PAS ≅ ∆PBQ [By SAS congruency]

⇒ PS = PQ [By C.P.C.T.]

Also, PS = QR and PQ = SR. [∵ opposite sides of parallelogram PQRS are equal]

So, PQ = QR = RS = SP

i.e., PQRS is a parallelogram having all of its sides equal.

Hence, PQRS is a rhombus.

ABCD is a trapezium in which AB ∥DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid point

of BC.

In ∆DAB, We know that E is the midpoint of AD and EG ∥ AB [∵ EF ∥ AB]

∴ Using the converse of midpoint theorem, we get, G is the midpoint of BD.

Again in ∆BDC, we have

G is the midpoint of BD and GF ∥ DC.

[∵ AB ∥ DC and EF ∥ AB and GF is a part of EF]

Using the converse of the midpoint theorem, we get, F is the midpoint of BC.

In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Since, the opposite sides of a parallelogram are parallel and equal.

∴ AB ∥ DC ⇒ AE ∥ FC …(1)

and AB = DC

From (1) and (2), we have

AE ∥ FC and AE = FC

∴ AECF is a parallelogram.

Now, in ∆ DQC, we have

F is the midpoint of DC and FP ∥ CQ

[∵ AF ∥ CE]

⇒ DP = PQ ...(3)

[By converse of mid-point theorem]

Similarly, in ∆ BAP, E is the mid-point of AB and EQ ∥ AP [∵ AF ∥ CE]

⇒ BQ = PQ …(4)

[By converse of mid-point theorem]

∴ From (3) and (4), we have

DP = PQ = BQ

So, the line segments AF and EC trisect the diagonal BD.

Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.

SOLUTION:Join PQ, QR, RS and SP. Let us also join PR and SQ.

Now, in ∆ABC, we have P and Q are the midpoints of its sides AB and BC respectively.

[By mid-point theorem]

∴ By (1) and (2), we get

PQ ∥ RS, PQ = RS

∴ PQRS is a parallelogram.

And the diagonals of a parallelogram bisect each other, i.e., PR and SQ bisect each other.

Thus, the line segments joining the mid-points of opposite sides of a quadrilateral ABCD bisect each other.

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

(i) D is the midpoint of AC

(ii) MD ⊥AC

(i) In ∆ACB, we have

M is the midpoint of AB. [Given]

MD ∥ BC [Given]

∴ Using the converse of mid-point theorem, D is the midpoint of AC.

(ii) Since, MD ∥ BC and AC is a transversal.

∴ ∠MDA = ∠BCA

[∵ Corresponding angles are equal]

As ∠BCA = 90° [Given]

∴ ∠MDA = 90°

⇒ MD ⊥ AC.

(iii) In ∆ADM and ∆CDM, we have ∠ADM = ∠CDM [Each equal to 90°]

MD = DM [Common]

AD = CD [∵ D is the midpoint of AC]

∴ ∆ADM ≅ ∆CDM [By SAS congruency]

⇒ MA = MC [By C.P.C.T.] …(1)

∵ M is the midpoint of AB [Given]