## MP Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.

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

Solution:

∠A = 3x, ∠B = 5x, ∠C = 9x, ∠D = 13x

In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°

(∴ Sum of all the angles of ♢ is equal to 360°)

3x + 5x + 9x + 13x = 360°

30x = 360°

x = \(\frac{360^{\circ}}{30^{\circ}}\)

Let angle in ratio be x then angles are x= 12°

∠A = 3 x 12° = 36°

∠B = 5 x 12° = 60°

∠C = 9 x 12° = 108°

∠D = 13 x 12° = 156°

Question 2.

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

Solution:

Given

ABCD is a parallelogram in which

AC =DB

To prove:

ABCD is a rectangle.

Proof

In ∆DAB and ∆CBA

DB = CA (given)

AB = BA (common)

AD = BC (∴ opposites sides of ∥^{gm} are equal)

∆DAB = ∆CBA (by SSS)

and so ∠DAB = ∠CBA

AD∥BC and AB is the transversal (by CPCT)

∴ ∠A + ∠B = 180° (CIA’s)

⇒ ∠A + ∠A = 180° (∴ ∠A = ∠B)

∴ ∠A = 90°

∠A = ∠C = 90°

and ∠B = ∠D = 90°

In ∥^{gm} ABCD, all the angles are right angles.

ABCD is a rectangle.

Question 3.

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

Solution:

Given

OA = OC, OB = OD and ∠AOD = 90°

To prove:

ABCD is a rhombus.

Proof:

In ∆AOD and ∆COB

OA = OC (given)

OD = OB (given)

∠1 = ∠2 (V.O.A.’s)

∴ ∆AOD = ∆COB (by SAS)

and so AD = CB (by CPC 7)

∠3 = ∠4 (by CPCT)

∠3 and ∠4 are A.I.A.’s and are equal

∴ AD ∥ PC (proved)

AD = BC

∴ ABCD is a parallelogram

In ∆AOD and ∆COD,

OA = OC (given)

OD = OD (common)

∠1 = ∠5 = 90°

(∴ ∠1 + ∠5 = 180° ⇒ 90° + ∠5 = 180° ∴ ∠5 = 90°)

∴ ∠AOD = ∠COD (by SAS)

and so AD = CD (by CPCT)

In ∥^{gm} APCD, all the sides are equal.

ABCD is rhombus.

Question 4.

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

Solution:

Given

ABCD is a square.

To prove:

- AC = BD
- OA = OC and OB = OD
- ∠AOD = 90°

Proof:

In ∆DAB and ACBA,

DA = CB (given)

AB = BA (common)

∠A = ∠B (each 90°)

∴ ∆DAB = ∆CBA (by SAS)

and so BD =AC (by CPCT)

2. In ∆AOD and ∆COB, AD = CB(given)

∠4 = ∠5 (A.I.A.’s)

∠6 = ∠7 (A.I.A. ’s)

∆AOD = ∆COB (by ASA)

and so OA=OC , (byCPCT)

OD = OB (byCPCT)

3. In ∆AOD and ∆COD,

AO = CO (proved)

OD = OD (common)

AD = CD (given)

∆AOD = ∆COD (by SSS)

and so ∠1 = ∠3 (byCPCT)

∠1 + ∠3 = 180° (LPA’S)

⇒ ∠1 + ∠1 = 180° (∠1 = ∠3)

⇒ 2∠1 = 180°

∴ ∠1 = \(\frac{180^{\circ}}{2}\)

Question 5.

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

Solution:

Give

ABCD is in which

AC = BD

OA = OC

and OB = OD

∠AOD = 90°

To prove

ABCD is a square

Proof:

In ∆AOD and ∆COB,

OA = OC (given)

OD = OB (given)

∠7 = ∠8 (V.CXA.’s)

∴ ∆AOD = ∠COB

and so AD = BC

and ∠3 = ∠1 (byCPCT)

∠3 and ∠1 are A.I.A.’s and are equal

∴ AD ∥ BC

Similarly, AB ∥ CD

∴ ABCD is a parallelogram.

In ∆AOD and ∆COD,

OA = OC (given)

OD = OD (common)

∠7 = ∠9 (each 90°)

∆AOD = ∠COD (by SAS)

AD = CD (byCPCT)

In ∥^{gm}ABCD, adjacent sidesAD = CD

∴ ABCD is a rhombus

In ∆DAB and ∆CBA,

DA = CB (proved)

AB = BA (common)

DB = CA (given)

∆DAB = ∆CBA (bySSS)

∠A = ∠B (by CPCT)

∠A + ∠B = 180° (CIA’s)

2∠A = 180°

∠A =90°

ABCD is a square.

Question 6.

Diagonals AC of a parallelogram ABCD bisects ∠A (see Fig). Show that

- it bisects ∠C also,
- ABCD is a rhombus.

Solution:

Given

ABCD is a parallelogram in which ∠1 = ∠2

To prove:

- ∠3 = ∠4
- ABCD is a rhombus.

Proof

1. ∠1 = ∠4 (A.I.A.’s) ….(i)

(∴ AD ∥ SC and AC is the transversal)

∠2 = ∠3 (A.I.A.’s) …(ii)

(∴ AB ∥ DC and AC is the transversal)

∠1 = ∠2 (given) …(iii)

From (i), (ii) and (iii), we get

∠4 = ∠3

(ii) From (ii) and (iii), we get,

∠1 = ∠3

In ∆ADC,

∠1 = ∠3

∴ AD = DC (In a A, sides opposites to equal angles are equal) and so ABCD is a rhombus

(∴ In a ∥^{gm}, if adjacent sides are equal then it is a rhombus)

Question 7.

ABCD is a rhombus, show that diagonal AC biusects ∠A as well as ∠C and diagonal BD biusects ∠B as well as ∠D.

Solution:

Given

ABCD is a rhombus.

To prove

∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8.

Proof:

In ∆ADC and ∆ABC.

AD = AB (Adj. sides of a rhombus)

DC = BC (Adj. sides of a rhombus)

AC = AC (common)

∴ ∆ADC = ∆ABC (by SSS)

so ∠1 = ∠2 (by CPCT)

and ∠3 = ∠4 (by CPCT)

∴ AC bisects ∠A and ∠C

Similarly, ∠5 = ∠6 and ∠7 = ∠8

BD bisects ∠B and ∠D.

Question 8.

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

- ABCD is a square
- diagonal BD bisects ∠B as well as ∠D.

Solution:

Given

ABCD is a rectangle.

∠1 = ∠2 and ∠3 = ∠4

To prove:

- ABCD is a square
- ∠5 = ∠6 and ∠7 = ∠8

Proof:

1. ∠A = ∠C

(∵ Rectangle is ∥^{gm} and in a ∥^{gm} opp. angles are equal.)

\(\frac{1}{2}\)∠A = \(\frac{1}{2}\)∠C

∠A = ∠C

∠2 = ∠4

In ∆ABC, ∠2 = ∠4

AB = BC

(In a A, sides opp. to equal angles are always equal)

ABCD is a rectangle in which adjacent sides are equal.

∴ ABCD is a square.

2. In ∆ABD,

AB = AD (∴ ABCD is a square)

∴ ∠5 = ∠7 (∴ In a A, angles opp. to equal sides are equal) ….(1)

AB ∥ DC and BD is the transversal

∴ ∠6 = ∠7 …(2)

AD ∥ BC and BD is the transversal.

∴ ∠5 = ∠8 …(3)

From (1) and (3), we get

∠7 = ∠8

From (1) and (2), we get

∠5 = ∠6

Diagonal BD bisects ∠B as well as ∠D.

Question 9.

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

- ∆APD = ∆CQB
- AP = CQ
- ∆AQB = ∆CPD
- AQ = CP A
- APCQ is a parallelogram

Given

ABCD is a parallelogram.

∴ AD = BC, AB= DC and DP = BQ

To prove:

- ∆APD = ∆CQB
- AP = CQ
- ∆AQB = ∆CPD
- AQ = CP
- APCQ is a parallelogram

Proof:

In ∆APD and ∆CQB

PD = QB (given)

AD = CB (given)

∠2 = ∠1 (AIA’s)

∆APD = ∆CQB (by SAS)

and so AP = CQ (by CPCT)

In ∆AQB and ∆CPD,

AB = CD (given)

∠3 = ∠4 (AIA’s)

BQ = DP (given)

∆AQB = ∆CPD (by SAS)

and so AQ = CP (by CPCT)

In quadrilaterals AQCP,

AP = CQ

AQ = CP

AQCP is a parallelogram.

Question 10.

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

- ∆APB = ∆CQD
- AP = CQ

Solution:

Given

ABCD is a ∥^{gm} in which Ap ⊥ BD and CQ ⊥ BD.

To prove:

- ∆APB = ∆CQD
- AP = CQ

Proof:

In ∆APB and ∆CQD,

∠P = ∠Q (each 90°)

∠1 = ∠2 (AIA’s)

AB = CD (given)

∆APB ≅ ∆CQD (byAAS)

and so AP = CQ (by CPCT)

Question 11.

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

- quadrilateral ABED is a parallelogram
- quadrilateral BEFC is a parallelogram
- AD ∥ CF and AD = CF
- quadrilateral ACFD is a parallelogram
- AC = DF
- ∆ABC = ∆DEF.

Solution:

Given

AB = DE and AB ∥ DE

BC = EF and BC ∥ EF

To prove

- ABED is a ∥
^{gm} - BEFC is a ∥
^{gm} - AD ∥ CF and AD – CF
- ACFD is a ∥
^{gm} - AC =DF
- ∆ABC ≅ ∆DEF

Proof:

1. AB = DE and AB ∥ DE (given)

ABED is ∥^{gm}

and so AD ∥ BE and AD = BE …(1)

2. BC = EF and BC ∥ FC (given)

BEFC is a ∥^{gm}

and so BE ∥ CF and BE = CF …..(2)

3. From (1) and (2), we get

AD ∥ CF and AD = CF

4. AD ∥ CF and AD = CF (proved)

ACFD is a ∥^{gm}

5. and so AC = DF

(In a parallelogram, opp. sides are equal)

6. In ∆ABC and ∆DEF,

AB = DE (given)

BC = EF (given)

AC = DF (proved)

∆ABC = ∆DEF (by SSS)

Question 12.

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

- ∠A = ∠B
- ∠C = ∠D
- ∆ABC ≅ ∆BAD
- Diagonal AC = diagonal BF)

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

Solution:

Given

AB ∥ CD, AD = BC

To prove:

- ∠A = ∠B
- ∠C = ∠D
- ∆ABC ≅ ∆BAD
- diagonal AC = diagonal BD

Construction:

Draw a line CE ∥ DA which intersect AB produced at E.

Proof:

1. In quadrilateral ADCE,

AD ∥ EC (by const)

and AE ∥ DC ( AB ∥ DC)

∴ ADCE is a parallelogram

and so AD = EC (opp. sides of a ∥ are equal)…(i)

AD = BC (given) ….(ii)

From (i) and (ii), we get

BC = EC

In ∆BCE BC = EC (proved)

∠4 = ∠3

(∴ In a ∆, angles opp. to equal sides are equal)

∠2 + ∠3 = 180° (LPA’s) …(iii)

∠1 + ∠4 = 180° (CIA’s) …(iv)

From (iii) and (iv), we get

∠2 + ∠3 = ∠1 + ∠4

∠2 = ∠1 (∠3 = ∠4)

i.e., ∠A = ∠B

2. ∠3 = ∠5 (AIA’s) …(v)

∠6 = ∠4

(∴ ADCE is a ∥^{gm} and in a ∥^{gm} opp. angles are equal) …(vi)

∠4 = ∠3 (proved) …(vii)

From (v), (vi) and (vii), we get

∠5 = ∠6

i.e., ∠C = ∠D

3. In ∆ABC and ∆BAD,

AB = BA (common)

BC = AD (given)

∠2 = ∠1 (proved)

∆ABC ≅ ∆BAD (by SAS)

and so AC =BD (by CPCT)

4. diagonal AC = diagonal BD (proved)

Mid Point Theorem:

The line segment joining the mid-points of the sides of a triangle is parallel to the third side and equal to half of it.

Given.

ABC is a A in which D and E are the mid-points of sides AB and AC respectively.

To prove.

DE ∥ BC and DE = \(\frac{1}{2}\) BC

Construction:

Extend DE uptoFsuch that DE = EF. Join CF.

Proof:

In ∆AED and ∆CEE

AE = CE (E is the mid – point of AC)

∠AED = ∠CEF (VOA’s)

DE = FE (By constriction)

∆AED = ∆CEF , (By SAS)

and so ∠DAE = ∠FCE (By CPCT)

AD = CF (By CPCT)

∠DAE and ∠FCE are alternate interior angles and are equal.

AD ∥ FC

⇒ DB ∥ FC

Now, AD = DB and AD = FC

DB = FC

In BCFD, DB ∥ FC and DF = BC

BCFD is a ∥^{gm}

and so DF ∥ BC and DF = BC

⇒ DF ∥ BC and 2DF = BC

DE ∥ BC and DE = \(\frac{1}{2}\) BC

Converse of mid point theorem:

The line drawn through the mid – point of one side of a triangle and parallel to another side, bisects the third side.

Given

ABC is a A in which D is the mid-point of AB and DE ∥ BC.

To Prove:

E is the mid – point of AC.

Construction:

Mark a point F on AC and join DF.

Proof:

Let E be not the mid – point of AC. Let us assume that F be the mid – point of AC.

Then by mid-point theorem

DF ∥ BC

DE ∥ BC (Given)

From (i) and (ii), we get

DE ∥ DF

But lines DE and DF are intersecting lines, intersecting at D. This is a contradiction. So our supposition is wrong. Hence E is the mid – point of AC.