RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.
The angles of a quadrilateral are in the ratio of 3:5:9:13. Find all the angles of the quadrilateral.
Answer:
Let the angles be (3x), (5x), (9x) and (13x)
Then, 3x + 5x + 9x + 13x = 360°
or 3x = 360°
or x = \(\frac{360^{\circ}}{30}\) = 12
∴ The angles are (3 × 12)°, (5 × 12)°, (9 × 12)° and (13 × 12)°, i.e., 36°, 60°, 108° and 156°.

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer:
A parallelogram ABCD in which AC = BD.

To show : ABCD is a rectangle.
In ∆ABC and ∆DCB, we have :
AB = DC (Opp. sides of a ∥gm)
BC = CB (Common)
and, AC = DB (Given)

∴ By SSS criterion of congruence, we have :
∆ABC = ∆DCB
So, ∠ABC = ∠DCB …………(1)
(Corresponding parts of congruent triangles are equal.) But AB ∥ DC and BC cuts them.
∴ ∠ABC + ∠DCB = 180° …………(2) (Sum of interior angles on the same side of the transversal is 180°.)
or 2∠ABC = 180° [From (1) and (2)]
⇒ ∠ABC = 90°
Thus, ∠ABC = ∠DCB = 90°
So, ABCD is a parallelogram one of whose angles is 90°.
Hence, ABCD is a rectangle. Hence proved.

Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer:
A quadrilateral ABCD in which the diagonals AC and BD intersect at O such that AO = OC, BO = OD and AC ⊥BD.
To show : ABCD is a rhombus.
The diagonals AC and BD of quadrilateral ABCD bisect each other at right angles.
∴ AC is the perpendicular bisector of the line segment BD.

A and C both are equidistant from B and D.
So, AB = AD and CB = CD
Also, BD is the perpendicular bisector of line segment AC.

B and D both are equidistant from A and C.
So, AB = BC and AD = DC (2)
From (1) and (2), we get AB = BC = CD = AD
Thus, ABCD is a quadrilateral whose diagonals bisect each other at right angles and all four sides are equal. •
Hence, ABCD is a rhombus.
Hence proved.

Alternate Method :
First we shall show that ABCD is a ∥gm.
In As AOD and COB, we have :
AO = OC (Given)
OD = OB (Given)
and ∠AOD = ∠COB (Vertically opp. angles)
∴ By SAS criterion of congruence,
∆AOD = ∆COB
⇒ ∠OAD = ∠OCB (1)
(Corresponding parts of congruent triangles are equal.)

Now, line AC intersects AD and BC at A and C respectively such that
∠OAD = ∠OCB [From (1))
i.e. alternate interior angles are equal.
AD ∥ BC
Similarly, AB ∥ CD
Hence, ABCD is a parallelogram.
Now, we shall show that ∥gm ABCD is a rhombus.

In As AOD and COD, we have :
OA = OC (Given)
∠AOD = ∠COD (Both are right angles)
OD = OD , (Common)

By SAS criterion of congruence, we have :
∆AOD = ∆COD
AD = CD …..(2)
(Corresponding parts of congruent triangles are equal.) (Shown above)
AB = CD and AD = BC (Opp. sides of a ∥gm are equal)
∴ AB = CD = AD = BC [Using (2)]
Hence, quadrilateral ABCD is a rhombus.
Hence proved.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Answer:
Given : A square ABCD.

To show : AC = BD, AC ⊥ BD and OA = OC, OB – OD.
Since ABCD is a square, therefore AB ∥DC and AD ∥BC.
Now, AB ∥DC and transversal AC intersects them at A and C respectively.
∴ ∠BAC = ∠DCA (Alternate interior angles are equal.)
⇒ ∠BAO = ∠DCO ………….. (1)

Again, AB ∥ DC and BD intersects them at B and D respectively.
∴ ∠ABD – ∠CDB (∵ Alternate interior angles are equal.)
⇒ ∠ABO = ∠CDO ………..(2)

Now, in As AOB and COD, we have :
∠BAO = ∠DCO
AB = CD
and, ∠ABO = ∠CDO
By ASA congruence criterion, we have:
∆AOB = ∆COD
So, OA = OC and OB = OD
(Corresponding parts of congruent As are equal.)
Hence, the diagonals bisect each other.

In As ADB and BCA, we have :
AD = BC (Sides of a square are equal.)
∠BAD = ∠ABC (Each angle equal to 90°)
and, AB = BA (Common)

∴ By SAS criterion of congruence, we have :
∆ADB = ∆BCA
So, AC =BD (∵ Corresponding parts of congruent As are equal.)
Hence, the diagonals are equal.

Now in As AOB and AOD, we have :
OB = OD (∵ Diagonals of a ∥gm bisect each other)
AB = AD (∵ Sides of a square are equal.)
and, AO = AO (Common)

∴ By SSS criterion of congruence, we have:
∆AOB = ∆AOD
So, ∠AOB = ∠AOD (Corresponding parts of congruent As are equal.)

But ∠AOB + ∠AOD = 180°
∴∠AOB = ∠AOD =90°
AO ⊥ BD AC ⊥ BD
Hence, diagonals intersect at right angles. Hence proved.

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer:
Given : A quadrilateral ABCD in which the diagonals AC = BD, AO = OC, BO = OD and AC ⊥ BD.

To show : Quadrilateral ABCD is a square.
First, we shall show that ABCD is a parallelogram.

In As AOD and COB, we have :
OA = OC (Given)
OD = OB (Given)
and ∠AOD = ∠COB (Vertically opp. angles)

By SAS criterion of congruence,
∆AOD = ∆COB
So, ∠OAD = ∠OCB (1)
(Corresponding parts of congruent triangles are equal.)

Now, line AC intersects AD and BC at A and C respectively such that
∠OAD = ∠OCB [from (1)]
i.e., alternate interior angles are equal.
∴ AD ∥ BC
Similarly, AB ∥ CD
Hence, ABCD is a parallelogram.
Now, we shall show that it is a square.

In As AOB and AOD, we have :
AO = BO (Common)
∠AOB = ∠AOD (Each = 90°, given)
and, OB =OD (v Diagonals of a ∥gm bisect each other)

∴ By SAS criterion of congruence, we have :
∆AOB = ∆AOD
So, AB =AD
(Corresponding parts of congruent triangles are equal.)
But AB = CD and AD = BC
AB = BC = CD = AD

Now, in As ABD and BAC, we have :
AB = BA AD =BC
and, BD = AC

By SSS criterion of congruence, we have :
∆ABD = ∆BAC
So, ∠DAB = ∠CBA
(Corresponding parts of congruent As are equal.)
But ∠DAB + ∠CBA = 180° (Adjacent angles of a parallelogram)
Similarly, other angles ∠ADC and ∠BCD are each equal to 90°.
Hence, ABCD is a square.
Hence Proved.

Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that:

(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Answer:
(i) Given : A parallelogram ABCD in which diagonal AC bisects ∠A.

To show: That AC bisects ∠C.
Since ABCD is a ∥gm, therefore AB ∥ DC.
Now, AB ∥ DC and AC intersects them.
∠1 = ∠3 [alternate interior angles] ……………(1)

Again, AD ∥ BC and AC intersects them.
∴ ∠2 = ∠4 [alternate interior angles] …………….(2)
But it is given that AC is the bisector of ∠A.
∴ ∠1 = ∠2 ………..(3)
From (1), (2) and (3), we have :
∠3 = ∠4 Hence, AC bisects ∠C.

(ii) To show: That ABCD is a rhombus.
From part (i): (1), (2) and (3) give ∠1 = ∠2 = ∠3 = ∠4
Now in ∆ABC, ∠1 = ∠4
So, AB = BC (Sides opp. to equal angles in a A are equal.)

Similarly, in AADC, we have :
AD = DC
Also, ABCD is a ∥gm.
∴ AB = CD and AD = BC.
Combining these, we get AB = BC = CD = DA
Hence, ABCD is a rhombus.
Hence proved.

Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Answer:
Given : A rhombus ABCD.
To show that: (i) Diagonal AC bisects ∠A as well ∠C.
(ii) Diagonal BD bisects ∠B as well as ∠D.
In ∆ADC, AD = DC
(Sides of a rhombus are equal) B
So, ∠DAC = ∠DCA …(1)
(Angles opp. to equal sides of a triangle are equal)

Now AB ∥ DC and AC intersects them.
∴ ∠BCA = ∠DAC (Alternate angles) ……………(2)
From (1) and (2), we have :
∠DCA = ∠BCA ⇒ AC bisects ∠C.

In ∆ABC, AB = BC (Sides of a rhombus are equal.)
⇒ ∠BCA = ∠BAC (3)
(Angles opp. to equal sides of a triangle are equal.)

From (2) and (3), we have:
∠BAC = ∠DAC ⇒ AC bisects ∠A.
Hence, diagonal AC bisects ∠A as well as ∠C.
Similarly, it can be shown that diagonal BD bisects ∠B as well as ∠D.
Hence proved.

Question 8.
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.
Answer:
Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

To show that: (i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.
(i) Since AC bisects ∠A as well as ∠C in the rectangle ABCD, therefore
∠1 = ∠2 = ∠3 = ∠4 [∵ Each = \(\frac{90^{\circ}}{2}\) = 45°]
In ∆ADC, ∠2 = ∠4
So, AD = CD (Sides opposite to equal angles of a triangle)
Thus, the rectangle ABCD is a square.

(ii) In a square, diagonals bisect the angles.
So, 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 figure). Show that:

(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Answer:
ABCD is a parallelogram. P and Q are points on the diagonal BD such that DP = BQ.

To show :
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

Construction : Join AC to meet BD in O,
(i) AD ∥ BC and BD is a transversal (ABCD is a ∥gm)
∠ADP = ∠CBQ (Alternate angles) …………. (1)

In As APD and CQB, we have :
AD = BC (Opposite sides of a ∥gm)
DP =BQ (Given)
and ∠ADP = ∠CBQ [From (1)]
∴ ∆APD = ∆CQB (SAS)

(ii) AP=CQ [CPCT in (i)]

(iii) AB ∥DC and BD is a transversal. (ABCD is a ∥gm)
∠ABQ = ∠CDP (Alternate angles)

In As AQB and CPD, we have :
AB = DC
BQ = DP
and ∠ABQ = ∠CDP
So, ∆AQB ≅ ∆CPD (SAS)
(iv) AQ = CP [CPCT in (iii)]
(v) OA = OC (AC andBD bisect each other) ………..(1)
and OB = OD
So, OB – BQ = OD – DP (BQ = DP, given) ………………. (2)
or OQ = OP [From (1) and (2)]
Thus, AC and PQ bisect each other So, APCQ is a ∥gm.

Aliter: We know that the diagonals of a parallelogram bisect each other. Therefore, AC and BD bisect each other at O.
OB = OD
But BQ = DP (Given)
OB – BQ = OD – DP or OQ = OP
Thus, in quadrilateral APCQ diagonals AC and PQ are such that OQ = OP and OA = OC. i.e., the diagonals AC and PQ bisect each other.
Hence, APCQ is a parallelogram, which prove the (v) part.

(i) In As APD and CQB, we have :
AD = CB (Opp. sides of a ∥gm ABCD)
AP = CQ (Opp. sides of a ∥gm APCQ)
DP = BQ (Given)

∴ By SSS criterion of congruence, we have :
∆APD = ∆CQB
(ii) AP = CQ (Corresponding parts of congruent triangles)

(iii) In As AQB and CPD, we have :
AB = CD AQ = CP BQ = DP

∴ By SSS criterion of congruence, we have :
∆AQB = ∆CPD

(iv) AQ = CP (Corresponding parts of congruent triangles)

(v) In As AQC and PCA, we have AQ = CP, CQ = AP and AC = AC
∆AQC ≅ ∆PCA (By SSS Rule)
So, ∠ACQ = ∠CAD (CPCT)
i.e., QA ∥ CP
Since QA = CP and QA ∥ CP i.e., one pair of opposite sides are equal and parallel.
∴ APCQ is a parallelogram.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure). Show that

(i) ∆APB ≅ ∆CQD
(ii) AP = CQ
Answer:
(i) Since ABCD is a parallelogram, therefore DC ∥ AB.
Now, DC ∥ AB and transversal BD intersects them at B and D.
∴ ∠ABD = ∠BDC (Alternate interior angles)
Now, in As APB and CQD, we have :
∠ABP = ∠QDC (∵∠ABD = ∠BDC)
∠APB = ∠CQD (Each = 90°)
and AB = CD (Opp. sides of a ∥gm)

By AAS criterion of congruence, we have :
∆APB ≅ ∆CQD
(ii) Since ∆APB = ∆CQD, therefore
AP =CQ (∵ Corresponding parts of congruent triangles are equal)

Question 11.
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
(vi) ∆ABC ≅ ∆DEF.
Answer:
Given : Two As ABC and DEF such that AB = DE and AB || DE. Also, BC = EF and BC || EF.
To 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
(vi) ∆ABC ≅ ∆DEF.

(i) Consider the quadrilateral ABED.
We have : AB = DE and AB ∥ DE.
That is, one pair of opposite sides are equal and parallel.
So, ABED is a parallelogram.
(ii) Now, consider quadrilateral BEFC. We have :
BC = EF and BC ∥ EF
That is, one pair of opposite sides are equal and parallel.
So, BEFC is a parallelogram.

(iii) Now, AD = BE and AD ∥ BE (∵ ABED is a ∥gm) (1)
and CF = BE and CF ∥BE (∵ BEFC is a ∥gm) (2)
From (1) and (2), we have :
AD = CF and AD ∥ CF.
(iv) Since AD = CF and AD ∥ CF, therefore one pair of opposite sides are equal and parallel So, ACFD is a parallelogram.

(v) Since ACFD is a parallelogram.
AC = DF (Opp. sides of a ∥gm ACFD)

(vi) In As ABC and DEF, we have :
AB = DE (Given)
BC = EF (Given)
and, CA = FD [Proved in (v)]
∴ By SSS criterion of congruence, we have :
∆ABC = ∆DEF.

Question 12.
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.
Answer:
Given : ABCD is a trapezium in which AB ∥ CD and AD = BC.

To show that:
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC = ∆BAD
(iv) diagonal AC = diagonal BD

Construction :
Produce AB and draw a line CE ∥ DA. Also, join AC and BD.
(i) Since AD ∥ CE and transversal AB cuts them at A and E respectively, therefore
∠A + ∠E = 180° (1)
Since AB ∥ CD and AD ∥ CE, therefore AECD is a parallelogram.
So, AD = CE
∴ BC = CE [v AD = BC (given)]

Thus, in ABCE, we have : BC = CE
∴∠CBE = ∠CEB
So, 180° – ∠B = ∠E
180° – ∠E = ∠B (2)

From (1) and (2), we get ∠A = ∠B
(ii) Since ∠A = ∠B, therefore ∠BAD = ∠ABC
∴ 180° – ∠BAD = 180° – ∠ABC
So, ∠ADC = ∠BCD
or ∠D = ∠C, i.e., ∠C = ∠D.

(iii) In As ABC and BAD, we have :
BC = AD (Given)
AB = BA (Common)
∠B = ∠A (Shown above)
∴ By SAS criterion of congruence, we have :
∆ABC = ∆BAD
∆ABC = ∆BAD, therefore AC =BD
(Corresponding parts of congruent triangles are equal.)