Class 9

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that:
(i) SR ∥ AC and SR = \(\frac{1}{2}\)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Answer:
Given: A quadrilateral ABCD in which P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA. Also AC is its diagonal.
To show :
(i) SR ∥ AC and SR = \(\frac{1}{2}\)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

(i) In ∆ACD, we have :
S is the mid-point of AD and R is the mid-point of CD.
Therefore, SR ∥ AC and SR = \(\frac{1}{2}\)AC (Mid-point theorem)

(ii) In ∆ABC, we have :
P is the mid-point of the side AB and Q is the mid-point of the side BC.
Therefore, PQ ∥ AC and PQ = \(\frac{1}{2}\)AC (Mid-point theorem)

Thus, we have shown that:

(iii) Since PQ = SR and PQ ∥ SR, therefore one pair-of opposite sides are equal and parallel.
∴ PQRS is a parallelogram.

Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Answer:
Given : ABCD is a rhombus in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.

To show: PQRS is a rectangle.

Construction: Join AC.
In ∆ABC, P and Q are respectively the mid-points of AS and BC.
PQ ∥ AC and PQ = \(\frac{1}{2}\)AC

Similarly, in ∆ADC, R and S are respectively the mid-points of CD and AD.
SR ∥ AC and SR = \(\frac{1}{2}\)AC ……………(2)
From (1) and (2), we get PQ ∥ RS and PQ = SR

Now, in quad. PQRS, its one pair of opposite sides PQ and SR is equal and parallel.
∴ PQRS is a parallelogram.
Also, AS = BC (Sides of a rhombus)

Now, SP ∥ RQ and PQ intersects them.
∴ ∠SPQ + ∠PQR = 180°
From (3) and (4), we get
∠SPQ = ∠PQR = 90°
Thus, PQRS is a parallelogram whose one angle ∠SPQ = 90°.
Hence, PQRS is a rectangle.

Question 3.
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus
Answer:
Given : ABCD is a rectangle in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quad. PQRS.

To show : PQRS is a rhombus.
Construction: Join AC.
In ∆ABC, P and Q are respectively the mid-points of sides AB and BC.
PQ ∥ AC and PQ = \(\frac{1}{2}\)AC ………..(1)

Similarly, in AADC, R and S are respectively the mid-points of sides CD and AD.
∴ SR ∥ AC and SR = \(\frac{1}{2}\)AC …………….(2)
From (1) and (2), we get PQ ∥ SR and PQ = SR …………….. (3)

Now, in quad. PQRS, its one pair of opposite sides PQ and SR is parallel and equal. [From (3)]
∴ PQRS is a parallelogram.
Now, AD = BC (Opp. sides of rect. ABCD) ….(4)
AP = BP ⇒ AS = BQ (∵ P is the mid-point of AB)
∠PAS = ∠PBQ (Each = 90°)
AS = BQ (Shown above)
∴ ∆APS = ∆BPQ (SAS Cong, axiom)
PS = PQ (CPCT) ………….(5)
From (4) and (5), we get PQRS is a rhombus.

Question 4.
ABCD is a trapezium in which AB ∥ DC, BD is a diagonal and E is the mid-point 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.

Answer:
Given : In trapezium ABCD, AB ∥ DC,
E is the mid-point of AD and EF ∥ AB.

To show: F is the mid-point of BC.
Construction: Join DB. Let it intersect EF in G.
In ADAB, E is the mid-point of AD (Given) and EG ∥ AB (∵ EF ∥AB)
∴ By converse of mid-point theorem, G is the midpoint of DB.
In ABCD, G is the mid-point of BD and GF ∥ DC (AB ∥ DC, EF ∥ AB ⇒ DC ∥ EF)

∴ By converse of mid-point theorem,
F is the mid-point of BC.

Question 5.
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.

Answer:
Given : E and F are the mid-points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
To-show : BQ = QP = PD.
ABCD is parallelogram. (Given)
∴ AB ∥ DC and AB = DC (Opp. sides of parallelogram)
E is the mid-point of AB. (Given)
AE = \(\frac{1}{2}\)AB
F is the mid-point of CD.
CF = \(\frac{1}{2}\)CD (∵ CD = AB)
So, AE = CF
From (1) and (2) AE ∥ CF (∵ AB ∥ DC)
Thus, a pair of opposite sides of a quadrilateral AECF are parallel and equal.
Quad. AECF is a parallelogram.
So, EC ∥ AF
∴ EQ ∥ AP and QC ∥ PF

In ABPA, E is the mid-point of BA (Given)
and EQ ∥ AP (Shown)
BQ = QP (Converse of mid-point theorem) ……….. (3)
Similarly by taking ACQD, we can prove that
DP = QP …………… (4)
From (3) and (4), we get
BQ = QP = PD
Hence, AF and CE trisect the diagonal BD.

Question 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer:
Given : A quad. ABCD, P, Q, R and S are respectively the mid-points of AB, BC, CD and DA. PR and QS intersect each other at O.
To show: OP = OR, OQ = OS.

Construction: Join PQ, QR, RS, SP, AC and BD
In ∆ABC, P and Q are mid-points of AB and BC respectively
∴ PQ ∥ AC and PQ = \(\frac{1}{2}\)AC
Similarly, we can show that
RS ∥ AC and RS = \(\frac{1}{2}\)AC
∴ PQ ∥SR and PQ = SR
Thus, a pair of opposite sides of a quadrilateral PQRS are parallel and equal.
∴ Quadrilateral PQRS is a parallelogram.
Since, the diagonals of a parallelogram bisect each other, therefore diagonals PR and QS of ∥gm PQRS, i.e., the line segments joining the mid-points of opposite sides of quadrilateral ABCD bisect each other.

Question 7.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac{1}{2}\)AB.
Answer:
Given : ∆ABC is right angled at C, M is the mid-point of hypotenuse AB. Also, MD ∥ BC.

To show that:
(i) D is the mid-point of AC.
(ii) MD ⊥ AC.
(iii) CM = MA = \(\frac{1}{2}\)AB.

(i) In ∆ABC, M is the mid-point of AB and MD ∥ BC. Therefore, D is the mid-point of AC.
i.e., AD = DC

(ii) Since MD ∥ BC, therefore
∠ADM = ∠ACB (Corresponding angles)
So, ∠ADM = 90° [v ∠ACB = 90° (given)]
But, ∠ADM + ∠CDM = 180°
[ v ∠ADM and ∠CDM are angles of a linear pair]
∴ 90° + ∠CDM = 180° or ∠CDM =90°
Thus, ∠ADM = ∠CDM =90° ……………….. (2)
So, MD ⊥ AC

(iii) In As AMD and CMD, we have :
AD = CD [From (1)]
∠ADM = ∠CDM [From (2)]
and MD = MD (common)
∴ By SAS criterion of congurence
∆AMD ≅ ∆CMD
So, MA = MC
(v Corresponding parts of congruent triangles are equal)
Also, MA = \(\frac{1}{2}\)AB, since M is the mid-point of AC.
Hence, CM = MA = \(\frac{1}{2}\)AB.

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.)

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A
Answer:

(i) In ∆ABC, we have: AB = AC
∴ ∠B = ∠C
(∵ Angles opposite to equal sides are equal.)
or \(\frac{1}{2}\)∠B = \(\frac{1}{2}\)∠C
i.e. ∠OBC = ∠OCB ……….. (1)
(∵ OB and OC bisect ∠s B and C respectively.
∴ ∠OBC = \(\frac{1}{2}\)∠B and ∠OCB = \(\frac{1}{2}\)∠C)
⇒ OB = OC …….. (2) (∵ Sides opp. to equal ∠s are equal.)

(ii) Now, in ∆s ABO and ACO, we have:

∴ By SAS criterion of congruence, we have:
∆ABO ≅ ∆ACO
∠BAO = ∠CAO (CPCT)
⇒ AO bisects ∠BAC.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ABC is an isosceles triangle in which AB = AC.
Answer:

In ∆s ABD and ACD, we have:
DB = DC (Given)
∠ADB = ∠ADC (∵ AD ⊥ BC, so each = 90°)
AD = AD (Common)
∴ By SAS criterion of congruence, we have:
∆ABD ≅ ∆ACD
So, AB = AC
(∵ Corresponding parts of congruent triangles are equal.)
Hence, ∆ABC is an isosceles.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Answer:
Let BE ⊥ AC and CF ⊥ AB.
In ∆s ABE and ACF, we have :
∠AEB = ∠AFC (∵ Each = 90°)
∠A = ∠A (Common)
and, AB = AC (Given)
∴ By AAS criterion of congruence, we have:
∆ABE ≅ ∆ACF
So, BE = CF
(∵ Corresponding parts of congruent triangles are equal.)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:

(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e. ABC is an isosceles triangle.
Answer:
(i) In ∆s ABE and ACF, we have:
∠AEB = ∠AFC (∵ Each = 90°)
∠BAE = ∠CAF (Common)
and, BE = CF (Given)
∴ By AAS criterion of congruence, we have:
∆ABE ≅ ∆ACF
So, AB = AC
(∵ Corresponding parts of congruent triangles are equal.)
Hence, ∆ABC is isosceles.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Answer:
In ∆ABC, we have: AB = AC
∠ABC = ∠ACB ……… (1)
(∵ Angles opposite to equal sides are equal.)
In ADBC, we have : BD = CD
∠DBC = ∠DCB …………… (2)
(∵ Angles opposite to equal sides are equal.)
Adding (1) and (2), we have :
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.
Answer:

In ∆ABC, we have : AB = AC
∴ ∠ACB = ∠ABC …………… (1)
(∵ Angles opp. to equal sides are equal.)
Now, AB = AD (Given)
∴ AD = AC (∵ AD = AC)
∠ACD = ∠ADC ………. (2)
(∵ Angles opp. to equal sides are equal.)
Adding (1) and (2), we get:
∠ACB + ∠ACD = ∠ABC + ∠ADC
or ∠BCD = ∠ABC + ∠BDC (∵ ∠ADC = ∠BDQ)
So, ∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD
(Adding ∠BCD on both sides)
2 ∠BCD = 180° (Angle sum property)
or ∠BCD = 90°
Hence, ∠BCD is a right angle.
Hence proved.

Question 7.
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C
Answer:
Given, AB = AC and ∠A = 90°
∴ ∠C = ∠B (Angles opposite to equal sides of a triangle)
In ∆ABC, ∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ 90° + ∠B + ∠B = 180°
⇒ 2 ∠B = 180° – 90°
⇒ 2 ∠B = 90°
⇒ ∠B = 45°
∴ ∠B = ∠C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:
Let us take ∆ABC is an equilateral triangle.
Then, AB = BC = AC
Now, AB = AC ⇒ ∠C = ∠B (Angle opposite to equal sides of a triangle)
Again, AC = BC ⇒ ∠B = ∠A (Angle opposite to equal sides of a triangle)
So we have ∠A = ∠B = ∠C
Now in AABC ∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ ∠A + ∠A + ∠A = 180°
⇒ 3 ∠A = 180°
⇒ ∠A = 60°
∴ ∠A = ∠B = ∠C = 60°
Hence, all interior angles in an equilateral triangle are of 60°.
Hence proved.

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.1

Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?
Answer:

Now, in ∆s ABC and ABD, we have :
AC = AD (Given)
∠CAB = ∠BAD (∵ AB bisects ∠∆)
and, AB = AB (Common)
∴ By SAS congruence criterion, we have:
∆ABC ≅ ∆ABD
⇒ BC = BD, i.e. they are equal.
(∵ Corresponding parts of congruent triangles are equal.)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that:
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Answer:

In ∆s ABD and BAC, we have :
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = BA (Common)
∴ By SAS criterion of congruence, we have
∆ABD ≅ ∆BAC, which proves (i)
From (i), (ii) BD = AC (C.P.C.T.)
and, (iii) ∠ABD = ∠BAC
(∵ Corresponding parts of congruent triangles (CPCT) are equal.)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Answer:

Since AB and CD intersect at O, therefore
∠AOD = ∠BOC ….(1)
(Vertically opp. angles)
In ∆s AOD and BOC, we have:
∠AOD = ∠BOC [From (1)]
∠DAO = ∠CBO (Each = 90°)
and, AD = BC (Given)
∴By AAS congruence criterion, we have :
∆AOD ≅ ∆BOC
⇒ OA = OB
(∵ Corresponding parts of congruent triangles (CPCT) are equal.)
i.e. O is the mid-point AB.
Hence, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure).
Show that ∆ABC ≅ ∆CDA.
Answer:

Since l and m are two parallel lines intersected by another pair of parallel lines p and q, therefore AD || BC and AB || CD.
AB || CD and AC is a transversal.
∠BAC = ∠DCA (Alternate angles) ……….. (1)
Again, BC || AD and AC is a transversal.
∠BCA = ∠DAC (Alternate angles) ……….. (2)
Now, in ∆s ABC and CDA, we have :
∠BAC = ∠DCA [Proved above in (1)]
∠BCA = ∠DAC [Proved above in (2)]
and AC = CA
∴ By ASA criterion of congruence,
∆ABC ≅ ∆CDA.

Question 5.
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B. to the arms of ∠A (see figure). Show that:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Answer:

In ∆s APB and AQB, we have:
∠APB = ∠AQB (∵ Each = 90°)
∠PAB = ∠QAB (∵ AB bisects ∠PAQ)
AB = AB (Common)
By AAS congruence criterion, we have:
∆APB ≅ ∆AQB, which proves (i)
⇒ BP = PQ
(∵ Corresponding parts of congruent triangles are equal.)
i.e. B is equidistant from the arms of ∠A, which proves (ii).

Question 6.
In the figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Answer:
In ∆s ABC and ADE, we have :
AB = AD (Given)
∠BAC = ∠DAE
∵ ∠BAD = ∠EAC => ∠BAD + ∠DAC =
∠EAC + ∠DAC => ∠BAC = ∠DAE
and, AC = AE (Given)
∴ By SAS criterion of congruence, we have :
∆ABC ≅ ∆ADE
⇒ BC = DE
(∵ Corresponding parts of congruent triangles are equal.)

Question 7.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD – ∠ABE and ∠EPA = ∠DPB (see figure). Show that:
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Answer:

We have: ∠EPA = ∠DPB
∴ ∠EPA + ∠DPE = ∠DPB + ∠DPE
or ∠DPA = ∠EPB
Now, in ∆s DAP and EBP, we have: …………. (1)
∠DPA = ∠EPB [From (1)]
AP = BP (Given)
and, ∠DAP = ∠EBP (Given)
So, by ASA criterion of congruence, we have:
∆DAP ≅ ∆EBP, which proves (i)
So, AD = BE, which proves (ii)
(∵ Corresponding parts of congruent triangles are equal.)

Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM – CM. Point D is joined to point B (see figure). Show that:
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle.
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac{1}{2}\)AB
Answer:

Given : ∠C = 90°, AM = BM, DM = CM
To prove:
(i) ∆AMC = ∆BMD
(ii) ∆DBC = 90°
(iii) ∆DBC = AACB
(iv) CM = \(\frac{1}{2}\)AB
Proof:
(i) In ∆AMC and ∆BMD
AM = BM
MC = MD
∠1 = ∠2
∴ ∆AMC ≅ ∆BMD (by SAS)
and so AC = DB and ∠4 = ∠3 (By CPCT).

(ii) ∠4 = ∠3 (Proved above)
∠4 = ∠3 But these are alternate interior angles.
∴ DB || AC
AS AC || BD and BC is the transversal.
∠C + ∠B = 180° (Cointerior angles)
⇒ ∠B = 180° – 90° = 90° (∵ ∠C = 90°)

(iii) In ∆DBC and ∆ACB
DB = AC (Proved)
∠B = ∠C (Each 90°)
BC = CB (Common)
∴ ∆DBC ≅ ∆ACB (By SAS)
and so DC =AB (By CPCT) (Proved)

(iv) DC = AB
\(\frac{1}{2}\)DC = \(\frac{1}{2}\)AB (Proved)
CM = \(\frac{1}{2}\)AB.

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 1.
In the figure, find the values of x and y and then show that AB ∥ CD.

Answer:
x and 50° form a linear pair,
∴ x + 50° = 180°
⇒ x = 180° – 50° = 130°………………. (1)
Also, y = 130° (Vertically opposite angles) ……. (2)
∴x = y [From (1) and (2)]
But x and y are alternate angles.
Therefore, AB ∥CD.
Hence Proved.

Question 2.
In the figure, if AB ∥ CD, CD ∥ EF and y : z = 3:7, find x.

Answer:
Since CD ∥ EF and transversal PQ intersects them at S and T respectively, therefore
∠CST = ∠STF . (Alternate angles)
⇒ 180° – y = z
(∵ ∠y + ∠CST = 180° being linear pair)
⇒ y + z = 180

Given y: z = 3:7.
Let y = 3a, z = 7a
3a + 7a = 180°
10a = 180° A B
⇒ a = 18°
y = 3 × 18 = 54°,
z = 7 × 18= 126°
y = 54°, z = 126°

Since AD ∥ CD and transversal PQ intersects them at R and S respectively, therefore
∠ARS + ∠RSC = 180° (Interior angles on the same side of the transversal are supplementary.)
or x + y = 180°
⇒ x = 180° – y = 180° – 54° = 126° (∵ y = 54°)
Hence, x = 126°.

Question 3.
In the figure, if AB ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer:
Here, AB ∥ CD and transversal GE cuts them at G and E respectively.
∠AGE ~ ∠GED (Alternate angles)
⇒ ∠AGE = 126° [∵ ∠GED = 126° (given)]
∠GEF = ∠GED – ∠FED = 126° – 90° = 36° C
and, ∠FGE = ∠GEC (Alternate angles)
⇒ ∠FGE – 90° – ∠GEF
= 90°- 36° = 54°
Hence, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°.

Question 4.
In the figure, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
(Hint: Draw a line parallel to ST through point R.)

Answer:
Produce PQ to intersect ST in a point M. Now, PM ∥ ST and transversal SM intersects them at M and S respectively.
∠SMQ = ∠TSM (Alternate angles)
⇒ ∠SMQ = 130°
⇒ ∠QMR = 180° – 130° = 50° (∵ ∠SMQ + ∠QMR = 180°, linear pair)

Now, ray QR stands at Q on PM.
∠PQR + ∠RQM = 180°
⇒ 110° + ∠RQM = 180°
⇒ ∠RQM = 70°
∠QRS = 180° – (70° + 50°) = 60° (∵ Sum of the angles of a triangle is 180°)

Question 5.
In the figure, if AB ∥ CD, ∠APQ = 50° ∠PRD = 127°, find x and y.

Answer:
Here, AB ∥ CD and transversal PQ intersects them at P and Q respectively.
∴ ∠PQR = ∠APQ (Alternate angles)
⇒ x = 50° [∵ ∠APQ = 50° (given)]

Also, AB ∥ CD and transversal PR intersects them at P and R respectively.
∴ ∠APR = ∠PRD (Alternate angles)
⇒ ∠APQ + ∠QPR = 127° (∵ ∠PRD = 127°)
⇒ 50° + y = 127° (∵∠APQ = 50°)
⇒ y = 127° – 50° = 77°
Hence, x = 50° and y = 77°.

Question 6.
In the figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB ∥ CD.

Answer:
Two plane mirrors PQ andBS are placed parallel to each other, i.e. PQ paths BC and CD.
Let BN and CM are the normals to the plane mirrors PQ and RS respectively. Since BN ⊥ PQ, CM ⊥ RS and PQ ∥ RS, therefore BN ⊥ RS ⇒ BN ∥ CM
Thus, BN and CM are two parallel lines and a transversal BC cuts them at B and C respectively.
∴ ∠2 = ∠3 (Alternate interior angles)

But, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)
∠1 + ∠2 – ∠2 + ∠2
and ∠3 + ∠4 – ∠3 + ∠3
⇒ ∠1 + ∠2 = 2(∠2)
and ∠3 + ∠4 = 2(∠3) ‘
⇒ ∠1 + ∠2 = ∠3 + ∠4
∠ABC = ∠BCD
Thus, lines AB and CD are intersected by transversal BC, such that ∠ABC = ∠BCD.
i.e., alternate interior angles are equal.
Therefore, AB ∥ CD.