Question-1 :- The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution :-
Let the common ratio between the angles be x.
Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360°,
3x + 5x + 9x + 13x = 360°
30x = 360°
x = 12°
Hence, the angles are
3x = 3 × 12 = 36°
5x = 5 × 12 = 60°
9x = 9 × 12 = 108°
13x = 13 × 12 = 156°
Question-2 :- If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution :-
Given that : Let ABCD be a parallelogram.
Prove that : ABCD is a rectangle or one of its interior angles is 90°.
Proof : In ∆ABC and ∆DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC ( Common)
AC = DB (Given)
∆ABC ≅ ∆DCB (By SSS Congruence rule)
By CPCT Rule,
∠ABC = ∠DCB
It is known that the sum of the measures of angles on the same side of transversal is 180°.
∠ABC + ∠DCB = 180° (AB || CD)
∠ABC + ∠ABC = 180°
2 ∠ABC = 180°
∠ABC = 90°
Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.
Question-3 :- : Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution :-Given that : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°. Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal. Proof : In ∆AOD and ∆COD, OA = OC (Diagonals bisect each other) ∠AOD = ∠COD (Given) OD = OD (Common) ∆AOD ≅ ∆COD (By SAS congruence rule) By CPCT Rule, AD = CD .....(1) Similarly, it can be proved that AD = AB and CD = BC .....(2) From equations (1) and (2), AB = BC = CD = AD Since opposite sides of quadrilateral ABCD are equal, It can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
Question-4 :- Show that the diagonals of a square are equal and bisect each other at right angles.
Solution :-
Given that : Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
Prove that : The diagonals of a square are equal and bisect each other at right angles
or AC = BD, OA = OC, OB = OD, and ∠AOB = 90°.
Proof : In ∆ABC and ∆DCB,
AB = DC (Sides of a square are equal to each other)
∠ABC = ∠DCB (All interior angles are of 90 )
BC = CB (Common side)
∆ABC ≅ ∆DCB (By SAS congruency)
AC = DB (By CPCT)
Hence, the diagonals of a square are equal in length.
In ∆AOB and ∆COD,
∠AOB = ∠COD (Vertically opposite angles)
∠ABO = ∠CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
∆AOB ≅ ∆COD (By AAS congruence rule)
AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.
In ∆AOB and ∆COB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
∆AOB ≅ ∆COB (By SSS congruency)
∠AOB = ∠COB (By CPCT)
However, AOB + COB = 180° (Linear pair)
2 ∠AOB = 180°
∠AOB = 90°
Hence, the diagonals of a square bisect each other at right angles.
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 :-
Given that : Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.
The diagonals of ABCD are equal and bisect each other at right angles.
Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = = 90°.
Prove that : ABCD is a square or ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90°.
Proof : In ∆AOB and ∆COD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite angles)
∆AOB ≅ ∆COD (SAS congruence rule)
By CPCT Rule,
AB = CD ... (1)
And, ∠OAB = ∠OCD
However, these are alternate interior angles for line AB and CD and
alternate interior angles are equal to each other only when the two lines are parallel.
AB || CD ... (2)
From equations (1) and (2), we obtain ABCD is a parallelogram.
In ∆AOD and ∆COD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Given that each is 90°)
OD = OD (Common)
∆AOD ≅ ∆COD (SAS congruence rule)
By CPCT Rule,
AD = DC ... (3)
However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)
AB = BC = CD = DA
Therefore, all the sides of quadrilateral ABCD are equal to each other.
In ∆ADC and ∆BCD,
AD = BC (Already proved)
AC = BD (Given)
DC = CD (Common)
∆ADC ≅ ∆BCD (SSS Congruence rule)
By CPCT Rule,
∠ADC = ∠BCD
However, ∠ADC + ∠BCD = 180° (Co-interior angles)
∠ADC + ∠ADC = 180°
2 ∠ADC = 180°
∠ADC = 90°
One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and
one of its interior angles is 90°.
Therefore, ABCD is a square.
Question-6 :- : Diagonal AC of a parallelogram ABCD bisects ∠ A. Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.
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.
Solution :-
Given that : ABCD is a rhombus.
Prove that : Diagonal AC bisects ∠ A as well as ∠ C and
Diagonal BD bisects ∠ B as well as ∠ D.
Proof : In ∆ABC,
BC = AB (Sides of a rhombus are equal to each other)
∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)
However, ∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)
∠2 = ∠3
Therefore, AC bisects C.
Also, ∠2 = ∠4 (Alternate interior angles for || lines BC and DA)
∠1 = ∠4
Therefore, AC bisects A.
Similarly, it can be proved that BD bisects B and D as well.
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.
Given that : ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C.
Prove that : (i) ABCD is a square
(ii) diagonal BD bisects ∠ B as well as ∠ D.
Proof : (i) It is given that ABCD is a rectangle.
∠A = ∠C
CD = DA (Sides opposite to equal angles are also equal)
However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)
AB = BC = CD = DA
ABCD is a rectangle and all of its sides are equal.
Hence, ABCD is a square.
(ii) Let us join BD.
In ∆BCD,
BC = CD (Sides of a square are equal to each other)
∠CDB = ∠CBD (Angles opposite to equal sides are equal)
However, ∠CDB = ∠ABD (Alternate interior angles for AB || CD)
∠CBD = ∠ABD
BD bisects B.
Also, ∠CBD = ∠ADB (Alternate interior angles for BC || AD)
∠CDB = ∠ABD
BD bisects D.
Question-9 :- : In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Given that : ABCD is a parallelogram and AC bisects A.
Prove that : (i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Proof : (i) In ∆APD and ∆CQB,
∠ADP = ∠CBQ (Alternate interior angles for BC || AD)
AD = CB (Opposite sides of parallelogram ABCD)
DP = BQ (Given)
∆APD ≅ ∆CQB (Using SAS congruence rule)
(ii) As we had observed that ∆APD ≅ ∆CQB,
AP = CQ (CPCT)
(iii) In ∆AQB and ∆CPD,
∠ABQ = ∠CDP (Alternate interior angles for AB || CD)
AB = CD (Opposite sides of parallelogram ABCD)
BQ = DP (Given)
∆AQB ≅ ∆CPD (Using SAS congruence rule)
(iv) As we had observed that ∆AQB ≅ ∆CPD,
AQ = CP (CPCT)
(v) From the result obtained in (ii) and (iv),
AQ = CP and AP = CQ Since
opposite sides in quadrilateral APCQ are equal
to each other, APCQ is a parallelogram.
Question-10 :- ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ
Given that : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
Prove that : (i) Δ APB ≅ Δ CQD
(ii) AP = CQ
Proof : (i) In ∆APB and ∆CQD,
∠APB = ∠CQD (Each 90°)
AB = CD (Opposite sides of parallelogram ABCD)
∠ABP = ∠CDQ (Alternate interior angles for AB || CD)
∆APB ≅ ∆CQD (By AAS congruency)
(ii) By using the above result
∆APB ≅ ∆CQD, we obtain
AP = CQ (By CPCT)
Question-11 :- In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively . 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) Δ ABC ≅ Δ DEF.
Given that : In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF.
Prove 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) Δ ABC ≅ Δ DEF.
Proof : (i) It is given that AB = DE and AB || DE.
If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.
Therefore, quadrilateral ABED is a parallelogram.
(ii) Again, BC = EF and BC || EF
Therefore, quadrilateral BCEF is a parallelogram.
(iii) As we had observed that ABED and BEFC are parallelograms, therefore
AD = BE and AD || BE (Opposite sides of a parallelogram are equal and parallel)
And, BE = CF and BE || CF (Opposite sides of a parallelogram are equal and parallel)
AD = CF and AD || CF
(iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD
are equal and parallel to each other, therefore, it is a parallelogram.
(v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal
and parallel to each other.
AC || DF and AC = DF
(vi) ∆ABC and ∆DEF, AB = DE (Given)
BC = EF (Given)
AC = DF (ACFD is a parallelogram)
∆ABC ≅ ∆DEF (By SSS congruence rule)
Question-12 :- : ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) Δ ABC ≅ Δ BAD
(iv) diagonal AC = diagonal BD
Given that : ABCD is a trapezium in which AB || CD and AD = BC .
Prove that : (i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) Δ ABC ≅ Δ BAD
(iv) diagonal AC = diagonal BD
Construction : Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E.
Proof : (i) AD = CE (Opposite sides of parallelogram AECD)
However, AD = BC (Given)
Therefore, BC = CE
∠CEB = ∠CBE (Angle opposite to equal sides are also equal)
Consider parallel lines AD and CE. AE is the transversal line for them.
∠A + ∠CEB = 180º (Angles on the same side of transversal)
∠A + ∠CBE = 180º (Using the relation ∠CEB = ∠CBE) ... (1)
However, B + ∠CBE = 180º (Linear pair angles) ... (2)
From equations (1) and (2), we obtain
∠A = ∠B
(ii) AB || CD
∠A + ∠D = 180º (Angles on the same side of the transversal)
Also, ∠C + ∠B = 180° (Angles on the same side of the transversal)
∠A + ∠D = ∠C + ∠B
However, ∠A = ∠B [Using the result obtained in (i)]
∠C = ∠D
(iii) In ∆ABC and ∆BAD,
AB = BA (Common side)
BC = AD (Given)
∠B = ∠A (Proved before)
∆ABC ≅ ∆BAD (SAS congruence rule)
(iv) We had observed that, ∆ABC ≅ ∆BAD
AC = BD (By CPCT)
