Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 8 Quadrilaterals Ex 8.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1

Question 1.

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

Answer:

Let, ABCD be a given quadrilateral.

∴ ∠A : ∠B : ∠C : ∠D = 3 : 5 : 9 : 13

Sum of ratios = 3 + 5 + 9 + 13 = 30

In quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360°

∴ ∠A = \(\frac{3}{30}\) × 360° = 3 × 12 = 36°

∴ ∠B = \(\frac{5}{30}\) × 360° = 5 × 12 = 60°

∴ ∠C = \(\frac{9}{30}\) × 360° = 9 × 12 = 108°

∴ ∠D = \(\frac{13}{30}\) × 360° = 13 × 12 = 156°

Thus the angles of the given quadrilateral are 36°, 60°, 108° and 156°.

Question 2.

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

Answer:

In parallelogram ABCD, diagonals are equal.

∴ AC = BD.

In ∆ DAB and ∆ CBA.

DA = CB (Theorem 8.2)

AB = BA (Common)

DB = CA (Given)

∴ ∆ DAB ≅ ∆ CBA sss rule)

∴ ∠ DAB = ∠CBA (CPCT)

In parallelogram ABCD, AD || BC and AB is their transversal.

∴ ∠ DAB + ∠ CBA = 180°

(Interior angles on the same side of transversal)

Thus, in parallelogram ABCD, two angles ∠A and∠B are right angles. Hence, all the angles are right angle.

Hence, the parallelogram ABCD having equal diagonals is a rectangle.

Question 3.

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

Answer:

In quadrilater ABCD. diagonals AC and BD bisect each other at M at right angles.

∴ AM = CM, BM = DM and

∠AMB = ∠CMB = ∠CMD = ∠AMD = 90°.

In ∆ AMB and ∆ CMB,

AM = CM

∠ AMB = ∠CMB

BM = BM (Common)

∴ ∆ AMB ≅ ∆ CMB (SAS rule)

∴ AB = CB (CPCT)

Similarly, proving ∆ BMC ≅ ∆ DMC and ∆ DMA ≅ ∆ BMA, we get BC = DC and DA = BA.

Thus, in quadrilateral. ABCD.

AB = BC CD = DA.

Therefore, quadrilateral ABCD is a rhombus.

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

Question 4.

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

Answer:

ABCD is a square in which diagonals AC and BD intersect at M.

Every square is a parallelogram.

∴ AC and BD bisect each other. …………… (1)

In ∆ DAB and ∆ CBA,

DA = CB (Sides of a square)

∠ DAB = ∠ CBA (Right angles in a square)

AB = BA (Common)

∴ ∆ DAB ≅ ∆ CBA (SAS rule)

∴ BD = AC (CPCT) ……………….. (2)

Now, in ∆ AMB and ∆ CMB,

AM = CM (BD bisects AC at M).

BM = BM (Common)

AB = CB (Sides of a square)

∴ ∆ AMB ≅ ∆ CMB (SSS rule)

∴ ∠ AMB = ∠CMB (CPCT)

But, ∠ AMB and ∠ CMB form a linear pair.

∴ ∠ AMB + ∠ CMB = 180°

Hence, ∠AMB = ∠ CMB = 90° (3)

(1), (2) and (3) taken together proves that the diagonals of a square are equal and 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.

Answer:

In quadrilateral ABCD, diagonals AC and BD are equal and bisect each other at right angles.

∴ AC = BD,

MA = MC = MB = MD = \(\frac{1}{2}\)AC = \(\frac{1}{2}\)BD and

∠AMB = ∠CMB= ∠DMC = ∠DMA= 90°.

In ∆ AMB and ∆ CMB,

AM = CM

∠ AMB = ∠ CMB (Right angles)

BM = BM (Common)

∴ ∆ AMB ≅ ∆ CMB (SAS rule)

∴ AB = CB (CPCT)

Similarly, we can prove that BC = DC and

DA = BA.

Thus, in quadrilateral ABCD,

AB = BC = CD = DA …………… (1)

Now, in ∆ DAB and ∆ CBA,

DA = C B

BD = AC (Given)

AB = BA (Common)

∴ ∆ DAB ≅ ∆ CBA (SSS rule)

∴ ∠DAB = ∠CBA (CPCT)

Thus, in quadrilateral ABCD, ∠A = ∠B.

Similarly, we can prove that ∠B = ∠C and ∠C = ∠D.

Thus, in quadrilateral ABCD,

∠A = ∠B = ∠C = ∠D.

Moreover. In quadrilateral ABCD,

∠A + ∠B + ∠C + ∠D = 360°

∴ ∠A = ∠B = ∠C = ∠D = \(\frac{360^{\circ}}{4}\) = 90° ……………… (2)

Thus, (1) and (2) taken together proves that in quadrilateral ABCD, all the sides are equal and all the angles are equal.

Therefore, quadrilateral ABCD is a square.

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

Question 6.

Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.

Answer:

Diagonal AC of parallelogram ABCD bisects ∠A.

∴ ∠DAC = ∠BAC …………… (1)

Now, ∠BAC and ∠DCA are alternate angles formed by transversal AC of AB || CD.

∴ ∠BAC = ∠DCA …………… (2)

Similarly, ∠DAC and ∠BCA are alternate angles formed by transversal AC of AD || BC.

∴ ∠DAC = ∠BCA ……………… (3)

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

∠DCA = ∠BCA.

But, ∠DCA + ∠BCA = ∠BCD (Adjacent angles)

∴ AC bisects ∠C also.

In parallelogram ABCD,

∠A = ∠C (Theorem 8.4)

∴ \(\frac{1}{2}\)∠A = \(\frac{1}{2}\)∠C

∴ ∠ DAC = ∠ DCA

∴ In ∆ DAC, DA = DC (Sides opposite to equal angles)

Moreover, in parallelogram ABCD,

AB = CD and BC = DA (Theorem 8.2)

∴ AB = BC = CD = DA

Thus. In parallelogram ABCD, all the sides are equal.

Hence, 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.

Answer:

ABCD is a rhombus

∴ AB || DC, BC || AD and AB = BC = CD = DA.

AB || DC and AC is their transversal.

∴ ∠CAB = ∠ACD (Alternate angles)

In, ∆ DAC, CD = DA

∴ ∠ACD = ∠CAD

Then, ∠CAB = ∠CAD

But, ∠CAB + ∠CAD = ∠ DAB (Adjacent angles)

∴ ∠ CAB = ∠CAD = \(\frac{1}{2}\) ∠DAB

This shows that AC bisects ∠A.

Again, BC || AD and AC is their transversal.

∴ ∠ BCA = ∠ DAC (Alternate angles)

In, ∆ DAC, DA = DC

∴ ∠ DAC = ∠ DCA

Then, ∠BCA = ∠DCA

But, ∠ BCA + ∠ DCA = ∠ DCB (Adjacent angles)

∴ ∠ BCA = ∠ DCA = \(\frac{1}{2}\)∠ DCB

This shows that AC bisects ∠C.

Thus, AC bisects ∠A as well as ∠C.

Similarly, taking BD as transversal of AB || DC, and BC || AD, it can be proved that BD bisects ∠B as well as ∠D.

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:

In rectangle ABCD, AB = CD, BC = AD, AB || CD and BC || AD.

AC bisects ∠A as well as ∠C.

∴ ∠DAC = ∠BAC = \(\frac{1}{2}\)∠A and

∠ DCA = ∠ BCA = \(\frac{1}{2}\)∠C

Now, AB || CD and AC is their transversal.

∴ ∠ BAC = ∠ DCA (Alternate angles)

∴ ∠ DAC = ∠ DCA

Thus, in ∆ DAC, ∠DAC = ∠DCA

∴ AD = CD (Sides opposite to equal angles)

From this, we get AB = BC = CD = DA.

Also, in rectangle ABCD,

∠A = ∠B = ∠C = ∠D = 90°

Hence, ABCD is a square. …..Result (i)

In ∆ BCD, BC = CD

∴ ∠ CBD = ∠ CDB

Moreover, AB || CD and BD is their transversal.

∴ ∠ CDB = ∠ ABD (Alternate angles)

∴ ∠ CBD = ∠ ABD

Now, ∠ CBD + ∠ ABD = ∠ ABC

∴ ∠ CBD = ∠ ABD = \(\frac{1}{2}\) ∠ ABC

Thus, BD bisects ∠B.

Similarly, diagonal BD bisects ∠ D.

Hence, diagonal BD bisects ∠B as well as ∠D …….. Result (ii)

Question 9.

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

∴ AD || BC and BD is their transversal.

∴ ∠ADB = ∠CBD (Alternate angles)

∴ ∠ADP = ∠CBQ …………… (1)

Similarly, CD || BA and BD is their transversal.

∴ ∠ ABD = ∠ CDB (Alternate angles)

∴ ∠ABQ = ∠CDP ……………… (2)

In ∆ APD and ∆ CQB,

AD = CB (Opposite sides of a parallelogram)

∠ ADP = ∠ CBQ [by (1)]

DP = BQ (Given)

∴ ∆ APD ≅ ∆ CQB (SAS rule) ……. Result (i)

∴ AP = CQ (CPCT) …… Result (ii)

In ∆ AQB and ∆ CPD,

AB = CD (Opposite sides of a parallelogram)

∠ ABQ = ∠ CDP [by (2)]

BQ = DP (Given)

∴ ∆ AQB ≅ ∆ CPD (SAS rule) …….. Result (iii)

∴ AQ = CP (CPCT) ………….. Result (iv)

Now, in quadrilateral APCQ, AP = CQ and AQ = CP

Hence, by theorem 8.3, APCQ is a parallelogram. ………. Result (v)

Question 10.

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

(i) ∆ APB ≅ ∆ CQD

(ii) AP = CQ

Answer:

In parallelogram ABCD, AB || CD and BD is their transversal.

∴ ∠ ABD = ∠ CDB (Alternate angles)

∴ ∠ABP = ∠CDQ ……………. (1)

Now, in ∆ APB and ∆ CQD,

AB = CD (Opposite sides of a parallelogram)

∠ ABP = ∠ CDQ [by (1)]

∠ APB = ∠ CQD (Right angles)

∆ APB ≅ ∆ CQD (AAS rule) ………… Result (i)

∴ AP = CQ (CPCT) ……….. Result (ii)

Question 11.

In ∆ ABC and ∆ DBF, AB = DE, AB || DE, j BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given 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:

In quadrilateral ∆ BED, AB = DE and AB || DE. Thus, in quadrilateral ABED, sides in one s pair of opposite sides are equal and parallel. Hence, by theorem 8.8, quadrilateral ABED is a parallelogram. …… Result (i)

Similarly, in quadrilateral BEFC, BC = EF and BC || EF.

Hence, by theorem 8.8, quadrilateral BEFC is a parallelogram. …………. Result (ii)

In parallelogram ABED, AD || BE and in parallelogram BEFC, BE || CE Thus, AD and CF both are parallel to BE.

∴ AD || CF ……….(1)

In parallelogram ABED, AD = BE and in parallelogram BEFC, BE = CF.

∴ AD = CF ……… (2)

Taking (1) and (2) together, we get

AD || CF and AD = CF ………. Result (iii)

In quadrilateral ACFD, AD || CF and AD = CF. Hence, by theorem 8.8, quadrilateral ACFD is a parallelogram. ………. Result (iv)

AC and DF are opposite sides of parallelogram ACFD.

∴ AC = DF ………….. Result (v)

Now, in ∆ ABC and ∆ DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF [by result (v)l

∴ ∆ ABC ≅ ∆ DEF (SSS rule) …… Result (vi)

Question 12.

ABCD is a trapezium in which AB || CD and AD = BC (see the given 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:

AB is extended to E, and AB || CD.

∴ AE || CD

In quadrilateral ADCE, AE || CD and by consturction CE || DA.

∴ Quadrilateral ADCE is a parallelogram.

∴ AD = CE

Moreover, AD = BC (Given)

∴ BC = CE

In ∆ BCE, BC = CE

∴ ∠CBE = ∠CEB

∴ ∠CBE = ∠CEA ………….. (1)

In parallelogram ADCE, AD || CE and AE is their transversal.

∴ ∠ DAE + ∠ CEA = 180° (Interior angles on the same side of transversal)

∴ ∠ DAE + ∠ CBE = 180° [by (1)]

∴ ∠ DAE = 180° – ∠ CBE …………… (2)

Moreover, ∠ ABC + ∠ CBE = 180° (Linear pair)

∴ ∠ ABC = 180° – ∠ CBE …………. (3)

From (2) and (3),

∠ DAE = ∠ ABC

∴ ∠A = ∠B ……… Result (i)

AB || CD and AD is their transversal.

∴ ∠A + ∠D = 180°

∴ ∠D = 180°- ∠A ………….. (4)

AB || CD and BC is their transversal.

∴ ∠B + ∠C = 180°

∴ ∠C = 180°- ∠B

∴ ∠C = 180° – ∠ A [by result (i)] ……… (5)

From (4) and (5),

∠C = ∠D …….. Result (ii)

Draw diagonals AC and BD.

In ∆ ABC and ∆ BAD,

BC = AD (Given)

∠ ABC = ∠ BAD [by result (i)]

AB = BA (Common)

∴ ∆ ABC ≅ ∆ BAD (SAS rule) ………. Result (iii)

∴ AC = BD (CPCT)

Thus, diagonal AC = diagonal BD … Result (iv)

Note: A trapezium in which non-parallel sides are equal is called an isosceles trapezium. As proved above, in an isosceles trapezium, the diagonals are equal and the angles on each parallel side are equal.