ML Aggarwal Rectilinear Figures Exe-13.1 Class 9 ICSE Maths Solutions Ch-13. Step by Step Solutions of Exercise-13.1 Rectilinear Figures of ML Aggarwal for ICSE Class 9th Mathematics Questions. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Rectilinear Figures Exe-13.1 Class 9 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-13 | Rectilinear Figures |

Topics | Solution of Exe-13.1 Questions |

Academic Session | 2024-2025 |

### Rectilinear Figures

ML Aggarwal Rectilinear Figures Exe-13.1 Class 9 ICSE Maths Solutions Ch-13.

**Question 1. ****If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.**

**Answer : **Sum of all four angles of a quadrilateral = 360°

Sum of two given angles = 40° + 110° = 150°

So, the sum of remaining two angles = 360° – 150° = 210°

Ratio in these angles = 3 : 4

Third angle = (210° ×3)/(3 + 4)

= (210° ×3)/7

= 90°

Fourth angle

= (210° ×4)/(3 + 4)

= (210° ×4)/7

= 120°

**Question 2. ****If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.**

**Answer : **In trapezium ABCD in which

∠A : ∠B : ∠C : ∠D = 1 : 2 : 3 : 4

The sum of angles of the quad. ABCD = 360°

∠A = (360° ×1)/10 = 36°

∠B = (360° ×2)/10 = 72°

∠C = (360° ×3)/10 = 108°

∠D = (360° ×4)/10 = 144°

∠A + ∠D = 36° + 114° = 180°

Since, the sum of angles ∠A and ∠D is 180° and these are co-interior angles

Thus, AB || DC

Hence, ABCD is a trapezium.

**Question 3. ****If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.**

**Answer : **Here ABCD is a parallelogram.

Let ∠A = x°

Then, ∠B = (2x/3)°

∠A + ∠B = 180° (As the sum of adjacent angles in a parallelogram is 180°)

Therefore,

∠A = 108°

∠B = 2/3 ×108^{o} = 2 ×36° = 72°

⇒ ∠B = ∠D = 72° (opposite angles in a parallelogram is same)

Also,

∠A = ∠C = 108° (opposite angles in a parallelogram is same)

Hence, angles of parallelogram are 108°, 72°, 108°^{ }and 72°.

**Question 4. ****(a) In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.****(b) In figure (2) given below, ABCD is a parallelogram. Find the angles of the AAOD.****(c) In figure (3) given below, ABCD is a rhombus. Find the value of x.**

**Answer : ****(a)** ABCD is a || gm

We have, AB || CD

∠ADB = ∠DBC **(**Alternate angles)

⇒ ∠ADB = 80° (Given, ∠DBC = 80°)

In ∆ADB, we have

∠A + ∠ADB + ∠ABD = 180° (Angle sum property of a triangle)

⇒ 70° + 80° + ∠ABD = 180°

⇒ 150^{o} + ∠ABD = 180^{o}

⇒ ∠ABD = 180^{o} – 150^{o} = 30^{o}

∠CDB = ∠ABD (Since, AB || CD and alternate angles)

∠CDB = 30^{o}

Hence, ∠ADB = 80^{o} and ∠CDB = 30^{o}.

**(b) ∠BOC = 35**^{o} and ∠CBO = 77^{o}

^{o}and ∠CBO = 77

^{o}

In ∆BOC,

∠BOC + ∠BCO + ∠CBO = 180^{o} (Angle sum property of a triangle)

⇒ ∠BOC = 180^{o} – 112^{o} = 68^{o}

Now, in || gm ABCD

We have,

∠AOD = ∠BOC

Hence, ∠AOD = 68^{o}.

**(c) ABCD is a rhombus**

So, ∠A + ∠B = 180^{o}

(Sum of adjacent angles of a rhombus is 180^{o})

⇒ 72^{o} + ∠B = 180^{o} (Given, ∠A = 72^{o})

⇒ ∠B = 180^{o} – 72^{o} = 108^{o}

Therefore, x = ½ B

= ½ x 108^{o}

= 54^{o}

**Question 5. **

**(a) In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.**

**(b) In figure (2) given below. ABCD is a parallelogram. Find the values of x and y.**

**(c) In figure (3) given below. ABCD is a rhombus. Find x and y.**

**Answer : ****(a)** Since, ABCD is a parallelogram

So, AB = CD and BC = AD

⇒ 3x = 2y + 2

⇒ 3x – 2y = 2 **…(i)**

AB + BC + CD + DA = 40

⇒ 3x + 2x + 2y + 2 + 2x = 40

⇒ 7x + 2y = 40 – 2

⇒ 7x + 2y = 38 **…(ii)**

Now, adding (i) and (ii) we get

(3x – 2y) + (7x + 2y) = 2 + 38

⇒ 10x = 40

⇒ x = 40/10 = 4

3(4) – 2y = 2

⇒ 12 – 2y = 2

⇒ 2y = 12 – 2

⇒ y = 10/2 = 5

Therefore, x = 4 and y = 5

**(b)** In parallelogram ABCD, we have

∠A = ∠C **(Opposite angles are same in || gm)**

⇒ 3x – 20^{o} = x + 40^{o}

⇒ 3x – x = 40^{o} + 20^{o}

⇒ 2x = 60^{o}

⇒ x = 60^{o}/2 = 30^{o} **…(i)**

Also,

∠A + ∠B = 180^{o} **(Sum of adjacent angles in || gm is equal to 180 ^{o})**

⇒ 3x – 20^{o} + y + 15^{o} = 180^{o}

⇒ 3x + y = 180^{o} + 20^{o} – 15^{o}

⇒ 3x + y = 185^{o}

⇒ 3(30^{o}) + y = 185^{o}

⇒ 90^{o} + y = 185^{o}

⇒ y = 185^{o} – 90^{o} = 95^{o}

Therefore,

x = 30^{o} and 95^{o}

**(c)** ABCD is a rhombus

So,

AB = CD

⇒ 3x + 2 = 4x – 4

⇒ 3x – 4x = -4 – 2

⇒ -x = -6

⇒ x = 6

Now, in ∆ABD we have

∠BAD = 60^{o} and AB = AD

∠ADB = ∠ABD

So,

∠ADB = (180^{o} – ∠BAD)/2

= (180^{o} – 60^{o})/2

= 120^{o}/2 = 60^{o}

As ∆ABD is an equilateral triangle, all the angles of the triangle are 60^{o}

Therefore, AB = BD

3x + 2 = y – 1

⇒ 3(6) + 2 = y – 1

⇒ 18 + 2 = y – 1

⇒ 20 = y – 1

⇒ y = 20 + 1

⇒ y = 21

Thus,

x = 6 and y = 21.

**Question 6. **The diagonals AC and BD of a rectangle > ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.

**Answer : **ABCD is a rectangle

AP = BP (in rectangle diagonal are equal and bisect each other)

∠PAB = ∠PBA (Equal sides have equal opposite angles)

⇒ ∠PAB = 50^{o }(Since, given ∠PBA = 50^{o})

Now, in ∆APB

∠APB + ∠ABP + ∠BAP = 180^{o}

⇒ ∠APB + 50^{o} + 50^{o} = 180^{o}

⇒ ∠APB = 180^{o} – 100^{o}

⇒ ∠APB = 80^{o}

Then,

∠DPB = ∠APB **(**Vertically opposite angles**)**

Therefore, ∠DPB = 80^{o}

**Question 7. **

**(a) In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.**

**(b) In figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of the ∆ AOB.**

**(c) In figure (3) given below, ABCD is rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3:2, find the angles of the ∆ AOD.**

**Answer : **EBC is an equilateral triangle, we have

EB = BC = EC **…(i)**

Also, ABCD is a square

So, AB = BC = CD = AD **…(ii)**

From (i) and (ii), we get

EB = EC = AB = BC = CD = AD **…(iii)**

in ∆ECD

∠ECD = ∠BCD + ∠ECB

= 90^{o} + 60^{o}

= 150^{o} **…(iv)**

EC = CD **[From (iii)]**

∠DEC = ∠CDE **…(v)**

∠ECD + ∠DEC + ∠CDE = 180^{o}

⇒ 150^{o} + ∠DEC + ∠DEC = 180^{o} **[Using (iv) and (v)]**

⇒ 2 ∠DEC = 180^{o} – 150^{o} = 30^{o}

⇒ ∠DEC = 30^{o}/2

⇒ ∠DEC = 15^{o} **…(vi)**

Now, ∠BEC = 60^{o}

⇒ ∠BED + ∠DEC = 60^{o}

⇒ x^{o} + 15^{o} = 60^{o} **[From (vi)]**

⇒ x = 60^{o} – 15^{o}

⇒ x = 45^{o}

Therefore, the value of x is 45^{o}.

**(b)** Given, ABCD is a rectangle

∠ECD = 146^{o}

As ACE is a straight line, we have

146^{o} + ∠ACD = 180^{o} **[Linear pair]**

⇒ ∠ACD = 180^{o} – 146^{o} = 34^{o }**…(i)**

And, ∠CAB = ∠ACD **…(ii) [Alternate angles] **

From (i) and (ii), we have

∠CAB = 34^{o}

⇒ ∠OAB = 34^{o} **…(iii)**

In ∆AOB

AO = OB [Diagonals of a rectangle are equal and bisect each other]

∠OAB = ∠OBA …(iv) [Equal sides have equal angles opposite to them]

From (iii) and (iv),

∠OBA = 34^{o} **…(v)**

Now,

∠AOB + ∠OBA + ∠OAB = 180^{o}

⇒ ∠AOB + 34^{o} + 34^{o }= 180^{o} **[Using (3) and (5)]**

⇒ ∠AOB + 68^{o} = 180^{o}

⇒ ∠AOB = 180^{o} – 68^{o} = 112^{o}

∠AOB = 112^{o}, ∠OAB = 34^{o} and ∠OBA = 34^{o}

**(c)** Here, ABCD is a rhombus and diagonals intersect at O and ∠OAB : ∠OBA = 3 : 2

Let ∠OAB = 2x^{o}

Then,

∠OBA = 2x^{o}

∠OAB = 90^{o}

Now, in ∆AOB

∠OAB + ∠OBA = 180^{o}

⇒ 90^{o} + 3x^{o }+ 2x^{o} = 180^{o}

⇒ 90^{o} + 5x^{o} = 180^{o}

⇒ 5x^{o} = 180^{o} – 90^{o} = 90^{o}

⇒ x^{o} = 90^{o}/5 = 18^{o}

∠OAB = 3x^{o} = 3 x 18^{o} = 54^{o}

⇒ OBA = 2x^{o} = 2×18^{o} = 36^{o} and ∠AOB = 90^{o}

**Question 8. **

**(a) In figure (1) given below, ABCD is a trapezium. Find the values of x and y.**

**(b) In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and.y.**

**(c) In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.**

**Answer :**

**(a)** ABCD is a trapezium

∠A = x + 20^{o}, ∠B = y, ∠C = 92^{o}, ∠D = 2x + 10^{o}

∠B + ∠C = 180^{o} **[Since AB || DC]**

⇒ y + 92^{o} = 180^{o}

⇒ y = 180^{o} – 92^{o} = 88^{o}

∠A + ∠D = 180^{o}

⇒ x + 20^{o} + 2x + 10^{o} = 180^{o}

⇒ 3x + 30^{o} = 180^{o}

⇒ 3x = 180^{o} – 30^{o} = 150^{o}

⇒ x = 150^{o}/3 = 50^{o}

the value of x = 50^{o} and y = 88^{o}.

**(b)** ABCD is an isosceles trapezium BC = AD

∠A = 2x, ∠C = y and ∠D = 3x

ABCD is a trapezium and AB || DC

∠A + ∠D = 180^{o}

⇒ 2x + 3x = 180^{o}

⇒ 5x = 180^{o}

⇒ x = 180^{o}/5 = 36^{o} **…(i)**

AB = BC and AB || DC

So, ∠A + ∠C = 180^{o}

⇒ 2x + y = 180^{o}

⇒ 2×36^{o} + y = 180^{o}

⇒ 72^{o} + y = 180^{o}

⇒ y = 180^{o} – 72^{o} = 108^{o}

Therefore, value of x = 72^{o} and y = 108^{o}.

**(c)** ABCD is a kite and diagonal intersect at O.

∠DAB = 112^{o} and ∠DCB = 64^{o}

As AC is the diagonal of kite ABCD, we have

∠DCO = 64^{o}/2 = 32^{o}

∠DOC = 90^{o} **[Diagonal of kites bisect at right angles]**

In ∆OCD, we have

∠ODC = 180^{o }– (∠DCO + ∠DOC)

= 180^{o} – (32^{o} + 90^{o})

= 180^{o} – 122^{o}

= 58^{o}

In ∆DAB, we have

∠OAB = 112^{o}/2 = 56^{o}

∠AOB = 90^{o} [Diagonal of kites bisect at right angles]

In ∆OAB,

∠OBA = 180^{o }– (∠OAB + ∠AOB)

= 180^{o} – (56^{o} + 90^{o})

= 180^{o} – 146^{o}

= 34^{o}

Therefore, ∠ODC = 58^{o} and ∠OBA = 34^{o}.

**Question 9. **

(i) Prove that each angle of a rectangle is 90°.

(ii) If the angle of a quadrilateral are equal, prove that it is a rectangle.

(iii) If the diagonals of a rhombus are equal, prove that it is a square.

(iv) Prove that every diagonal of a rhombus bisects the angles at the vertices.

**Answer : ****(i)** ABCD is a rectangle

**To prove**: Each angle of rectangle = 90^{o}

**Proof: **

In a rectangle opposite angles of a rectangle are equal

So, ∠A = ∠C and ∠B = ∠C

∠A + ∠B + ∠C + ∠D = 360^{o} [**Sum of angles of a quadrilateral]**

⇒ ∠A + ∠B + ∠A + ∠B = 360^{o}

⇒ 2(∠A + ∠B) = 360^{o}

⇒ (∠A + ∠B) = 360^{o}/2

⇒ ∠A + ∠B = 180^{o}

But, ∠A = ∠B **[Angles of a rectangle]**

∠A = ∠B = 90^{o}

∠A = ∠B = ∠C = ∠D = 90^{o}

Therefore, each angle of a rectangle is 90°.

**(ii)** In quadrilateral ABCD, we have

∠A = ∠B = ∠C = ∠D

**To prove:** ABCD is a rectangle

**Proof:**

∠A = ∠B = ∠C = ∠D

⇒ ∠A = ∠C and ∠B = ∠D

But these are opposite angles of the quadrilateral.

ABCD is a parallelogram

And, as ∠A = ∠B = ∠C = ∠D

Hence, ABCD is a rectangle.

**(iii)** ABCD is a rhombus in which AC = BD

**To prove:** ABCD is a square

**Proof: Join AC and BD.**

Now, in ∆ABC and ∆DCB we have

∠AB = ∠DC **[Sides of a rhombus]**

∠BC = ∠BC **[Common]**

∠AC = ∠BD** [Given]**

∆ABC ≅ ∆DCB by S.S.S axiom of congruency

∠ABC = ∠DBC [By C.P.C.T]

But these are made by transversal BC on the same side of parallel lines AB and CD.

So, ∠ABC + ∠DBC = 180^{o}

∠ABC = 90^{o}

Therefore, ABCD is a square.

**(iv)** ABCD is rhombus.

**To prove:** Diagonals AC and BD bisects ∠A, ∠C, ∠B and ∠D respectively

**Proof:**

In ∆AOD and ∆COD, we have

AD = CD [sides of a rhombus are all equal]

OD = OD [Common]

AO = OC [Diagonal of rhombus bisect each other]

So, ∆AOD ≅ ∆COD by S.S.S axiom of congruency

Thus,

∠AOD = ∠COD **[By C.P.C.T]**

So, ∠AOD + ∠COD = 180^{o} **[Linear pair]**

∠AOD = 180^{o}

∠AOD = 90^{o}

And, ∠COD = 90^{o}

OD ⊥ AC ⇒ BD ⊥ AC

Also, ∠ADO = ∠CDO **[By C.P.C.T]**

OD bisect ∠D

BD bisect ∠D

we can prove that BD bisect ∠B and AC bisect the ∠A and ∠C.

**Question 10. ****ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:**

(i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC ⊥ BD.

**Answer : **In parallelogram ABCD in which diagonal AC bisects ∠A

**To prove:** (i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC ⊥ BD

**Proof:**

**(i)** As AB || CD, we have [Opposite sides of a || gm]

∠DCA = ∠CAB

Similarly,

∠DAC = ∠DCB

But, ∠CAB = ∠DAC [Since, AC bisects ∠A]

Therefore,

∠DCA = ∠ACB and AC bisects ∠C.

**(ii)** As AC bisects ∠A and ∠C

And, ∠A = ∠C

ABCD is a rhombus.

**(iii)** Since, AC and BD are the diagonals of a rhombus and

AC and BD bisect each other at right angles

Therefore, AC ⊥ BD

**Question 11. **

**(i) Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.**

**(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.**

**(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.**

**Answer : ****(i)** Given AM bisect angle A and BM bisects angle of || gm ABCD.

**To prove:** ∠AMB = 90^{o}

**Proof: **We have,

∠A + ∠B = 180^{o} **[AD || BC and AB is the transversal]**

⇒ ½ (∠A + ∠B) = 180^{o}/2

⇒ ½ ∠A + ½ ∠B = 90^{o}

⇒ ∠MAB + ∠MBA = 90^{o} **[Since, AM bisects ∠A and BM bisects ∠B]**

in ∆AMB

∠AMB + ∠MAB + ∠MBA = 180^{o} **[Angles sum property of a triangle]**

⇒ ∠AMB + 90^{o} = 180^{o}

⇒ ∠AMB = 180^{o} – 90^{o} = 90^{o}

Therefore, bisectors of any two adjacent angles of a parallelogram are at right angles.

**(ii)** A || gm ABCD in which bisector AR of ∠A meets DC in R and bisector CQ of ∠C meets AB in Q

∠A = ∠C **[Opposite angles of || gm are equal]**

½ ∠A = ½ ∠C

∠DAR = ∠BCQ **[Since, AR is bisector of ½ ∠A and CQ is the bisector of ½ ∠C]**

Now, in ∆ADR and ∆CBQ

∠DAR = ∠BCQ **[Proved above]**

AD = BC **[Opposite sides of || gm ABCD are equal]**

So, ∆ADR ≅ ∆CBQ, by A.S.A axiom of congruency

Then by C.P.C.T, we have

∠DRA = ∠BCQ

∠DRA = ∠RAQ **[Alternate angles since, DC || AB]**

∠RAQ = ∠BCQ

But these are corresponding angles,

Hence, AR || CQ.

**(iii)** In quadrilateral ABCD, diagonals AC and BD are equal and bisect each other at right angles

**To prove:** ABCD is a square

**Proof: **In ∆AOB and ∆COD, we have

AO = OC **[Given]**

BO = OD **[Given]**

∠AOB = ∠COD **[Vertically opposite angles]**

So, ∆AOB ≅ ∆COD, by S.A.S axiom of congruency

By C.P.C.T, we have

AB = CD

and ∠OAB = ∠OCD

AB || CD

ABCD is a parallelogram

In a parallelogram, the diagonal bisect each other and are equal

Therefore, ABCD is a square.

**Question 12.**

**(i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.**

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

**Answer : ****(i)** ABCD is a rectangle and its diagonals AC bisects ∠A and ∠C

**To prove:** ABCD is a square

**Proof: **As AC bisects ∠A and ∠C

So, ∠1 = ∠2 and ∠3 = ∠4

But, ∠A = ∠C = 90^{o}

∠2 = 45^{o} and ∠4 = 45^{o}

And, AB = BC **[Opposite sides of equal angles]**

AB = CD and BC = AD

So, AB = BC = CD = DA

Hence, ABCD is a square.

**(ii)** In quadrilateral ABCD diagonals AC and BD are equal and bisect each other at right angle

**To prove:** ABCD is a square

**Proof: **In ∆AOB and ∆BOC, we have

AO = CO

OB = OB

∠AOB = ∠COB **[Each 90 ^{o}]**

So, ∆AOB ≅ ∆BOC, by S.A.S axiom

By C.P.C.T, we have

AB = BC **…(i)**

Similarly, in ∆BOC and ∆COD

OB = OD

OC = OC

∠BOC = ∠COD **[Each 90 ^{o}]**

So, ∆BOC ≅ ∆COD, by S.A.S axiom

By C.P.C.T, we have

BC = CD **…(ii)**

From (i) and (ii), we have

AB = BC = CD = DA

Therefore, ABCD is a square.

**Question 13. ****P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.**

**Answer : **ABCD is a parallelogram, P and Q are the points on AB and DC. Diagonals AC and BD intersect each other at O.

**To prove: **Diagonals of || gm ABCD bisect each other at O

So, AO = OC and BO = OD

Now, in ∆AOP and ∆COQ we have

AO = OC and BO = OD

Now, in ∆AOP and ∆COQ

AO = OC

∠OAP = ∠OCQ

∠AOP = ∠COQ

So, ∆AOP ≅ ∆COQ by S.A.S axiom

Thus, by C.P.C.T

OP = OQ

Therefore, O bisects PQ.

**Question 14. **

**(a) In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that:**

**(i) the triangles ABX and QCX are congruent;**

**(ii)DC = CQ = QP**

**(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.**

**Answer : ****(a) **ABCD is parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q and ABPQ is a || gm.

**To prove: **(i) ∆ABX ≅ ∆QCX, (ii) DC = CQ = QP

**Proof: **In ∆ABX and ∆QCX, we have

BX = XC

∠AXB = ∠CXQ

∠XCQ = ∠XBA

So, ABX ≅ ∆QCX by A.S.A axiom of congruence

Now, by C.P.C.T

CQ = AB

But, AB = DC and AB = QP **[As ABCD and ABPQ are ||gms]**

Hence, DC = CQ = QP

**(b)** In || gm ABCD, P and Q are points on AB and CD respectively, PQ and AC intersect each other at O and AP = CQ

**To prove:** AC and PQ bisect each other i.e. AO = OC and PO = OQ

**Proof: **In ∆AOP and ∆COQ

AP = CQ **[Given]**

∠AOP = ∠COQ

∠OAP = ∠OCP

So, ∆AOP ≅ ∆COQ by A.A.S axiom of congruence

Now, by C.P.C.T

OP = OQ and OA = OC

Hence proved.

**Question 15. ****ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP=DQ, prove that AP and DQ are perpendicular to each other.**

**Answer : **ABCD is a square. P is any point on BC and Q is any point on AB and these points are taken such that AP = DQ

To prove: AP ⊥ DQ

**Proof: **In ∆ABP and ∆ADQ, we have

AP = DQ **[Given]**

AD = AB **[Sides of square ABCD]**

∠DAQ = ∠ABP **[Each 90 ^{o}]**

So, ∆ABP ≅ ∆ADQ by R.H.S axiom of congruency

by C.P.C.T

∠BAP = ∠ADQ

But, ∠BAD = 90^{o}

∠BAD = ∠BAP + ∠PAD **…(i)**

⇒ 90^{o} = ∠BAP + ∠PAD

⇒ ∠BAP + ∠PAD = 90^{o}

⇒ ∠BAP + ∠ADQ = 90^{o}

Now, in ∆ADM we have

(∠MAD + ∠ADM) + ∠AMD = 180^{o}

⇒ 90^{o} + ∠AMD = 180^{o} **[From (i)]**

⇒ ∠AMD = 180^{o} – 90^{o} = 90^{o}

So, DM ⊥ AP

⇒ DQ ⊥ AP

Therefore, AP ⊥ DQ

**Question 16. ****If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.**

**Answer : **ABCD is a || gm is which BP = PQ = QD

To prove: CQ || AP

**Proof:**

In || gm ABCD, we have

AB = CD **[Opposite sides of a || gm are equal]**

And BD is the transversal

So, ∠1 = ∠2 **…(i) [Alternate interior angles] **

Now, in ∆ABP and ∆DCQ

AB = CD **[Opposite sides of a || gm are equal]**

∠1 = ∠2 **[From (i)]**

BP = QD **[Given]**

So, ∆ABP ≅ ∆DCQ by S.A.S axiom of congruency

Then by C.P.C.T, we have

AP = QC

Also, ∠APB = ∠DQC **[By C.P.C.T]**

-∠APB = -∠DQC **[Multiplying both sides by -1]**

⇒ 180^{o} – ∠APB = 180^{o} – ∠DQC

⇒ ∠APQ = ∠CQP

Therefore, AP || QC

⇒ CQ || AP.

**Question 17. ****A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B ; the four bisectors form a quadrilateral ABCD. Prove that**

(i) ABCD is a rectangle.

(ii) CD is parallel to the original parallel lines.

**Answer : **LM || PQ and AB is the transversal line cutting ∠M at A and PQ at B

AC, AD, BC and BD is the bisector of ∠LAB, ∠BAM, ∠PAB and ∠ABQ respectively.

AC and BC intersect at C and AD and BD intersect at D.

A quadrilateral ABCD is formed.

**To prove:** (i) ABCD is a rectangle

(ii) CD || LM and PQ

**Proof: ****(1)** ∠LAB + ∠BAM = 180^{o}

⇒ ½ (∠LAB + ∠BAM) = 90^{o}

⇒ ½ ∠LAB + ½ ∠BAM = 90^{o}

⇒ ∠2 + ∠3 = 90^{o} **[AC and AD is bisector of ∠LAB & ∠BAM respectively]**

⇒ ∠CAD = 90^{o}

⇒ ∠A = 90^{o}

**(2)** Similarly, ∠PBA + ∠QBA = 180^{o}

⇒ ½ (∠PBA + ∠QBA) = 90^{o}

⇒ ½ ∠PBA + ½ ∠QBA = 90^{o}

⇒ ∠6 + ∠7 = 90^{o} **[ BC and BD is bisector of ∠PAB & ∠QBA respectively]**

⇒ ∠CBD = 90^{o}

⇒ ∠B = 90^{o}

**(3)** ∠LAB + ∠ABP = 180^{o} **[Sum of co-interior angles is 180**^{o} and given LM || PQ]

^{o}and given LM || PQ]

⇒ ½ ∠LAB + ½ ∠ABP = 90^{o}

⇒ ∠2 + ∠6 = 90^{o} **[Since, AC and BC is bisector of ∠LAB & ∠PBA respectively]**

**(4) **In ∆ACB,

∠2 + ∠6 + ∠C = 180^{o}** **

⇒ (∠2 + ∠6) + ∠C = 180^{o}

⇒ 90^{o} + ∠C = 180^{o}** [using (3)]**

⇒ ∠C = 180^{o} – 90^{o}

⇒ ∠C = 90^{o}

**(5)** ∠MAB + ∠ABQ = 180^{o} **[Sum of co-interior angles is 180**^{o} and given LM || PQ]

^{o}and given LM || PQ]

⇒ ½ ∠MAB + ½ ∠ABQ = 90^{o}

⇒ ∠3 + ∠7 = 90^{o} **[Since, AD and BD is bisector of ∠MAB & ∠ABQ respectively]**

**(6)** In ∆ADB,

∠3 + ∠7 + ∠D = 180^{o} **[Angles sum property of a triangle]**

⇒ (∠3 + ∠7) + ∠D = 180^{o}

⇒ 90^{o} + ∠D = 180^{o} **[using (5)]**

⇒ ∠D = 180^{o} – 90^{o}

⇒ ∠D = 90^{o}

**(7)** ∠LAB + ∠BAM = 180^{o}

∠BAM = ∠ABP **[From (1) and (2)]**

⇒ ½ ∠BAM = ½ ∠ABP

⇒ ∠3 = ∠6 **[Since, AD and BC is bisector of ∠BAM and ∠ABP respectively]**

∠2 = ∠7

**(8)** In ∆ABC and ∆ABD,

∠2 = ∠7 **[From (7)]**

AB = AB **[Common]**

∠6 = ∠3 **[From (7)]**

∆ABC ≅ ∆ABD by A.S.A axiom of congruency

by C.P.C.T we have

AC = DB

CB = AD

**(9)** ∠A = ∠B = ∠C =∠D = 90^{o} **[From (1), (2), (3) and (4)]**

AC = DB **[Proved in (8)]**

CB = AD **[Proved in (8)]**

Hence, ABCD is a rectangle.

**(10)** ABCD is a rectangle **[From (9)]**

OA = OD **[Diagonals of rectangle bisect each other]**

**(11)** In ∆AOD, we have

OA = OD** [From (10)]**

∠9 = ∠3

**(12)** ∠3 = ∠4 **[AD bisects ∠MAB]**

**(13)** ∠9 = ∠4 **[From (11) and (12)]**

But these are alternate angles.

OD || LM

⇒ CD || LM

∠10 = ∠8

But these are alternate angles,

So, OD || PQ

⇒ CD || PQ

**(14)** CD || LM **[Proved in (13)]**

CD || PQ **[Proved in (13)]**

**Question 18****. ****In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that**

(i) BE bisects ∠B

(ii) ∠AEB = a right angle.

**Answer : ****(1)** In || gm ABCD

∠1 = ∠2 **[AD bisects angles ∠A]**

**(2)** AB || DC and AE is the transversal

∠2 = ∠3 **[Alternate angles]**

**(3)** ∠1 = ∠2 **[From (1) and (2)]**

**(4)** In ∆ADE, we have

∠1 = ∠3 **[Proved in (3)]**

DE = AD

⇒ AD = DE

**(5)** AB = 2 AD **[Given]**

AB/2 = AD

AB/2 = DE **[using (4)]**

DC/2 = DE **[AB = DC, opposite sides of a || gm are equal]**

So, E is the mid-point of D.

⇒ DE = EC

**(6) **AD = BC **[Opposite sides of a || gm are equal]**

**(7)** DE = BC **[From (4) and (6)]**

**(8)** EC = BC **[From (5) and (7)]**

**(9)** In ∆BCE, we have

EC = BC **[Proved in (8)]**

∠6 = ∠5

**(10)** AB || DC and BE is the transferal

∠4 = ∠5 **[Alternate angles]**

**(11)** ∠4 = ∠6 **[From (9) and (10)]**

So, BE is bisector of ∠B

**(12)** ∠A + ∠B = 180^{o} **[Sum of co-interior angles is equal to 180**^{o}, AD || BC]

^{o}, AD || BC]

⇒ ½ ∠A + ½ ∠B = 180^{o}/2

⇒ ∠2 + ∠4 = 90^{o} **[AE is bisector of ∠A and BE is bisector of ∠B]**

**(13)** In ∆APB,

∠AEB + ∠2 + ∠4 = 180^{o}

⇒ ∠AEB + 90^{o} = 180^{o}

Therefore, ∠AEB = 90^{o}

**Question 19. ****ABCD is a parallelogram, bisectors of angles A and B meet at E which lie on DC. Prove that AB**

**Answer : **ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC in E

**To prove:** AB = 2 AD

**Proof: **In parallelogram ABCD, we have AB || DC

∠1 = ∠5 **[Alternate angles, AE is transversal]**

∠1 = ∠2** [AE is bisector of ∠A, given]**

Thus, ∠2 = ∠5 **…(i)**

Now, in ∆AED

DE = AD** [Sides opposite to equal angles are equal]**

∠3 = ∠6 **[Alternate angles]**

∠3 = ∠4** [Since, BE is bisector of ∠B (given)]**

∠4 = ∠6 **…(ii)**

In ∆BCE, we have

BC = EC **[Sides opposite to equal angles are equal]**

AD = BC **[Opposite sides of || gm are equal]**

AD = DE = EC **[From (i) and (ii)]**

AB = DC **[Opposite sides of a || gm are equal]**

AB = DE + EC

= AD + AD

AB = 2 AD

**Question 20. ****ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO =AP, prove that 3 ∠POB = ∠AOP.**

**Answer : **ABCD is a square and the diagonals intersect at O. P is the point on AB such that AO = AP

**To prove:** 3 ∠POB = ∠AOP

**Proof: ****(1)** In square ABCD, AC is a diagonal

∠CAB = 45^{o}

∠OAP = 45^{o}

**(2)** In ∆AOP,

∠OAP = 45^{o} **[From (1)]**

AO = AP **[Sides opposite to equal angles are equal]**

Now,

∠AOP + ∠APO + ∠OAP = 180^{o}

⇒ ∠AOP + ∠AOP + 45^{o} = 180^{o}

⇒ 2∠AOP = 180^{o} – 45^{o}

⇒ ∠AOP = 135^{o}/2

**(3)** ∠AOB = 90^{o} **[Diagonals of a square bisect at right angles]**

∠AOP + ∠POB = 90^{o}

⇒ 135^{o}/2 + ∠POB = 90^{o} **[From (2)]**

⇒ ∠POB = 90^{o} – 135^{o}/2

= (180^{o} – 135^{o})/2

= 45^{o}/2

3∠POB = 135^{o}/2** [Multiplying both sides by 3]**

Therefore,

∠AOP = 3 ∠POB **[From (2) and (3)]**

**Question 21. ****ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square**.

**Answer :**

ABCD is a square in which E, F, G and H are points on AB, BC, CD and DA

Such that AE = BF = CG = DH

EF, FG, GH and HE are joined

**To prove:** EFGH is a square

**Proof: **Since, AE = BF = CG = DH

EB = FC = GD = HA

In ∆AEH and ∆BFE

AE = BF **[Given]**

AH = EB **[Proved]**

∠A = ∠B **[Each 90 ^{o}]**

∆AEH ≅ ∆BFE by S.A.S axiom of congruency

by C.P.C.T we have

EH = EF

And ∠4 = ∠2

But ∠1 + ∠4 = 90^{o}

⇒ ∠1 + ∠2 = 90^{o}

Thus, ∠HEF = 90^{o}

Therefore, EFGH is a square.

**Question 22. ****(a) In the Figure (1) given below, ABCD and ABEF are parallelograms. Prove that**

(i) CDFE is a parallelogram

(ii) FD = EC

(iii) Δ AFD = ΔBEC.

**(b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC**

**Answer : ****To prove: **

(i) CDFE is a parallelogram

(ii) FD = EC

(iii) Δ AFD = ΔBEC

**Proof: **

**(1)** DC || AB and DC = AB **[ABCD is a || gm]**

**(2)** FE || AB and FE = AB **[ABEF is a || gm]**

**(3)** DC || FE and DC = FE **[From (1) and (2)]**

Thus, CDFE is a || gm

**(4)** CDEF is a || gm

So, FD = EC

**(5)** In ∆AFD and ∆BEC, we have

AD = BC** [Opposite sides of || gm ABCD are equal]**

AF = BE **[Opposite sides of || gm ABEF are equal]**

FD = BE **[From (4)]**

Hence, ∆AFD ≅ ∆BEC by S.S.S axiom of congruency

**(b) **ABCD is a || gm, ADEF and AGHB are two squares

To prove: FG = AC

**Proof: **(1) ∠FAG + 90^{o} + 90^{o} + ∠BAD = 360^{o} **[At a point total angle is 360 ^{o}]**

⇒ ∠FAG = 360^{o} – 90^{o} – 90^{o }– ∠BAD

⇒ ∠FAG = 180^{o} – ∠BAD

**(2)** ∠B + ∠BAD = 180^{o} **[Adjacent angle in || gm is equal to 180**^{o}]

^{o}]

⇒ ∠B = 180^{o} – ∠BAD

**(3)** ∠FAG = ∠B [From (1) and (2)]

**(4)** In ∆AFG and ∆ABC, we have

AF = BC **[FADE and ABCD both are squares on the same base]**

AG = AB

∠FAG = ∠B **[From (3)]**

∆AFG ≅ ∆ABC by S.A.S axiom of congruency

Therefore, by C.P.C.T

FG = AC

**Question 23. ****ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.**

**Answer : **Let each side of the rhombus ABCD be a

∠A = 60^{o}

ABD is an equilateral triangle

⇒ BD = AB = a

We know that, the diagonals of a rhombus bisect each other at right angles

So, in right triangle AOB, we have

AO^{2} = AB^{2} – OB^{2}

= a^{2} – (½ a)^{2}

= a^{2} – a^{2}/4

= 3a^{2}/4

AO = √(3a^{2}/4) = √3a/2

But, AC = 2 AO = 2 x 3a/2 = 3a

Therefore,

AC : BD = √3a : a = √3 : 1

— : End of ML Aggarwal Rectilinear Figures Exe-13.1 Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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