ML Aggarwal Rectilinear Figures Chapter Test Class 9 ICSE Maths Solutions Ch-13. Step by Step Solutions of Ch-Test questions on Rectilinear Figures of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Rectilinear Figures Chapter Test Class 9 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-13 | Rectilinear Figures |

Topics | Solution of Ch-Test Questions |

Academic Session | 2024-2025 |

### Chapter Test Questions with Solutions on Rectilinear Figures

ML Aggarwal Rectilinear Figures Chapter Test Class 9 ICSE Maths Solutions Ch-13

**Question 1. ****In the given figure, ABCD is a parallelogram. CB is produced to E such that BE=BC. Prove that AEBD is a parallelogram.**

**Answer : **ABCD is a || gm in which CB is produced to E such that BE = BC

BD and AE are joined

**To prove: **AEBD is a parallelogram

**Proof: **In ∆AEB and ∆BDC

EB = BC** [Given]**

∠ABE = ∠DCB **[Corresponding angles]**

AB = DC **[Opposite sides of ||gm]**

Thus, ∆AEB ≅ ∆BDC by S.A.S axiom

So, by C.P.C.T

But, AD = CB = BE **[Given]**

As the opposite sides are equal and ∠AEB = ∠DBC

But these are corresponding angles

Therefore, AEBD is a parallelogram.

**Question 2. ****In the given figure, ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC and CD || BA. Show that**

(i) ∠DAC=∠BCA

(ii) ABCD is a parallelogram.

**Answer : **In isosceles triangle ABC, AB = AC. AD is the bisector of ext. ∠PAC and CD || BA

**To prove: **(i) ∠DAC = ∠BCA

(ii) ABCD is a || gm

**Proof: **In ∆ABC

AB = AC **[Given]**

∠C = ∠B

Since, ext. ∠PAC = ∠B + ∠C

= ∠C + ∠C

= 2 ∠C

= 2 ∠BCA

So, ∠DAC = 2 ∠BCA

∠DAC = ∠BCA

Thus, AD || BC

But, AB || AC

Therefore, ABCD is a || gm.

**Question 3. ****Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.**

**Answer : **ABCD is an isosceles trapezium in which AB || DC and AD = BC

P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.

**To prove: **PQRS is a rhombus

**Construction: **Join AC and BD

**Proof: **ABCD is an isosceles trapezium

Its diagonals are equal

AC = BD

in ∆ABC

P and Q are the mid-points of AB and BC

PQ || AC and PQ = ½ AC **…(i)**

in ∆ADC

S and R mid-point of CD and AD

So, SR || AC and SR = ½ AC **…(ii)**

From (i) and (ii), we have

PQ || SR and PQ = SR

PQRS is a parallelogram.

Now, in ∆APS and ∆BPQ

AP = BP **[P is the mid-point]**

AS = BQ **[Half of equal sides]**

∠A = ∠B **[As ABCD is an isosceles trapezium]**

So, ∆APS ≅ ∆BPQ by SAS Axiom of congruency

by C.P.C.T we have

PS = PQ

But there are the adjacent sides of a parallelogram

So, sides of PQRS are equal

Hence, PQRS is a rhombus

**Question 4. ****Find the size of each lettered angle in the Following Figures.**

**Answer : ****(i)** As CDE is a straight line

∠ADE + ∠ADC = 180^{o}

⇒ 122^{o} + ∠ADC = 180^{o}

⇒ ∠ADC = 180^{o} – 122^{o} = 58^{o} **…(i)**

⇒ ∠ABC = 360^{o} – 140^{o} = 220^{o} **…(ii) (At any point the angle is 360 ^{o})**

∠ADC + ∠BCD + ∠BAD + ∠ABC = 360^{o}

⇒ 58^{o} + 53^{o} + x + 220^{o} = 360^{o} (**Using (i) and (ii))**

⇒ 331^{o} + x = 360^{o}

⇒ x = 360^{o} – 331^{o}

⇒ x = 29^{o}

**(ii)** As DE || AB

∠ECB = ∠CBA (**Alternate angls])**

⇒ 75^{o} = ∠CBA

⇒ ∠CBA = 75^{o}

Since, AD || BC we have

(x + 66^{o}) + 75^{o} = 180^{o}

⇒ x + 141^{o} = 180^{o}

⇒ x = 180^{o }– 141^{o}

⇒ x = 39^{o} **…(i)**

in ∆AMB

x + 30^{o} + ∠AMB = 180^{o} **(Angles sum property of a triangle)**

⇒ 39^{o} + 30^{o} + ∠AMB = 180^{o}** [From (i)]**

⇒ 69^{o} + ∠AMB + 180^{o}

⇒ ∠AMB = 180^{o} – 69^{o} = 111^{o} **…(ii)**

Since, ∠AMB = y

⇒ y = 111^{o}

Hence, x = 39^{o} and y = 111^{o}

**(iii)** In ∆ABD

AB = AD** [Given]**

∠ABD = ∠ADB

⇒ ∠ABD = 42^{o}** (Since, given ∠ADB = 42 ^{o})**

∠ABD + ∠ADB + ∠BAD = 180^{o} **(****Angles sum property of a triangle)**

⇒ 42^{o} + 42^{o} + y = 180^{o}

⇒ 84^{o} + y = 180^{o}

⇒ y = 180^{o} – 84^{o}

⇒ y = 96^{o}

⇒ ∠BCD = 2 ×26^{o} = 52^{o}

In ∆BCD,

As BC = CD

∠CBD = ∠CDB = x

∠CBD + ∠CDB + ∠BCD = 180^{o}

⇒ x + x + 52^{o} = 180^{o}

⇒ 2x + 52^{o} = 180^{o}

⇒ 2x = 180^{o} – 52^{o}

⇒ x = 128^{o}/2

⇒ x = 64^{o}

Therefore, x = 64^{o} and y = 90^{o}.

**Question 5. ****Find the size of each lettered angle in the following figures :**

**Answer : ****(i)** AB || CD and BC || AD

ABCD is a || gm

y = 2× ∠ABD

⇒ y = 2× 53^{o} = 106^{o} **…(1)**

Also, y + ∠DAB = 180^{o}

⇒ ∠DAB = 180^{o} – 106^{o}

= 74^{o}

x = ½ ∠DAB (**As AC bisects ∠DAB)**

x = ½ × 74^{o} = 37^{o}

and ∠DAC = x = 37^{o} **…(ii)**

Also, ∠DAC = z **…(iii) (Alternate angles)**

From (ii) and (iii), z = 37^{o }

Hence, x = 37^{o}, y = 106^{o} and z = 37^{o}

**(ii)** As ED is a straight line, we have

60^{o} + ∠AED = 180^{o}

⇒ ∠AED = 180^{o} – 60^{o}

⇒ ∠AED = 120^{o} **…(i)**

as CD is a straight line

50^{o} + ∠BCD = 180^{o}

⇒ ∠BCD = 180^{o} – 50^{o}

⇒ ∠BCD = 130^{o} **…(ii)**

In pentagon ABCDE, we have

∠A + ∠B+ ∠AED + ∠BCD + ∠x = 540^{o} **(Sum of interior angles in pentagon is 540 ^{o})**

⇒ 90^{o} + 90^{o} 120^{o} + 130^{o} + x = 540^{o}

⇒ 430^{o} + x = 540^{o}

⇒ x = 540^{o} – 430^{o}

⇒ x = 110^{o}

Hence, value of x = 110^{o}

**(iii)** In given figure, AD || BC

60^{o} + y = 180^{o} and x + 110^{o} = 180^{o}

⇒ y = 180^{o} – 60^{o} and x = 180^{o} – 110^{o}

⇒ y = 120^{o} and x = 70^{o}

CD || AF

∠FAD = 70^{o} **…(i)**

In quadrilateral ADEF,

∠FAD + 75^{o} + z + 130^{o} = 360^{o}

⇒ 70^{o} + 75^{o} + z + 130^{o} = 360^{o}

⇒ 275^{o} + z = 360^{o}

⇒ z = 360^{o} – 275^{o} = 85^{o}

Therefore, x = 70^{o}, y = 120^{o} and z = 85^{o}

**Question 6. ****In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find**

(i) ∠DAG

(ii) ∠FEG

(iii) ∠GAC

(iv) ∠AGC.

**Answer :**

AB = BC = DC = AD **…(i)**

DC = EF = FC = EF **…(ii)**

From (i) and (ii), we have

AB = BC = DC = AD = EF = FC = EF **…(iii)**

∠ABC = 56^{o}

∠ADC = 56^{o} **(Opposite angle in rhombus are equal)**

So, ∠EDA = ∠EDC + ∠ADC = 90^{o} + 56^{o} = 146^{o}

In ∆ADE, DE = AD (**From (iii))**

∠DEA = ∠DAE **(Equal sides have equal opposite angles)**

∠DEA = ∠DAG = (180^{o} – ∠EDA)/2

= (180^{o} – 146^{o})/2

= 34^{o}/2

= 17^{o}

⇒ ∠DAG = 17^{o}

Also, ∠DEG = 17^{o}

∠FEG = ∠E – ∠DEG

= 90^{o} – 17^{o}

= 73^{o}

In rhombus ABCD,

∠DAB = 180^{o} – 56^{o} = 124^{o}

⇒ ∠DAC = 124^{o}/2 (**Since, AC diagonals bisect the ∠A)**

⇒ ∠DAC = 62^{o}

∠GAC = ∠DAC – ∠DAG

= 62^{o} – 17^{o}

= 45^{o}

In ∆EDG, ∠D + ∠DEG + ∠DGE = 180^{o}

⇒ 90^{o} + 17^{o} + ∠DGE = 180^{o}

⇒ ∠DGE = 180^{o} – 107^{o} = 73^{o} **…(iv)**

∠AGC = ∠DGE **…(v) (Vertically opposite angles)**

Therefore, from (iv) and (v), we have

∠AGC = 73^{o}

**Question 7. ****If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.**

**Answer : **Each side of rhombus ABCD is 8cm

AB = BC = CD = DA = 8cm

Let ∠A = 60^{o}

∆ABD is an equilateral triangle

Then, AB = BD = AD = 8cm

the diagonals of a rhombus bisect each other at right angles

AO = OC, BO = OD = 4 cm and ∠AOB = 90^{o}

in right ∆AOB

AB^{2} = AO^{2} + OB^{2}

⇒ 8^{2} = AO^{2} + 4^{2}

⇒ 64 = AO^{2} + 16

⇒ AO^{2} = 64 – 16 = 48

⇒ AO = √48 = 4√3cm

AC = 2 AO

Therefore, AC = 2 × 4√3 = 8√3cm.

**Question 8. ****Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.**

**Answer : ****Construction:**

(i) Draw AB = 5.0cm

** **

(ii) Draw BAP = 45^{o} on side AB

(iii) Take A as centre and radius 2.5cm cut the line AP at D

(iv) Take D as centre and radius 5.0cm draw an arc

(v) Take B as centre and radius equal to 2.5cm cut the arc of step (iv) at C

(vi) Join BC and CD

(vii) ABCD is the required parallelogram

(viii) Draw the bisector of ∠BAD, which cuts the DC at E

(ix) Join EB

(x) Measure ∠AEB which is equal to 90^{o}.

— : End of ML Aggarwal Rectilinear Figures Chapter Test Class 9 ICSE Maths Solutions :–

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