## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 13 Rectilinear Figures Chapter Test

P.Q.

The interior angles of a polygon add upto 4320°. How many sides does the polygon have ?

Answer:

Sum of interior angles of a polygon

= (2n – 4) × 90°

⇒ 4320° = (2n – 4) × 90°

⇒ \(\frac{4320^{\circ}}{90^{\circ}}\) = (2n – 4) ⇒ \(\frac{432}{9}\) = 2n – 4

⇒ 48 = 2n – 4 ⇒ 48 + 4 = 2n => 52 = 2 n

⇒ 2n = 52 ⇒ n = \(\frac{52}{2}\) = 26

Hence, the polygon have 26 sides.

P.Q.

If the ratio of an interior angle to the exterior angle of a regular polygon is 5 : 1, find the number of sides.

Answer:

The ratio of an interior angle to the exterior angle of a regular polygon = 5 : 1

Hence, number of sides of regular polygon = 12.

P.Q.

In a pentagon ABCDE, BC || ED and ∠B: ∠A : ∠E = 3 : 4 : 5. Find ∠A.

Answer:

∵ BC||ED

∴ ∠C +∠D= 180° (Co-interior angles)

But∠A + ∠B + ∠C + ∠D + ∠E = 540°

∴ ∠A + ∠B + ∠E ≠ 180° = 540°

⇒ ∠A + ∠B + ∠E = 540° – 180° = 360°

But ∠B: ∠A = ∠E = 3 : 4 : 5

Let ∠B = 3x, ∠A = 4x and ∠E = 5x

∴ 3x + 4x + 5x = 360° ⇒ 12x = 360°

⇒ x = \(\frac{360^{\circ}}{12}\) = 30°

∴ A = 4x = 4 × 30°= 120°

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.

Solution:

In the figure, ABCD is a ||gm side 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)

∴ ∆AEB = ∆BDC (SAS axiom)

∴ AE = DB (c.p.c.t.)

But AD = CB = BE (Given)

∵ The opposite sides are equal and ∠AEB = ∠DBC (c.p.c.t.)

But these are corresponding angle

∴ 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.

Solution:

Given: In isosceles ∆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 (Angles opposite to equal sides)

∵ Ext. ∠PAC = ∠B + ∠C

= ∠C + ∠C = 2∠C = 2∠BCA

∴ 2∠DAC = 2∠BCA

∠DAC = ∠BCA

But these are alternate angles

∴ AD || BC

But AB || AC (Given)

∴ ABCD is a||gm

Question 3.

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

Answer:

Given. 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.

Constructions. Join AC and BD.

Proof, ∵ ABCD is an isosceles trapezium

∴ Its diagnoals are equal

∴ AC = BD

Now in ∆ABC,

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

∴ PQ || AC and PQ = \(\frac{1}{2}\) AC …(i)

Similarly in ∆ADC,

S and R mid-points of CD and AD

∴ SR || AC and SR = \(\frac{1}{2}\) AC …(ii)

from (i) and (ii)

PQ | | SR and PQ = SR

∴ PQRS is a parallelogram

Now in ∆APS and ∆BPQ,

AP = BP (P is mid-point of AB)

AS = BQ Half of equal sides)

∠A = ∠B ( ∵ ABCD is isosceles trapezium)

∴ ∆APS ≅ BPQ

∴ PS = PQ

But there are the adjacent sides of a parallelogram

∴ Sides of PQRS are equal

Hence PQRS is a rhombus.

Hence proved.

Question 4.

Find the size of each lettered angle in the following figures:

Answer:

(i) ∵ CDE is a st. line

∴ ∠ADE + ∠ADC = 180°

122°+ ∠ADC = 180°

∠ADC = 180° – 122°°

∠ADC = 58° ….(1)

∠ABC = 360° – 140° = 220°

(At any point the angle is 360°) …(2)

Now, in quadrilateral ABCD,

∠ADC + ∠BCD + ∠B AD + ∠ABC = 360°

⇒ 58° + 53° + x + 220° = 36o° [using (1) and (2)]

⇒ 331° + x = 360° ⇒ x = 360° – 331°

⇒ x = 29° Answer:

(ii) ∵ DE || AB (given)

∴ ∠ECB = ∠CBA (Alternate angles)

⇒ 75° = ∠CBA

∴ ∠CBA =75°

∵ AD || BC (given)

∴ (x + 66°) + (75°) =180° (co-interior angles are supplementary)

⇒ x + 66° + 75° = 180° ⇒ x + 141° = 180°

⇒ x = 180°- 141°

A = 39° …(1)

Now, in ∆AMB,

x + 30° + ∠AMB = 180° (sum of all angles in a triangle is 180°)

⇒ 39° + 30° + ∠AMB = 180° [From (1)]

⇒ 69° + ∠AMB = 180°

⇒ ∠AMB = 180°- 69°

⇒ ∠AMB = 111° ….(2)

∵ ∠AMB = y (vertically opposite angles)

⇒ 111° = y [From (2)]

∴ y = 111°

Hence, A = 39° and y =111°

(iii) In ∆ABD

AB = AD (given)

∠ABD = ∠ADB (∵ equal sides have equal angles opposite to them)

⇒ ∠ABD = 42° [∵ ∠ADB = 42° (given)]

∵ ∠ABD + ∠ADB + ∠BAD = 180°

(Sum of all angles in a triangle is 180°)

⇒ 42° + 42° + y = 180° ⇒ 84° + y = 180°

⇒ y = 180° – 84° ⇒ y = 96°

∠BCD = 2 × 26° = 52°

In ∠BCD

∵ BC = CD (given)

∴ ∠CBD = ∠CDB = x [equal side have equal angles opposite to them]

∴ ∠CBD + ∠CDB + ∠BCD = 180°

⇒ x + x + 52° = 180° ⇒ 2A = 180° – 52°

⇒ 2x = 128° ⇒ x = \(\frac{128^{\circ}}{2}\) ⇒ x = 64°

Hence, x = 64° and y = 90°

Question 5.

Find the size of each lettered angle in the following figures :

Answer:

(i) Here AB || CD and BC II AD (given)

∴ ABCD is a || gm

∴ y = 2 × ∠ABD

⇒ y = 2 × 53° = 106° ….(1)

Also, y + ∠DAB = 180°

⇒ 106° + ∠DAB = 180°

⇒ ∠DAB = 180° – 106° ⇒ ∠DAB = 74°

∴ x = \(\frac{1}{2}\) ∠DAB (∵ AC bisect ∠DAB)

⇒ x = \(\frac{1}{2}\) × 74° = 37°

and ∠D AC = x = 37 ….(2)

∴ ∠DAC = z (Alternate angles) ….(3)

From (2) and (3),

z = 37°

Hence,x = 37°, y = 106°, z = 37°

(ii) ∵ ED is a st. line

∴ 60° + ∠AED = 180° (linear pair)

⇒ ∠AED = 180° – 60°

⇒ ∠AED = 120° …(1)

∵ CD is a st. line

∴ 50° +∠BCD = 180° (linear pair)

⇒ ∠BCD = 180° – 50°

⇒ ∠BCD = 130° ….(2)

In pentagon ABCDE

∠A + ∠B + ∠AED + ∠BCD + x = 540° (Sum of interior angles in pentagon is 540°)

⇒ 90° + 90° + 120° + 130° + x = 540°

⇒ 430° + x = 540° ⇒ x = 540° – 430°

⇒ x = 110°

Hence, value of x = 110°

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

∴ 60° + y = 180°and x + 110° = 180°

⇒ y = 180° – 60° and x = 180° – 110°

⇒ y = 120° and x = 70°

∵ CD||AF (given)

∴ ∠FAD = x (Alternate angles)

⇒ ∠FAD = 70° ….(1)

In quadrilateral ADEF,

∠FAD + 75° + z + 130° = 360°

⇒ 70° + 75° + z + 130° = 360° [using(1)]

⇒ 275° + z = 360° ⇒ z = 85°

Hence, x = 10°, y = 120° and z = 85°

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:

Here ABCD and DCFE is a rhombus and square respectively.

∴ AB = BC = DC = AD ….(1)

Also DC = EF = FC = EF ….(2)

From (1) and (2),

AB = BC = DC = AD = EF = FC = EF ….(3)

∠ABC =56° (given)

∠ADC = 56° (opposite angle in rhombus are equal)

∴ ∠EDA = ∠EDC +∠ADC = 90° + 56° = 146°

In ∆ADE,

DE = AD [From (3)]

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

⇒ DEA = DAE = \(\frac{180^{\circ}-\angle \mathrm{EDA}}{2}\)

= \(\frac{180^{\circ}-146^{\circ}}{2}=\frac{34^{\circ}}{2}\) = 17°

⇒ ∠DAG = 17°

Also, ∠DEG =17°

∴ ∠FEG = ∠E – ∠DEG

= 90°- 17° = 73°

In rhombus ABCD,

∠DAB = 180° – 56° = 124°

∠DAC = \(\frac{124^{\circ}}{2}\) (∵ AC diagonals bisect the ∠A)

∠DAC =62°

∴ ∠GAC = ∠DAC – ∠DAG

= 62° – 17° = 45°

In ∆EDG,

∠D + ∠DEG + ∠DGE = 180° (Sum of all angles in a triangle is 180°)

⇒ 90° + 17° + ∠DGE = 180°

⇒ ∠DGE = 180° – 107° = 73° ….(4)

Hence, ∠AGC = ∠DGE ….(5) (vertically opposite angles)

From (4) and (5)

∠AGC = 73°

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 8 cm. AB = BC = CD = DA = 8cm.

Let ∠A = 60°

∴ ∆ABD is an equilateral triangle

∴ AB = BD = AD = 8cm.

∵ Diagonals of a rhombus bisect each other eight angles.

∴ AO = OC, BO = OD=4 cm.

and ∠AOB = 90°

Now in right ∆AOB,

AB = AO^{2} + OB^{2} (Pythagoras Theorem)

⇒ (8)^{2} = AO^{2} + (4)^{2}

⇒ 64 = AO^{2} + 16 ,

⇒ AO^{2} = 64 – 16 = 48 = 16 + 3

∴ AO = \(\sqrt{16 \times 3}=4 \sqrt{3}\) cm.

But AC = 2 AO

∴ AC = 2 x \(4 \sqrt{3}=8 \sqrt{3}\) cm

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:

Given : AB = 5 cm, AD = 2.5 cm and ∠BAD =45°.

Required : (i) To construct a parallelogram ABCD.

(ii) If the bisector of ∠BAD meets DC at E then prove that ∠AEB = 90°.

Steps of Construction:

- Draw AB = 5.0cm.
- Draw ∠BAP = 45° on side AB.
- Take A as centre and radius 2.5 cm cut the line AP at D.
- Take D as centre and radius 5.0 cm draw an arc.
- Take B as centre and radius equal to 2.5 cm cut the arc of step (4) at C.
- Join BC, and CD.
- ABCD is the required parallelogram.
- Draw the bisector of ∠BAD, which cuts the DC at E.
- Join EB.
- Measure the ∠AEB which is equal to 90°. (Q.E.D.)