ML Aggarwal Solution Class 9 Chapter 13 Rectilinear Figures Test

 Test

Question 1

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

Sol :
Sum of interior angles of a polygon
=(2n-4)×90°
⇒4320°=(2n-4)×90°
⇒$\frac{4320^{\circ}}{90^{\circ}}=(2 n-4)$
⇒$\frac{432}{9}=2 n-4$
⇒48=2n-4
⇒48+4=2n
⇒52=2n
⇒2n=52
⇒$n=\frac{52}{2}=26$

Hence , the polygon have 26 sides


Question 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.

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

⇒$\frac{(2 n-4) \times 90^{\circ}}{n}: \frac{360}{n}=5: 1$

⇒(2n-4)×90° : 360=5 : 1

⇒$\frac{(2 n-4) \times 90^{\circ}}{360}=\frac{5}{1}$

⇒$\frac{2 n-4}{4}=\frac{5}{1}$

⇒2n-4=5×4

⇒2n-4=20

⇒2n=20+4

⇒2n=24

⇒$n=\frac{24}{2}$

⇒n=12

Hence, number of sides of regular polygon =12


Question P.Q.

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

Sol :
∵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.













Sol :

In the figure , ABCD is a parallelogram side CB is produced to E such that BE=BC

BD and AE are joined

To prove : AEBD is a parallelogram 

Proof : In ΔAEB and ΔBDC

⇒EB=BC  (given)

⇒∠ABE=∠DCB  (corresponding angles)

⇒AB=DC (opposite sides of parallelogram)

∴ΔAEB≅ΔBDC  (S.A.S 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.

Sol :
Given : In isosceles ΔABC, AB=AC.
AD is the bisector of ext.∠PAC and CD||BA






To prove : (i) ∠DAC=∠BCA

(ii) ABCD is a parallelogram 

Proof : In ΔABC

∵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 parallelogram


Question 3

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

Sol :
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 diagonals are equal

∴AC=BD

⇒Now in ΔABC,

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

∴PQ||AC and $\mathrm{PQ}=\frac{1}{2} \mathrm{AC}$ ..(i)

Similarly in ΔADC,

⇒S and R mid points of CD and AD

∴SR||AC and $\mathrm{SR}=\frac{1}{2} \mathrm{AC}$...(ii)

From (i) and (ii)

⇒PQ||SR and PQ=SR

∴PQRS is a parallelogram

Now in ΔAPS and ΔBPQ

⇒AP=BP  (P is mid point of AB)

⇒AS=BQ (Half of equal sides)

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

⇒∴ΔAPS≅ΔBPQ

∴PS=PQ

But there are the adjacent sides of a parallelogram

∴Sides of PQRS is a rhombus

Hence PQRS is a rhombus

Hence proved


Question 4

Find the size of each lettered angle in the Following Figures.






















Sol :

(i) ∵CDE is a straight line

∴∠ADE+∠ADC=180°

⇒122°+∠ADC=180°

⇒∠ADC=180°-122°

⇒∠ADC=58°...(1)

⇒∠ADC=360°-140°=220° 

(At any point the angle is 360°)...(2)

Now, in quadrilateral ABCD

⇒∠ADC+∠BCD+∠BAD+∠ABC=360°

⇒58°+53°+x+220°=360°  [using (1) and (2)]

⇒331°+x=360°

⇒x=360°-331°=29°


(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°

∴x=39°...(1)

Now , in ΔAMB,

⇒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°=111°...(2)

∵∠AMB=y  (vertically opposite angles)

⇒111°=y  [From (2)]

∴y=111°

Hence , x=39° and y=111°


(iii) In ΔABD,

⇒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°=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°

⇒2x=180°-52°

⇒2x=180°

⇒$x=\frac{128^{\circ}}{2}$

⇒x=64°

Hence ,x =64° and y=90°


Question 5

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





















Sol :

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

∴ABCD is a parallelogram

∴y=2×∠ABD

⇒y=2×53°=106°...(1)

Also, y+∠DAB=180°

⇒106°+∠DAB=180°

⇒∠DAB=180°-106°

⇒∠DAB=74°

∴$x=\frac{1}{2} \angle \mathrm{DAB}$  (∵AC bisects ∠DAB )

⇒$x=\frac{1}{2} \times 74^{\circ}=37^{\circ}$

and ∠DAC=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 straight line

∴60°+∠AED=180° (linear pair)

⇒∠AED=180°-60°=120°...(1)

∵CD is a straight 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°=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||AD  (given)

∴∠FAD=x  (alternate angles)

⇒∠FAD=70° ...(1)


In quadrilateral ADEF,

⇒∠FAD+70°+z+130°=360°

⇒70°+75°+z+130°=360° [using (1)]

⇒275°+z=360°

⇒z=85°

Hence ,x=70°, 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.







Sol :

Here ABCD and DCFE is a rhombus and square respectively
∴AB=BC=DC=AD ...(1)
From (1) and (2)
⇒AB=BC=DC=AD=EF=FC=EF..(3)
⇒∠ABC=56°  (given)
⇒∠ADC=56° (opposite angles in rhombus are equl)

∴∠EDA=∠EDC+∠ADC=90°+56°=146°

In ΔADE,

⇒DE=AD

⇒∠DEA=∠DAE

(equal sides have equal opposite angles)

⇒∠DEA=∠DAG=$\frac{180^{\circ}-\angle \mathrm{EDA}}{2}$

⇒$=\frac{180^{\circ}-146^{\circ}}{2}=\frac{34^{\circ}}{2}=17^{\circ}$

⇒∠DAG=17°

Also, ∠DAG=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 ΔEDG

⇒∠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.

Sol :
Each side of rhombus ABCD is 8cm
∴AB=BC=CD=DA=8cm






Let ∠A=60°

∴ΔABD is an equilateral triangle

∴AB=BD=AD=8 cm

∵Diagonals of a rhombus bisect each other eight angles.

∴AO=OC, BO=OD=4 cm and ∠AOB=90°

Now in right ΔAOB

⇒AB2=AO2+OB2 (Pythagoras theorem)

⇒(8)2=AO2+(4)2 

⇒64=AO2+16 

∴AO$=\sqrt{16 \times 3}=4 \sqrt{3}$ cm

But AC=2AO

∴AC$=2 \times 4 \sqrt{3}=8 \sqrt{3}$ 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.

Sol :
Given : AB=5 cm, AD=2.5 cm and ∠BAD=45°
Required :
(1) To construct a parallelogram ABCD
(2) If the bisector of ∠BAD meets DC at E then prove that ∠AEB=90°








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

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