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Q.1. Find the value of  x in each of the following figures:p1

Answer:

i)

\angle ABC=\angle BAC=35^{\circ} \ (since \ BD=DA)

\angle BDA = 180 - 35 - 35 = 110^{\circ}

Similarly, \angle ECA=\angle AEC=28^{\circ} (since EC=EA)

\angle AEC=180-28-28=124^{\circ}

 x+ \angle ADE+ \angle AED=180

 x+(180-110)+(180-124)=180

\Rightarrow x=54^{\circ}

ii)

 \angle OBC= \angle OCB=40^{\circ}  (Since \ OB=OC)

 \angle BOC=180-40-40=100

 \angle OCA= \angle OAC=30^{\circ}  (Since \ OA=OC)

 \angle AOC=180-30-30=120^{\circ}

We know  \angle BOC+ \angle COA+ \angle AOB=360

 \Rightarrow \angle AOB=(360-100-120)=140^{\circ})

 \Rightarrow  2x+140=180

\Rightarrow x=20^{\circ}

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Q.2. State giving reasons, whether it is possible to construct a triangle or not with sides of lengths:

  • 3 \ cm, \ 4 \ cm,\ 7 \ cm
  • 9 \ cm, \ 8 \ cm,\ 16 \ cm
  • 7 \ cm, \ 9.2 \ cm, \ 6.7 \ cm
  • 3 \ cm, \ 6.2 \ cm, \ 10.8 \ cm

Answer (Note: The sum of any two sides of a triangle is always greater than the third side)

i) Not Possible to construct the triangle because (3+4 not greater than 7

ii) Possible to construct the triangle as sum of any two sides is greater then the third side.

iii) Possible to construct the triangle as sum of any two sides is greater then the third side.

iv) Not possible to construct a triangle as sum of 4.3+6.2<10.8

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Q.3. In \Delta ABC, \angle B=60^{\circ}    \ and\  \angle C=45^{\circ} . Find \angle A . Name i) the largest side of \Delta ABC , ii) the smallest side of \Delta ABC iii) write the sides of \Delta ABC in ascending order of their lengths.

Answer:

\angle A=180-60-45=75^{\circ}   .   Therefore

i) the largest side of \Delta ABC=BC

ii) the smallest side of \Delta ABC=AB

iii) AB < AC < AB

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Q.4. In \Delta XYZ, \angle X=53^{\circ}    \ and\  \angle Y=67^{\circ}    .  Name i) the smallest side of \Delta XYZ , ii) the largest side of \Delta XYZ iii) write the sides of \Delta XYZ in ascending order of their lengths.

Answer:

\angle Z=180-53-67=60^{\circ}    .   Therefore

i) the largest side of \Delta XYZ=XZ

ii) the smallest side of \Delta XYZ=YZ

iii) YZ < XY <XZ

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Q.5. In \Delta PQR, \angle P= \angle R=64{\circ}    .  Name the smallest side and the equal sides.

Answer

\angle Q=180-64-64=52^{\circ} 

Smallest side =PR

Equal Sides =QR \ \&\  PQ

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Q.6 In \Delta ABC, \angle A= \angle B \ and\  \angle C=100^{\circ} . Name the largest side and the equal sides.

Answer:

x+x+100=180 (let \ \angle A= \angle B=x)

\Rightarrow x=40^{\circ} 

Largest side =AB  , Equal Sides =AC \ \&\  BC

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Q.7. In \Delta DEF, \angle D \colon \angle E \colon \angle F=7 \colon 3\colon 5 . Name the smallest side and the largest sides.

Answer:

7x+3x+5x=180 \Rightarrow x=12

\angle D=84^{\circ}   , \angle E=36^{\circ}    \ and\  \angle F=60^{\circ} 

Largest side =EF , Smallest Side =DF

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Q.8. In the adjoining figure, AC=DC,  \angle CAD=50^{\circ}    \ and\  \angle BAD=23^{\circ}  . Find \angle ADC, \angle ACD, \angle ADB \ and\  \angle ABD . Also show that AD>AC, \ AD>BD \ and\  AB>BC p8

Answer:

AD=AD \Rightarrow  \angle ADC= \angle DAC=50^{\circ} 

\angle ACD=180-50-50=80^{\circ} 

Therefore \angle ABD=180-50=130^{\circ} 

\Rightarrow  \angle ABD=180-130-23=27^{\circ} 

Since \angle ACD> \angle ADC \Rightarrow  AD>AC

Since \angle  ABD> \angle BAD \Rightarrow AD>BD

Since \angle ACB> \angle BAC\Rightarrow AB>BC

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Q.9. In the adjoining figure, BA \parallel CD \ and\ AB=BC . If \angle BAC=56^{\circ} \ and \ \angle COD=85^{\circ} , find the values of x, \ y \ and \ z .p9

Answer:

Since AB=BC, \angle BAC= \angle ACB=56^{\circ} 

x+y=180-112=68 ...i)

\angle AOB= \angle DOC=85^{\circ}    (opposite angles)

Therefore x+56+85=180 \Rightarrow x=39

Which implies that y=68-39=29

Becasue AB \parallel CD, \angle ABC=\angle DCX=x+y=39+29=68^{\circ} 

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Q.10. In the adjoining figure, BA \parallel CD, BA=BC and CE=DE . If \angle ABC=66^{\circ}   , \angle DCE=48^{\circ}   \ and\  \angle DAC=63^{\circ}   , find the value of x, \ y \ and\  z .p10

Answer:

\angle BAC= \angle BCA=y

2y+66=180 \Rightarrow y=57^{\circ} 

\angle ECD= \angle EDC=48

x+48+48=180 \Rightarrow x=84^{\circ} 

Since BA \parallel CD, \angle BAC= \angle ACD=y=57

Therefore 63+57+z=180 \Rightarrow z=60^{\circ} 

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Q.11. In the adjoining figure, AD=BD=CD and \angle CAD=40^{\circ} . Show that AC>AD, AB>AD, AC>AB .p11

Answer:

Since AD=CD, \angle DAC= \angle DCA=40^{\circ} 

\angle ADB=180-(180-40-40)=80^{\circ} 

Since AD=BD, \angle DAB= \angle DBA

\angle ADC=100 \Rightarrow  \angle ADB=80^{\circ} 

\angle DAB= \angle DBA=50^{\circ} 

Since\angle ADC> \angle  ACD \Rightarrow AC>AD

Since \angle  ADB> \angle  ABD \Rightarrow AB>AD

Since \angle  ABD> \angle  ACB\Rightarrow AC>AB

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Q.12. If D is a point on side BC of \Delta ABC , prove that: i) \ AB+BD>AD \ ii)\  AC+CD>AD \ iii)\  AB+BC+AC>2AD  p12

Answer:

In \Delta ABD, because it is a valid triangle,  AB+BD>AD

In \Delta ADC, because it is a valid triangle,  AC+CD>AD

In \Delta ABC , because it is a valid triangle,

Add the above two relations

AB+(BD+CD)+AC>2 AD

\Rightarrow AB+BC+AC>2AD

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Q.13. In the adjoing figure, ABCD is a quadrilateral. Prove that: i)AD+DC>AC ii)AB+BC>AC iii) AB+BC+AD+DC>2AC iv) AD+AB+BC>CD p13

Answer:

In \Delta ADC, AD+DC>AC

In \Delta ABC AB+BC>AC

Adding the above two expresions we get

AB+BC+AD+DC>2AC

AB+BC>AC and AD+AC>CD . Add these two expressions we get

AD+AB+BC>CD

E Board: Suggested Books     ICSE Board:  Foundation Mathematics 
Class 8: Reference Books     Class 8: NTSE Preparation 
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