class 7 math quiz

• Figure is scalene triangle ABC.

• It's angles sum equal to 180°.

• It's made by using pencil compass.

• It's made by using set Square and protactor.

• Click if it's a correct answer.

  क) Solution:

  <AOB+<AOC=180°[° .° Being Supplementary angle's sum]

  or,2x+3x+45°=180°

  or,5x+45°=180°

  or,5x=180°-45°

  or,5x=135°

  or,x=135°/5

  x=27°

• Click if it's a correct answer.

  क) Solution:

  <AOB+<AOC=180°[° .° Being Supplementary angle's sum]

  or,2x+3x+45°=180°

  or,5x+45°=180°

  or,5x=180°-45°

  or,5x=135°

  or,x=135°/5

  x=17°

• Click if it's a correct answer.

  ग) Solution:<AOD +<AOB+<BOC+<COD=360°[Sum of angles formed on a point=360°]

  or,7x+50°+3x+110°=360°

  or,10x+160°=360°

  or, 10x=360°-160°

  or,10x=200°

  or,x=200°/10

  x=20°

• Click if it's a correct answer.

  घ) Solution:

  Interior<AOB+ exterior <AOB=360°

  or,80°+7y=360°

  or,7y=360°-80°

  or,7y=280°

  or,y=280°/7

  y=50°

Answer explanation

घ) Solution:

Interior<AOB+ exterior <AOB=360°

or,80°+7y=360°

or,7y=360°-80°

or,7y=280°

or,y=280°/7

y=40°

क) Solution:

<AOB+<AOC=180°[° .° Being Supplementary angle's sum]

or,2x+3x+45°=180°

or,5x+45°=180°

or,5x=180°-45°

or,5x=135°

or,x=135°/5

x=27°

Find the value of x , y and a from given figure:

(answer are given below click which are correct.)

• ( II) <AEC+<AED=180°[Sum of angle formed on straight line.]

  Or,3a+y=180°

  or, 3×40°+y=180°

  or 120°+y=180°

  or,y=180°-120°

  y=60°

• (III) <BEC = <AED [Vertically Opposite Angle ]

  or, x = y

  or, x= 60°

  [So,a=40°,y=60° and x=60°]

• (III) <BEC = <AED [Vertically Opposite Angle ]

  or, x = y

  or, x= 60°

• Line AB and DC are Perpendicular line.

  Line ADand BC are Perpendicular line.

  Line AD and AB are Parallel line.

  Line BC and AB are Parallel line.

  Line AD and AB are Parallel line.

  Line BC and CD are Parallel line.

• Line AB and DC are Parallel line.

  Line AD and BC are Parallel line.

  Line AD and AB are Perpendicular line.

  Line BC and AB are Perpendicular line.

  Line AD and AB are Perpendicular line.

  Line BC and CD are Perpendicular line.

See figure,Click if answer is correct.

• ख) Solution:

  <POR+<ROQ=180°[° .° Being Supplementary

  angle's sum]

  or,70°+2y+10°=180°

  or,2y+80°=180°

  or,2y=180°-80°

  or,2y=100°

  or,y=100°/2

  y=50°

  Continue

• ख) Solution:( middle part)

  Again,<POR=<QOS [Being vertically

  opposite angle]

  or,70°=b

  b=70°

  Continue

• ख) Solution:(last part)

  Now,<POS=<ROQ [Being vertically

  opposite angle]

  x=2y+10°

  or,x=2×50°+10°

  or,x=100°+10°

  x=110°

• ख) Solution:

  <POR+<ROQ=180°[° .° Being Supplementary

  angle's sum]

  or,70°+2y+10°=180°

  or,2y+80°=180°

  or,2y=180°-80°

  or,2y=100°

  or,y=100°/2

  y=60°

  Again,<POR=<QOS [Being vertically

  opposite angle]

  or,70°=b

  b=75°

  Now,<POS=<ROQ [Being vertically

  opposite angle]

  x=2y+10°

  or,x=2×50°+10°

  or,x=100°+10°

  x=120°

Answer explanation

ख) Solution:

<POR+<ROQ=180°[° .° Being Supplementary

angle's sum]

or,70°+2y+10°=180°

or,2y+80°=180°

or,2y=180°-80°

or,2y=100°

or,y=100°/2

y=50°

Again,<POR=<QOS [Being vertically

opposite angle]

or,70°=b

b=70°

Now,<POS=<ROQ [Being vertically

opposite angle]

x=2y+10°

or,x=2×50°+10°

or,x=100°+10°

x=110°

Click if given below statement are correct .

• In letter X and L

  AD and BC are intersecting line.

  AD and BC are not parallel line.

  AC and CZ are perpendicular line.

• In letter X and L

  AD and BC are parallel line.

  AD and BC are not intersecting line.

  AC and CZ are not perpendicular

• Click if Answers are correct.

  क) उत्तर,

  यहा,त्रिभुज ABC को भुजा (AB )=12 cm

  त्रिभुज ABC को भुजा (BC)=5 cm

  त्रिभुज ABC को भुजा (AC )=13cm

  त्रिभुज ABC को परिमिति (perimeter)=?

  अब, P =AB+BC+AC

  =12cm+5cm+13cm

  =30cm

• Click if Answers are correct.

  ख) उत्तर,

  यहा,त्रिभुज XYZको भुजा (XY)=3 cm

  त्रिभुज XYZको भुजा (YZ)=4 cm

  त्रिभुज XYZ को भुजा (XZ)=3.9cm

  त्रिभुज XYZ को परिमिति (perimeter)=?

  अब, P =XY+YZ+ZX

  =3cm+4cm+3.9cm

  =10.9cm

• Click if Answers are correct.

  ग) उत्तर,

  यहा,त्रिभुज PQR को भुजा (PQ )=5.5 cm

  त्रिभुज PQR को भुजा (QR)=8.5 cm

  त्रिभुज PQR को भुजा (PR )=5cm

  त्रिभुज PQR को परिमिति (perimeter)=?

  अब, P =PQ+QR+QR

  =5.5cm+8.5cm+5cm

  =19cm

• Click if Answers are correct.

  क) उत्तर,

  यहा,त्रिभुज ABC को भुजा (AB )=12 cm

  त्रिभुज ABC को भुजा (BC)=5 cm

  त्रिभुज ABC को भुजा (AC )=13cm

  त्रिभुज ABC को परिमिति (perimeter)=?

  अब, P =AB+BC+AC

  =12cm+5cm+13cm

  =20cm

• Click if Answers are correct.

  ग) उत्तर,

  यहा,त्रिभुज PQR को भुजा (PQ )=5.5 cm

  त्रिभुज PQR को भुजा (QR)=8.5 cm

  त्रिभुज PQR को भुजा (PR )=5cm

  त्रिभुज PQR को परिमिति (perimeter)=?

  अब, P =PQ+QR+QR

  =5.5cm+8.5cm+5cm

  =29cm

• [Click if Answer is correct]

  Solution:

  In Here,

  In Triangle ABC,

  Side BC=?

  In Triangle AED,

  SideAD=?

  There's same symbols in both Triangles so,

  AD=AC=4cm

  BC=ED=4.5cm

• Here,

  In Triangle ABC

  Triangle's side(AB)=4.3cm

  Triangle's side(BC)=4.5cm

  Triangle's side(AC)=4cm

  Triangle's Perimter(P)=?

  Now, P = AB+BC+AC

  =4.3cm+4.5cm+4cm

  =12.8cm

• Here,

  In Triangle AED

  Triangle's side(AE)=3.7cm

  Triangle's side(ED)=4.5cm

  Triangle's side(AD)=4cm

  Triangle's Perimter(P)=?

  Now, P = AE+ED+DA

  =3.7cm+4.5cm+4cm

  =12.2cm

• Here,

  In Triangle ABC

  Triangle's side(AB)=4.3cm

  Triangle's side(BC)=4.5cm

  Triangle's side(AC)=4cm

  Triangle's Perimter(P)=?

  Now, P = AB+BC+AC

  =4.3cm+4.5cm+4cm

  =22.8cm

• Here,

  In Triangle AED

  Triangle's side(AE)=3.7cm

  Triangle's side(ED)=4.5cm

  Triangle's side(AD)=4cm

  Triangle's Perimter(P)=?

  Now, P = AE+ED+DA

  =3.7cm+4.5cm+4cm

  =13.2cm

If AB//CD, Find the value of x, y and z from given figure.

click,which are correct answer?

• Solution:

  In triangle EFG,

  <EFG+<FGE+<GEF=180°[Sum of three angles of triangle]

  or,x°+2x-5°+2x+5°=180°

  or,5x°=180°

  or,x°=180°/5

  x°=36°

  Continue

• <AEF=<EFG[alternate angle]

  or,Z=2x-5

  Or, z=2×36-5

  or,z=72-5

  or,z=67°

• <BEG=<EGF[alternate angle]

  or,y= 2x+5°

  or,y=2×36+5

  or,y=72+5

  or,y=77

• <BEG=<EGF[alternate angle]

  or,y= 2x+5°

  or,y=2×36+5

  or,y=72+5

  or,y=87°

• Click if Answer is correct.

  क)Solution:

  AB//CD and EC is transversal line

  □ <AFC=<EFB[Vertically opposite angle]

  y°=49°

  □ <FCD=<EFB[Corresponding angle]

  x°=49°

• Click if Answer is correct.

  ख) Solution

  □ <UVW+<SVW=180°

  [Supplementary angle]

  or,y°+100°=180°

  or,y°=180°-100°

  y°=80°

  □ <BQV=<UVW [corresponding angle ] or, x°=y°

  x°=80°.

  Continue

• Click if Answer is correct.

  □ <PBU=<BQV[Corresponding angle]

  or, a°=x°

  a°=80°

  □ <TUV=<WVS[corresponding angle]

  b°=100°

• Click if Answer is correct.

  □ <TUV+<WVU=180°[Co-interior angle]

  or,b°+y°=180°

  or,b°+80°=180°

  or,b°=180-80°

  b°=100°

• Click if Answer is correct.

  □ <TUV+<WVU=180°[Co-interior angle]

  or,b°+y°=180°

  or,b°+80°=180°

  or,b°=180+80°

  b°=260°