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°

Find the value of a, x ,y and z. Click below Answers ,Which are correct.?

Solution: MN //PQ ,CD is transversal line. □ <EFG=<CEN [Corresponding angle] Z°=40° □<EGF=<QGV[Vertically Opposite Angle] a°=50°

In Triangle EFG <EFG+<EGF+<GEF=180°[Sum of three angles of triangle] Z°+a°+y°=180° or,40°+50°+y=180° or,90°+y°=180° or,y°=180°-90° y°=90°

<UEC=<FEG[Vertically Opposite Angle] or, x°=y° x°=90°

<UEC=<FEG[Vertically Opposite Angle] or, x°=y° x°=50°

Find the value of x, y and z. Which are correct answers?click

Solution:]] AB //CD ,EB is a transversal line. <CGE=<ABG [corresponding angle] z°=38°

<CGB+<ABG=180° [Co-interior angle] or y°+38°=180° or,y°=180°-38° y°=142°

EB//FD. CD is transversal line. <GDF=<CGE[Corresponding angle] or,x°=z° x°=38°

CGB+<ABG=180° [Co-interior angle] or y°+38°=180° or,y°=180°-38° y°=152°

Find the value of a,x,y and z. Which are correct answers? Click.

AB//CD is given, a parallel line MN is drawn between them. So AB // MN // CD <EUA+<EUB=180°[Supplementary angle's sum] or, a°+50°=180° or,a°=180°-50° or,a=130°

□ <DVF=<CVT[Vertically Opposite Angle] Y°=45° <FVC+<DVF=180°[Supplementary angle's sum] or, z°+y°=180° z°+45°=180° or,z°=180°-45° z°=135°

AB//MN • <UTN=<EUB[corresponding angle] <UTN=50° • <NTV=<TVC[alternate angle] <NTV=45° • < UTV=<UTN+<VTN[whole part of Axioms] or,x°=50°+45° x°=95°

<EUA+<EUB=180°[Supplementary angle's sum] or, a°+50°=180° or,a°=180°-50° or,a=120°

Find the value of a ,x ,y and z from given figure alone side. Which answers are correct?click

Solution:AB//CD//EF and GI is transversal line. <DJK=FKI[corresponding angle] x°=75° <BIJ=<FKI [corresponding angle] a°=75°

<JKF+<IKF=190°[Supplementary angle's sum] or, z°+75°=180° or,z°=180°-75° or,z°=105° <IJD=<JKF[corresponding angle] or, y°=z° or,y°=105°

<BIJ=<DJK=<FKI[corresponding angle] a°=x°=75° <IJD=<JKF[corresponding angle] or, y°=z° or,y°=105°

<JKF+<IKF=190°[Supplementary angle's sum] or, z°+75°=180° or,z°=180°-75° or,z°=15°

Find the value of a, x ,y and z. Which are correct answers?

Solution: <AIJ=<EIK [Vertically Opposite Angle. ] x°=115° <FJC=<JIA[corresponding angle] or,y°=x° y°=115°

<IKL=<GKB[vertically opposite angle] z°=58° <KLD= <LkI[alternate angle] or,a°=z° a°=58°

<IKL=<GKB[vertically opposite angle] z°=58° <KLD= <GKB[corresponding angle] a°=58

<IKL=<GKB[vertically opposite angle] z°=48° <KLD= <LkI[alternate angle] or,a°=z° a°=48°

Find the value of x, y and z . Which are correct answers?

Solution: triangle CDE is isosceles triangle. <CDE=<CED[base angle of isosceles triangle] y°=x° <ACD=<CED+<CDE[Exterior angle of a triangle equal to sum of its opposite non adjacent interior angles ] 120°=x°+y° or, 120°=x°+x° or,2x°=120° x°=120°/2 : . x°=60° Continue

Y°=X °=60° <CED+<DEB=180°[adjacent angle formed on Straight line. or, Z °+ X°=180° or, Z°+60°=180° Z°=180°-60° Z°=120°

<CED+<DEB=180°[adjacent angle formed on Straight line. or, Z °+ X°=180° or, Z°+60°=180° or,Z°=180°+60° Z°=240°

<ACD=<CED+<CDE[Exterior angle of a triangle equal to sum of its opposite non adjacent interior angles ] 120°=x°+y° or, 120°=x°+x° or,3x°=120° x°=120°/3 : . x°=40°

Find the value of x and y. Which are correct? Click.

Solution: <AGF+<IGF=180°[adjacent angle formed on Straight line] or, 60° + x°= 180° or,x=180°-60° x=120°

<GHi+<HIG=<FGI[Exterior angle of a triangle is equal to sum of opposite non adjacent interior angles ] or,45°+y°=x° or,y°=x°-45° or,y°=120°-45° y°=75°

Solution: <AGF+<IGF=170°[adjacent angle formed on Straight line] or, 60° + x°= 170° or,x=170°-60° x=110°

<GHi+<HIG=<FGI[Exterior angle of a triangle is equal to sum of opposite non adjacent interior angles ] or,45°+y°=x° or,y°=x°-45° or,y°=120°-45° y°=85°

Find the value of (i) p b - c × p a - b × p c - a (ii) x y- x × x x- y × x z - x (iii)x r- p × x p- q × x r - p

(i)Solution: p b - c x p a - b x p c - a =p b - c +a-b+ c-a =p o =1

(ii) Solution: x y- x × x x- y ×x z - x = x y- x+ x- y+ z- x = $$x^o$$ =1

(iii)Solution: x r- p × x p- q × x r - p =x r- p +p -q + r-p = $$x^o$$ =1

(iii)Solution: x r- p × x p- q × x r - p =x r- p +p -q + r-p = $$x^o$$ =0

Q)find the area of given rectangle. Q) if l= (5a-b) m,b=(2a+b)m find area of rectangle.

Solution: A= lXb = (5a-b) (2a+b) =5a(2a+b)-b(2a+b) $$10a^2+5ab-2ab-b^2$$ = $$10a^2+3ab-b^2$$

Solution: A= lXb = (5a-b) (2a+b) =5a(2a+b)+b(2a+b) $$10a^2+5ab+2ab+b^2$$ = $$10a^2+3ab+b^2$$

class 7 math quiz

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class 7 math quiz

Quiz

•

Mathematics

•

7th Grade

•

Practice Problem

•

Hard

Dor Bahadur Aryal

Used 53+ times

FREE Resource

36 questions

Show all answers

MULTIPLE SELECT QUESTION

1 min • 5 pts

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.

MULTIPLE SELECT QUESTION

2 mins • 5 pts

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°

MULTIPLE SELECT QUESTION

45 sec • 5 pts

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°

MULTIPLE SELECT QUESTION

45 sec • 5 pts

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.

MULTIPLE SELECT QUESTION

45 sec • 5 pts

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°

MULTIPLE SELECT QUESTION

45 sec • 5 pts

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

MULTIPLE SELECT QUESTION

2 mins • 5 pts

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

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class 7 math quiz

36 questions

घ) Solution:Interior<AOB+ exterior <AOB=360°or,80°+7y=360°or,7y=360°-80°or,7y=280°or,y=280°/7y=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°/5x=27°

Find the value of x , y and a from given figure:(answer are given below click which are correct.)

See figure,Click if answer is correct.

Click if given below statement are correct .

If AB//CD, Find the value of x, y and z from given figure.click,which are correct answer?

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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.)

If AB//CD, Find the value of x, y and z from given figure.
click,which are correct answer?