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°

Solution:

here,

(I) $$\angle AEC=\angle DEB\left[\because Vertically\ opposite\ angle\right]$$   $$or,\ 3a=\ 2a+40^o$$  

or,3a-2a=40°

a=40°

( 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.

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°

In letter E

AX is parallel to BY.( or, AX//BY)

AX is parallel to CZ.( or, AX//CZ)

AX is perpendicular to AB.( or, AX $$\perp$$  AB)

YB is perpendicular to AC.( or, YB $$\perp$$  AC)

CZ is perpendicular to AC.( or, CZ $$\perp$$  AC)

In letter H

YX is perpendicular to AB.( or, YX $$\perp$$  AB)

XY is perpendicular to CD.( or, XY $$\perp$$  CD)

AB is parallel to CD.( or, AB//CD)

In letter N and T

AB is parallel to CD.( or, AB//CD)

DB is perpendicular to AC.( or, DB $$\perp$$  AC)

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

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°

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°

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°

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°

Click if Answer is correct.

Solution:

CD //IJ ,AB is transversal line.

□ <FKI= <AFC Corresponding angle]

y°=50°

□ <HKB=<FKG[Vertically Opposite Angle]

x°=90°

<IKF+<FKG+<GKJ=180°[sum of Straight angles]

or,y+90°+z°=180°

or,50°+90°+z°=180°

or,140°+z°=180°

or,z°=180°-140°

Z=40°

<IKF+<FKG+<GKJ=180°[sum of Straight angles]

or,y+90°+z°=180°

or,50°+90°+z°=180°

or,140°+z°=180°

or,z°=180°-140°

Z=50°

Click if Answer is correct.

Solution:

CD //IJ ,AB is transversal line.

□ <FKI= <AFC Corresponding angle]

y°=40°

□ <HKB=<FKG[Vertically Opposite Angle]

x°=80°

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°

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°

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°

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°

Click if Answers are correct.

क)उत्तर, सोमबार 30 वटा गणित पुस्तक विक्री भएछन्।

ख)उत्तर, शुक्रवार 10 वटा गणित पुस्तक विक्री भएछन्।

ग)उत्तर, सोमबार 30 र मङ्गलबार 30 वटा गणित पुस्तक विक्री भएकोले बराबर सङ्ख्यामा विक्री भएछ।

घ)उत्तर, आइतबार सबै भन्दा बढी 40 वटा गणित पुस्तक विक्री भएछ।

क)उत्तर, सोमबार 20 वटा गणित पुस्तक विक्री भएछन्।

ख)उत्तर, शुक्रवार 30 वटा गणित पुस्तक विक्री भएछन्।

ग)उत्तर, सोमबार 40 र मङ्गलबार 40 वटा गणित पुस्तक विक्री भएकोले बराबर सङ्ख्यामा विक्री भएछ।

घ)उत्तर, आइतबार सबै भन्दा बढी 50 वटा गणित पुस्तक विक्री भएछ।

(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

Above Answer is correct.

Above answer is incorrect.

Above answer is correct.

I got quadrilateral ABCD.

I got a square ABCD.

Above answer is incorrect.

Coordinate of points are C(-4,-4)andD(-3,4)

E(6,0),F(6,-4),I(0,3)

J(-5,6)

Above answer are correct.

Above answer are incorrect.

All Answers are correct.

All answers are not correct.

In above Question Numerical solved are correct but Venn Diagram are mistakes.

In second options all answers are correct.

क)उत्तर, A={8 र 8 भन्दा साना जोर सङ्ख्याहरू }={2,4,6,8} भएकोले,

Aको पुरक समूह=U-A={}-{}

ख)उत्तर, B={ विजोर सङ्ख्या < 8} ={ 8भन्दा साना विजोर सङ्ख्या }={1,3,5,7}

ग)उत्तर, C ={ 8 भन्दा साना रूढ सङ्ख्याहरू }={2,3,5,7}

घ)उत्तर, D={ जोर सङ्ख्या रूढ सङ्ख्या}={2}

All answers are correct.

All answers are not correct.

Yes

No

Yes

No

Above answer is correct.

Above answer is not correct.

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)

$$10a^2+5ab+2ab+b^2$$ = $$10a^2+3ab+b^2$$

Yes

No

Yes

No

Yes

No