(i) In \(\triangle\) AOD and \(\triangle\)BOC,
OA =OB (Given)
OD = OC (Given)
\(\angle\)AOD = \(\angle\)BOC
(Vertically opposite angles)
So, by SAS criteria,
\(\triangle AOD\cong \triangle BOC\)
(ii) \(\angle\)CBA = \(\angle\)DAB (By c.p.c.t.)
AD and BC are two lines intersected by AB such that \(\angle\)CBA = \(\angle\)DAB and they form a pair of alternate angles.
Hence, AD II BC

In \(\triangle\)BED and \(\triangle\)CFD,
\(\angle\)DEB = \(\angle\)DFC = 90° 
BD = DC (D is the mid-point)
ED = FD (Given)
\(\therefore \triangle BED\cong \triangle CFD\) (By RHS) 
\(\Rightarrow\) \(\angle\)B = \(\angle\)C (By c.p.c.t.) Proved.

\(\frac { 77 }{ 12 } { m }^{ 3 }\)

28 cm

\(36^{ 0 },54^{ 0 },90^{ 0 }\)

50

\(\frac{V_1}{V_2}=\frac{\pi (\frac{14}{2})^210}{\pi (\frac{14}{2})^220}=\frac{1}{2}=1:2\)

For a pillar
Radius (r) = \(\frac { 0.5 }{ 2 } m=0.25m\)
Height (h) = 4 m
\(\therefore \)  Curved surface area = \(2\pi rh\)
= 2 \(\times\) 3.14 \(\times\) 0.25 \(\times\) 4
= 6.28 m²
\(\therefore \)  Curved surface area of 20 pillars
= 6.28 \(\times\) 20 m² = 125.6 m²
\(\therefore \)  Cost of cleaning them at Rs 3 per m²
= 125.6 \(\times\) 3 = Rs 376.80

Given: ABCD is a parallelogram and E is the mid-point of side BC. DE and AB on producing meet at F.
To Prove: AF = 2AB
Proof: In  \(\Delta \)FAD,
\(\therefore\) E is the mid-point of BC        I Given
and EB II DA   | Opposite sides of a parallelogram are parallel
\(\therefore\) B is the mid-point of AF     I By converse of mid-point theorem
AB = BF = \(1\over2\) AF ⇒ AF = 2AB

Given: ABC is an equilateral triangle whose medians are AD, BE and CF.
To prove: AD = BE = CF
Proof: In \(\triangle ADC\) and \(\triangle BEC\)

AC = BC
\(\angle ACD=\angle BCE\)
AD is a median DC = DB = 1/2 BC
BE is a median EA = EC = 1/2 AC
AC = BC
DC = EC
\(\triangle ADC\cong \triangle BEC\) | SAS congruence rule
AD = BE  | C.P.C.T
Similarly, we can prove that
BF = CF ... (2)
CF = AD ....(3)
From (1), (2) and (3)
AD = BE = CF

Construction Join BD and extend upto E
\(x=\angle ADC\)
 =\(x=\angle ADE+\angle CDE\)
=\(x=\angle DAB+\angle ABD+(\angle DBC+\angle DCB\)
|Exterior Angle Theoram
\(=30^{ 0 }+\angle ABD)+\angle DBC+50^{ 0 }\)
=\(35^{ 0 }+\angle ABD)+\angle DBC+50^{ 0 }\)
= \(35^{ 0 }+45^{ 0 }+50^{ 0 }=130^{ 0 }\) 

Since AB||CD
\(\angle BAC=\angle DCE=72^o\) (Corresponding angles)
\(\angle ACE=\angle CEF+\angle CFE\) (Exterior angle)
\(\Rightarrow 72^o=40^o+\angle CFE\)
\(\therefore \angle CFE=32^o\)

\(\angle A=57^o\)
and \(\angle ACD=22^o+35^o=57^o\)
\(\angle A=\angle ACD\)
But these are alternate angles
AB||CD
Again \(\angle FEC+\angle ECD=145^o+35^o=180^o\)
EF||CD
Now AB||CD and EF||CD
AB||EF

A parallelogram PQRS in which the bisectors of LP and LQ meet SR at O.
To Prove: ㄥPOQ=90°
Now, since PQRS is a parallelogram. Therefore,
PSIIQR
Now, PS II QR and transversal PQ intersects them.
ㄥP+ㄥQ=180°
(∵ Sum of consecutive interior angles is 180°)
\(\frac{1}{2}\)ㄥP+\(\frac{1}{2}\)ㄥQ=90°
⇒ ㄥ1+ㄥ2=90° ( OP is bisector of ㄥP...(i) and OQ is bisector of ㄥQ.
ㄥ1=\(\frac{1}{2}\) ㄥP and ㄥ2=\(\frac{1}{2}\) ㄥQ)
Now, in ΔPOQ
ㄥ1+ㄥPOQ+ㄥ2=180°
⇒ 90°+ㄥPOQ=180°
⇒ ㄥPOQ=90°

(i) Slant height (l) = 14 cm 
Curved surface area = 308 cm²
\(\Rightarrow \pi rl=308\)
\(\\ \Rightarrow \frac { 22 }{ 7 } \times r\times 14=308\)
\(\\ \Rightarrow r=\frac { 308\times 7 }{ 22\times 14 } \)
\(\\ \Rightarrow r=7cm\)
Hence, the radius of the base is 7 cm.
(ii) Total surface area of the cone = \(\pi r\left( l+r \right) \)
\(=\frac { 22 }{ 7 } \times 7\times \left( 14+7 \right)\)
\( \\ =\frac { 22 }{ 7 } \times 7\times 21=462{ cm }^{ 2 }\)
Hence, the total surface area of the cone is 462 cm².

(a) 60⁰
(b) 60⁰

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.

(i) I and m are two parallel lines intersected byanother pair of parallel lines p and q.
AD II BC
and AB II CD.
\(\Rightarrow\) ABCD is a parallelogram.
i.e., AB = CD
and BC = AD
Now in \(\triangle\)ABC and \(\triangle\)CDA,we have
AB = CD (Prop. of II^gm)
BC =AD
and AC = AC (Common)
\(\therefore\) \(\triangle ABC\cong \triangle CDA\)
(By SSS criterion of congruence) 
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

Proof: QO is bisector of \(\angle\)PQR
\(\angle\)OQR = \(\frac{1}{2}\)\(\angle\)PQR = \(\frac{1}{2}\) =\(\angle\)Q
RO is bisector \(\angle\)ORQ
\(\therefore\) \(\angle\)ORQ =\(\frac{1}{2}\) \(\angle\)PRQ = \(\frac{1}{2}\) \(\angle\)R
In \(\angle\)OQR
\(\angle\)QOR + \(\angle\)OQR + \(\angle\)ORQ = 180°
(Angle sum property)
\(\angle\)QOR + \(\frac{1}{2}\) \(\angle\)Q + \(\frac{1}{2}\) \(\angle\)R = 180°
\(\angle\)QOR = 180°- \(\frac{1}{2}\)(\(\angle\)Q + \(\angle\)R)
But in \(\angle\)PQR
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180°
\(\angle\)Q + \(\angle\)R = 180°- \(\angle\)P
\(\angle\)QOR = 180°- \(\frac{1}{2}\) (180°- \(\angle\)P)
= 180°-90° + \(\frac{1}{2}\)\(\angle\)P
= 90° + \(\angle\)P Hence Proved.
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We are given that AB = BC and have to prove that
DE = EF.
Let us join A to E intersecting m at G.
Let trapezium ACFD is divided into two triangles, namely ΔACF and ΔAFD.
In ΔACF, it is given that B is the mid-point of AC(AB = BC) and BG II CF (Since m || n)
So, G is the mid-point of AF (By the converse of midpoint theorem)
Now in ΔAFD, we can apply the sam argument as G is the mid-point of AF, GE IIAD so E is the mid-point of DF
i.e., DE = EF
In otherwords l, m and n cut off equal intercepts on q also.