In \(\Delta\)PCB,
\(\angle\)PCB + \(\angle\)PBC = \(\angle\)APB (exterior angle of a \(\Delta\) is equal to the sum of two opposite angles)
\(\angle\)PCB + 15° = 120°
\(\therefore\) \(\angle\)PCB = 105°
or, \(\angle\)ACB = 105°  
\(\Rightarrow\)\(\angle\)ADB = \(\angle\)ACB = 105°  [Angle in same sagment]

(i) Total number of cases=100
Number of cases favourable to an odd number (1,3,5) = 20 + 20 + 20 = 60
P(odd number)\(=\frac{60}{100}=\frac{3}{5}\)
(ii) Number of cases favourable to the event a prime number (2,3,5) = 15 + 20 + 20 = 55
P(Prime number) = \(\frac{55}{100}=\frac{11}{20}\)

Prime numbers lying between 6 and 20 are 7,11,13,17,19
Mean = \(\frac{7+11+13+17+19}{5}=\frac{67}{5}\)
= 13.4

Euclid's axioms
(i) Things which are equal to the same thing are equal to one another.
(ii) If equals are added to equals, the wholes are equal.

2x-1 = 0
\(\Rightarrow \ x=\frac { 1 }{ 2 } \)
By remainder theorem, if f(x) is divided by 2x-1, the remainder is \(f\left( \frac { 1 }{ 2 } \right) \)
\(\therefore \quad f\left( \frac { 1 }{ 2 } \right) =4{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }-12{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+14\left( \frac { 1 }{ 2 } \right) -3\)
\(=4\times \frac { 1 }{ 8 } -12\times \frac { 1 }{ 4 } +14\times \frac { 1 }{ 2 } -3\)
\(=\frac { 1 }{ 2 } -3+7-3\)
\(=\frac { 1 }{ 2 } +1\)
\(=\frac { 3 }{ 2 } \)
Hence required remainder is \(\frac { 3 }{ 2 } \).

6.615 cm²

96 cm²

m=7

II

\(a=\frac { 138 }{ 59 } ,b=0\)

Required probability=\(1-\frac { 79 }{ 200 } \)

\(\frac { 2 }{ 3 } \times \left( 6\times 5\times 4 \right) 80{ m }^{ 3 }\)

See Hero's formula.

\(\angle QOT=50°\) 
\(\therefore\angle OQT+\angle OTQ=130°\)
But \(\angle OQT=\angle OTQ\)
∴ \(\angle OQT=\angle OTQ=65°\)
\(x+\angle OQT=180°\)

Trapezium ABCD and trapezium CDEF do not lie between the same parallels.

\(\angle \)AGF +\(\angle \) GAE = \(​​180°\)
\(\Rightarrow \) \(\angle \)AGF+\(​​60°\)=\(​​180°\)
\(\Rightarrow \)\(\angle \)AGF=\(​​120°\)

ㄥB > ㄥC
∴ AC > AB

\((30-a)^{ 0 }+(125+2a)^{ 0 }=180^{ 0 }\)

x+3=0
\(\Rightarrow\)  x=-3
\(\therefore\)  Remainder=p(-3)

y=180^o-(25^o+125^o)(Linear pair)
\(\Rightarrow\) y=30^o
y+z=110^o (Exterior angle)
z=110^o-30^o=80^o
x+25^o=z (Exterior angle)
x=80^o-25^o=55^o

Proof: In \(\triangle\)ABF and \(\triangle\)ACE,
AB =AC
\(\angle\)A =\(\angle\)A
AF = AE
\(\therefore\) \(\triangle\)ABF = \(\triangle\)ACE (by SAS cong.)
\(\therefore\) By c.p.c.t BF = CE.
Alternative Method:
AB = AC\(\Rightarrow \frac { AB }{ 2 } =\frac { AC }{ 2 } \)
\(\Rightarrow \) AE = AF,
sice E and F are the mid-points of AB and AC
In \(\triangle\)ABF and \(\triangle\)ACE,
AB = AC (Given) 
\(\angle\)A = \(\angle\)A (Common) 
AF = AE (Proved) 
\(\therefore \ \triangle ABF\cong \triangle ACE\) (By SAS cong.) 
\(\therefore\) BF = CE. (By c.p.c.t) 

\((i)\frac { 21 }{ 40 }\) 
\((ii)\frac { 29 }{ 40 }\) 
\((iii)\frac { 29 }{ 40 } \)

88

8800 cm³

Given: Two circles with centres O and P intersecting at A and B.
Prove: OP is the perpendicular bisector of AB.
Construction: Join OA, OB, PA and PB. Let OP intersect AB at M.
Proof: In \(\Delta \) OAP and \(\Delta \) OBP,
OA = OB         | Radii of a circle
PA = PB      I Radii of a circle
OP = OP     I Common
∴  \(\Delta \ OAP\cong \Delta \ OBP\) I SSS Rule
∴ \(\angle AOP=\angle BOP\)     I CPCT
⇒\(\angle AOM=\angle BOM\)                       ...(1)
In \(\Delta \) AOM and \(\Delta \) BOM,
OA = OB  I Radii of a circle
 \(\angle AOM=\angle BOM\)    | From (1)
OM = OM   I Common
∴\(\Delta AOM\cong \Delta BOM\)    I SAS Rule
∴ AM = BM                  ...(2)
I CPCT
and \(\angle AMO=\angle BMO\)          ... (3)
I CPCT
But ​​​​​​​ \(\angle AMO+\angle BMO=180°\) I Linear Pair Axiom
∴ \(\angle AMO+\angle BMO=90°\)                ...(4)
∴ OM, i.e., OP is the perpendicular bisector of AB. I From (2) and (4)

Given: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
To Prove: \(ar(\Delta APB)\times ar(\Delta CPD)=ar(\Delta APD)\times ar(\Delta BPC)\)
Construction: From A and C, draw perpendiculars AE and CF respectively to BD.

\(ar(\Delta APB)\times ar(\Delta CPD)\)=\(\frac { (PB)(AE) }{ 2 } \times \left( \frac { DP\times CF }{ 2 } \right) \)
|Area of \(\Delta\)=\(\frac { Base\times Corresponding\ altitude }{ 2 } \)
\(=\frac { 1 }{ 4 } (PB)(AE)(DP)(CF)\quad \quad ....(1)\)
\(\\ ar(\Delta APD)\times ar(\Delta BPC)\)
\(\\ =\frac { (DP)(AE) }{ 2 } \times \frac { (PB)(CF) }{ 2 }\)
\( \\ =\frac { 1 }{ 4 } (PB)(AE)(DP)(CF)\quad \quad ....(2)\)
From (1) and  (2),
\(ar(\Delta APD)\times ar(\Delta BPC)\)

Yes! These postulates contain two undefined terms: Point and Line. Yes! These postulates are consistent because they deal with two different situations
(i) says that given two points A and B, there is a point C lying on the line in between them,
(ii) says that given A and B, we can take C not lying on the line through A and B. These 'postulates' do not follow from Euclid's postulates, however, they follow from Axiom 'Given two distinct lines, there is a unique line that passes through them.

x=3y
x-3y+0=0
Comparing with ax+by+c=0, we get
a=1, b=-3, c=0.

20

\(128\sqrt { 2 } \)

Construction: Draw RM II SP meeting PQ M.
Proof: PQ||SR     (Given)
PS||MR   (Construction)

\(\therefore\)PMRS is a parallelogram
\(\therefore\) \(\angle\)S = \(\angle\)PMR 
In \(\triangle\)RMQ, RM = RQ (as RM = SP, opp. sides of a parallelogram and PS = QR)
\(\therefore\) \(\angle\)RMQ=\(\angle\)Q 
But \(\angle\)RMQ = 180° - \(\angle\)PMR (linear pair)
= 180°- \(\angle\)S. (proved) 
\(\therefore\) \(\angle\)Q = 180°- \(\angle\)S
Hence \(\angle\)Q + \(\angle\)S = 180°
Since one pair of opposite angles is supplementary, PQRS is cyclic. Proved.

Let E be the event the batsman hits a boundary.
Then \(\overline E\) is the event the batsman does not hit a boundary.
P(E) = \(\frac{12}{48}=\frac{1}{4}\)
\(\Rightarrow P(\overline E)=1-P(E)\)
\(=1-\frac{1}{4}=\frac{3}{4}\)

DB is the transversal because DC||AB

because \(\angle CDB=\angle ABD=90°\)(form a pair of alternate angles)
DC || AB and DC = AB
\(\therefore\) ABCD is a ||^gm
ar(ABCD) = b\(\times\)h
= 2.5 \(\times\)4 = 10 cm².

We have, \(\angle\)DCA = \(\angle\)ECB
\(\angle\)DCA + \(\angle\)DCE = \(\angle\)DCE + \(\angle\)ECB
(Adding \(\angle\)DCE on both the sides)
\(\Rightarrow\)\(\angle\)ACE =\(\angle\)BCD ...(i) 
Now, in \(\triangle\)s BDC and \(\triangle\)AEC,
\(\angle\)BCD = \(\angle\)ACE [From (i)]
BC = AC (Given)
and \(\angle\)DBC = \(\angle\)EAC (Given) 
So, by ASA criterion of congruence, we have
\(\angle\)DBC = \(\angle\)EAC
\(\therefore\) BD = AE and DC == EC
(\(\because\)  Corresponding parts of congruent triangles are equal)

Arranging the data in ascending order, we have,
17,17,17,18,19,23,23,23,23,25,26
Here, 23 occurs most frequency (4 times)
\(\therefore\)  Mode = 23
IF = f 4 is subtracted from each observation, then the new observations are
13,13,13,14,15,19,19,19,19,21,22
Here, 19 occurs most frequency (4 times)
\(\therefore\) Mode = 19

Let the base radius of the cylinder be r cm. Then,
\(2\pi r=22\)
\(\\ \Rightarrow \ 2\times \frac { 22 }{ 7 } \times r=22\)
\(\\ \Rightarrow r=\frac { 7 }{ 2 } \)
\( h=3m\)
\(\therefore \)   Total surface area 
 \(=2\pi rh+\pi { r }^{ 2 }\)
\(\\ =2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times 3+\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \)
\(\\ =66+38.5\)
\(\\ =104.5{ m }^{ 2 }\)

\(\therefore \)  Side = 8 cm
\(\therefore \) Perimeter = 6\(\times \)8 = 48 cm
Area of equilateral triangle OAB = \(\frac { \sqrt { 3 } }{ 4 } \)  (side)²
 
= \(\frac { \sqrt { 3 } }{ 4 } \)(8)² = 16 \(\sqrt { 3 } \) cm²
\(\therefore \) Area of the regular hexagon = 6 Area of equilateral triangle OAB
= 6\(\times \)16\(\sqrt { 3 } \) = 96\(\sqrt { 3 } \) cm².

Steps of Construction
1. Draw BC = 6 cm.
2. With B as centre and 6 cm as radius, draw an arc on one side of BC.
3. With C as centre and 6 cm as radius, draw another arc on the same side of BC to intersect the former arc at A.
4. Join AB and AC. Then, \(\Delta \) ABC is the required equilateral triangle.

Given: In a quadrilateral ABCD, the line segments bisecting \(\angle \)C and \(\angle \)D meet at E.
  To Prove:   \(​​\angle \)A+\(​​\angle \)B = 2 \(\angle \)CED
I Angle sum property of a quadrilateral
In \(\Delta \) CED,
\(\angle \)CED +\(\angle \)EDC+\(\angle \)ECD = \(180°\)   | Angle sum property of a triangle
⇒ \(\angle \)CED+\(1\over2\)\(\angle \)D+\(1\over2\)\(\angle \)C=\(180°\)
 ⇒ 2\(\angle \)CED +\(\angle \) D+\(\angle \)C =\(360°\)    ...(2)
From (1) .and (2),
2\(\angle \)CED+\(\angle \)D+C=A+B+C+D
 ⇒ 2 \(\angle \)CED =\(\angle \)A+\(\angle \)B
 ⇒ \(\angle \)A+\(\angle \)B=2\(\angle \)CED

We have
AB = BC
\(\Rightarrow \) AB - BX = BC - BX
|If equals are subtracted from equals, the remainders are equal (Euclid's Axiom (iii))
AB - BX = BC - BY  \(|\)\( \because\)  BX = BY
AB - BX coincides with AX;
BC - BY coincides with CY
[Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]

Let the present ages of father and son be x years and y year respectively.
Then, Age of father 5 years ago =(x-5)years
Age of his son 5 years ago(y-5) years
According to the question,
x-5=7(y-5)+2
x-y=7y-35+2
x-7y+28=0
which is the required linear equation in two variables.

(i) (3,2)
(ii) (-3,2)
(iii) (-3,-2)
(iv)(3,-2)

L.H.S=\((a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)\)
\(=\left\{ (a+b)+(b+c)+(c+a) \right\} [(a+b)^2+(b+c)^2+(c+a)^2-(a+b)(b+c-(b_c)(c+a)-(c+a)(a+b)]\)
\(=2(a+b+c)[a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+c^2-ab-ac-b^2-bc-bc-ba-c^2-ca-ca-cb-a^2-ab]\)
Using Identity I
\(=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
\(=2(a^3+b^3+c^3-3abc)\)Using Identity VIII

Mark the distance 4.5 units from a fixed point A on a given line to obtain a point B such that AB=4.5 units.From B, mark a distance of 1 unit and mark the new point as C.Find the mid-point of AC and mark that point as O.Draw a semi-circle with centre O and radius OC.Draw a semi-circle with Centre O and radius OC.Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D.Then BD=\(\sqrt { 4.5 } \). By measurement,
BD = 2.2 units
\(\sqrt { 4.5 } \)= 2.2
Now, let us treat the line BC as the number line, with B as zero, C as 1 and so on.Draw an arc with centre B and radius BD, which intersects the number line in E.

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) Let the butter-chords of the biscuit be AB and CD; and the centre of the biscuit be O.
Join each of A, B, C, D to O.

In \(\triangle\)OAB and In \(\triangle\)OAB and \(\triangle\)OCD,
AB = CD (given)
OA = OC (each equal to radius)
OB =OD
(each equal to radius)
\(\therefore \triangle OAB\cong \triangle OCD\) [S.S.S.]
\(\Rightarrow\) \(\angle\)AOB = \(\angle\)COD (c.p.c.t)
Therefore, the butter-chords subtend equal angles at the centre of the biscuit
\(\Rightarrow\) \(\angle\)AOB = \(\angle\)COD (c.p.c.t)
Therefore, the butter-chords subtend equal angles at the centre of the biscuit.
(ii) We are given the length of either chord is greater than the radius and less than diameter of the circle.
Let length of either chord = l, radius = r, and angle subtended by either butter-chord = 0
Two cases arise:
Case I. If I = r
In this case, the chord and the corresponding radius
form an equilateral triangle with side r
\(\therefore\) \(\theta\)=60°
Case II. If l = 2r 
In this case, the butter-chord passes through the centre.
\(\therefore\) \(\theta\) = 180°
Consequently, we arrive at the following inequality :
60° < 8 < 180°
(Asr < 1< 2r)
Thus, the required range is excluding both.
(iii) (a) Congruency of triangles.
(b) c.p.c.t. (Corresponding parts of congruent triangles are equal.)
(c) Equilateral triangle and its angles
(iv) Industrialist, Thoughtfulness, Self-confidence, Rationality.

In \(\triangle\)ABY and \(\triangle\)ACX,
AB = AC   (Given)
AY = AX     (Given)
\(\angle\)A =\(\angle\)A   (Common)
\(\therefore\) By SAS, \(\triangle ABY\cong \triangle ACX\)

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.

\(\frac { 4 }{ { \left( 216 \right) }^{ -\frac { 2 }{ 3 } } } +\frac { 1 }{ { \left( 216 \right) }^{ -\frac { 3 }{ 4 } } } +\frac { 2 }{ { \left( 216 \right) }^{ -\frac { 1 }{ 5 } } } \)
\(=\frac { 4 }{ { \left( { 6 }^{ 3 } \right) }^{ -\frac { 2 }{ 3 } } } +\frac { 1 }{ { \left( { 4 }^{ 4 } \right) }^{ -\frac { 3 }{ 4 } } } +\frac { 2 }{ { \left( { 3 }^{ 5 } \right) }^{ -\frac { 1 }{ 5 } } } \)
\(=\frac { 4 }{ { 6 }^{ -2 } } +\frac { 1 }{ { 4 }^{ -3 } } +\frac { 2 }{ { 3 }^{ -1 } } \)
\(=\frac { 4 }{ { 36 }^{ -1 } } +\frac { 4 }{ { 64 }^{ -1 } } +\frac { 2 }{ { 3 }^{ -1 } } \)
= 4 x 36 + 1 x 64 + 2 x 3
= 214