(i) Remaining cards = 52 - 3 = 49
p(heart) =\(\frac {13}{14}\)
(ii) Remaining cards = 52 - 3 = 49
p(king) =\(\frac {3}{49}\)

\(\frac { \cos { { 45 }^{ ° } } }{ \sec { { 30 }^{ ° } } } +\frac { 1 }{ \sec { { 60 }^{ ° } } } =\frac { \frac { 1 }{ \sqrt { 2 } } }{ \frac { 2 }{ \sqrt { 3 } } } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ \sqrt { 2 } } \times \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } }{ 4 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } +2 }{ 4 } \)

x²-2ax-(4b²-a²)=0
Given equation can be written as
x²-2ax-+a²-4b²=0
or (x-a)²(2b)²=0
(x-a+2b)(x-a-2b)=0
x=a-2b,x=a+2b

Let p(x) = 3x² + 11x - 4
= 3x² + 12x - x - 4
= 3x(x + 4) - 1(x + 4)
= (3x - 1)(x + 4)
So, zeroes are \(m=\frac{1}{3}\) and n = - 4
Now, \(\frac { m }{ n } +\frac { n }{ m } =\frac { \left( \frac { 1 }{ 3 } \right) }{ -4 } +\frac { -4 }{ \left( \frac { 1 }{ 3 } \right) } =\frac { 1 }{ -12 } -12\)
\(=\frac{-145}{12}\)

Diameter of solid ball =Length of edge of cubical box = a
\(\therefore\) Volume of remaining space inside the box=Volume of cubical box-Volume of solid ball.
\(\sqrt { 3 } :\frac { a^{ 3 } }{ 6 } (6-\pi )\)

Area of path = (Area of the field including the path) - (Area of the field encluding the path)
1.5 m.

Let O be the centre of a circle of radius 10 cm.
Here, chord AB = 16 cm
Since OP bisects chord AB
ஃ AL = LB = 8 cm
Consider rt, \(\angle \)ed \(\triangle\)OLB, by using Pythagoras Theorem,
we have OL² = OB² - LB²
= 10² - 8² = 100 - 64 = 36
OL = 6 cm
Let PL be x cm
ஃ   OP = OL + PL = (6+x)cm
In rt ed BLP, we have
PB² = BL² + PL²
PB² = 8² + x²       ........(i)
⇒  x² = PB² - 6⁴
Again, in rt, \(\angle \)ed \(\triangle\)OBP, we have
PB² = OP² - OB²
PB² = (6 x)² - 10²   ......(ii)
From (i) and (ii), we obtain
(6+x)² - 100 = 64 + x² 
36 + x² + 12x - 100 - 64 - x² = 0
12x - 128 = 0
\(x={128\over12}={32\over3}cm\)
From (i), we have
PB² = 64 + \(({32\over3})^2\)
PB² = \(64+{1024\over9}={576+1024\over9}={1600\over9}\)
PB = \(40\over3\) cm

From P, draw PQ⊥AB, join PA and PB,
In ΔAPQ and Δ BPQ, we have
AP = BP = 6cm
PQ = PQ
ㄥAQP = ㄥBQP=90⁰
⇒ ΔAPQ ~ Δ BPQ
Therefore, \(AQ = BQ={1\over 2}AB\)
\(⇒\ AQ=BQ=3cm\)
Let ㄥPBQ=θ

∴ In rt. Δ QBP, ㄥQ=90⁰
\(⇒\ cos\theta={QB\over PB}\)
\(cos\theta={3\over 6}={1\over 2}\)
⇒ cos θ = cos60⁰
⇒ θ=60⁰
Also, \({PQ\over PB}=sin60^0\)
\(PQ=6\times{\sqrt3\over2}=5.196cm...(i)\)
Now, area of sector BPA=\({60^0\over360^0}\times{22\over 7}\times6\times6\)
=18.857cm²
Area of the portion APB
= 2 X [Area of sector BPA - Area of ΔBPQ]
= 2 x [18.857 - 7.7941= 22.126cm²
Area of the shaded portion = 2 x [Area of quadrant ABC — Area of the portion APB]
\(=2\times\left[{90^0\over 360^0}\times{22\over 7}\times6\times6-22.126\right]\)
=2 x [28.286-22.126]
=12.32cm²
Hence, the required area of the shaded portion is 12.32cm²

\(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

Area of square field = 484 m² 
∴ Side x Side = 22 x 22 
⇒ Side = 22 m 
∴ Perimeter of square field = 4 X 22 = 88 m 
Since circumference of circular field = Perimeter of square field
\(\Rightarrow \pi d=88\Rightarrow \frac { 22 }{ 7 } d=88\)
\(\Rightarrow d=88\times \frac { 7 }{ 22 } =28m\)
Hence, the length of the diameter of the circular field is 28 m.

Area of the required \(\triangle\)ABC
\(=\frac { 1 }{ 2 } \left| 3(0-4)+7(4-0)+8(0-0) \right| \\ =\frac { 1 }{ 2 } \left| 28-12 \right| =8sq.units\)

\(\angle\)BPQ = \(\angle\)ABP
\(\angle\)BPQ = 60^o
\(\Rightarrow\) \(\angle\)CPQ + \(\angle\)BPC = 60^o
\(\Rightarrow\) \(\angle\)CPQ + 30⁰ = 60^o
\(\Rightarrow\) \(\angle\)CPQ = 60⁰ - 30⁰ = 30^o

(i) Let S = 1 + 2 + 3 + . . . + 1000
Using the formula \(\mathrm{S}_{n}=\frac{n}{2}(a+l)\) for the sum of the first n terms of an AP, we have 
\(S_{1000}=\frac{1000}{2}(1+1000)=500 \times 1001=500500\)
So, the sum of the first 1000 positive integers is 500500.
(ii) Let S_n = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore,   \(\mathrm{S}_{n}=\frac{n(1+n)}{2} \text { or } \mathrm{S}_{n}=\frac{n(n+1)}{2}\)
So, the sum of first n positive integers is given by
\(\mathrm{S}_{n}=\frac{n(n+1)}{2}\)

0

probability that it will not rain tomorrow
 = 1 - Probability that it will rain tomorrow = 1 - 0.85 = 0.15

no real root

Now,  AB + CD = BC + AD
⇒12 + 14 = 15 + AD
⇒ AD = 11 cm

Here, we have   
\(a_{2}-a_{1}=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}\),
\(a_{3}-a_{2}=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\)
\(a_{4}-a_{3}=\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\) and so on.
Since, the difference of any two consecutive terms is same. therefore, the given list of numbers forms an AP and its common difference (d) is \(\frac{1}{2}\).
Now, next three terms of this AP are,
a₅ = a₄ + d = \(\frac{7}{2}+\frac{1}{2}\)
[\(\because\) a₅ = a + 4d = a + 3d + d = a₄ + a₅]
\(=\frac{7+1}{2}=\frac{8}{2}=4\)
\(a_{6}=a_{5}+d=4+\frac{1}{2}=\frac{9}{2}\)
and \(a_{7}=a_{6}+d=\frac{9}{2}+\frac{1}{2}=\frac{9+1}{2}=\frac{10}{2}=5\)

Mean = A+ \(\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
= 45 + \(45+\left( \frac { -37 }{ 150 } \right) \times 10\)
= 42.5 approx

Let \(\triangle ABC\) be an equilateral triangle, then
AB = BC = AC = 2a
AD \(\bot\) BC
Also AD is median,
CD = a = BD
\(\angle A=\angle B=\angle C\) = 60°
In \(\triangle ADC\),   AC = 2a
DC = a
\(\therefore \quad AD=\sqrt { { \left( 2a \right) }^{ 2 }-{ \left( a \right) }^{ 2 } } =\sqrt { { 3a }^{ 2 } } \)
\(=\sqrt { { 3a } } \)
\(\therefore \quad \sin { { 60 }^{ ° } } =\frac { AD }{ AC } =\frac { \sqrt { { 3a } } }{ 2a } \)
\(\Rightarrow \quad \sin { { 60 }^{ ° } } =\frac { \sqrt { { 3 } } }{ 2 } \)

\(\frac { AP }{ AB } =\frac { 3.5 }{ 10.5 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \frac { AQ }{ AC } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
In \(\triangle\)ABC, \( \frac { AP }{ AB } =\frac { AQ }{ AC } \)and \(\angle\)A is common
\(\triangle\)APQ~\(\triangle\)ABC     (By SAS)
\(\therefore \frac { AP }{ AB } =\frac { AQ }{ BC } \)
\(\Rightarrow \frac { 1 }{ 3 } =\frac { 4.5 }{ BC } \Rightarrow BC=13.5 cm.\)

For x + 2y = 3
a₁ = 1, b₁ = 2 ,c₁ = - 3
For (k-1)x + (k +1)y= (k + 2)
a₂ = (k - 1), b₂= (k + 1), c₂ = - (k + 2)
For no solution, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { c_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { 1 }{ k-1 } =\frac { 2 }{ k+1 } \neq \frac { 3 }{ k+2 } \)
           I           II          III
From I and II, \(\frac { 1 }{ k-1 } =\frac { 2 }{ k+1 } \)
\(\Rightarrow\)  k + 1 = 2k - 2
\(\therefore\)  k = 3
From II and III, \(\frac { 2 }{ k+1 } \neq \frac { 3 }{ k+2 } \)
\(\Rightarrow\) 2(k + 2) \(\neq\) 3(k + 1)
\(\Rightarrow\) 2k+4\(\neq\)3k+3
\(\therefore\)  k \(\neq\) 1
From I and III, \(\frac { 1 }{ k-1 } \neq \frac { 3 }{ k+2 } \)
\(\Rightarrow \quad k+2\neq 3k-3\)
\(\therefore \quad k\neq -\frac { 5 }{ 2 } \)
Hence, k = 3 but k \(\neq\)1 and \(\\ \\ k\neq -\frac { 5 }{ 2 } \)

HCF of 324 and 252 = 36
HCF of 36 and 180 = 36
∴ HCF of 180, 252 and 324 is 36

Area of the square ABCD=(BC)²=10²=100 cm²
Now, the area of the square PQRS=(side)²=(5)²=25 cm²
P(the point will be chosen from the shaded part)
=\(\frac { Area \ of \ the \ square \ PQRS }{ Area \ of \ the \ square \ ABCD } =\frac { 25 }{ 100 } =0.25\)

Given A circle of centre O. Two tangents AP and AQ are drawn to the circle from an external point A.

To prove ㄥPAQ = 2ㄥOPQ
Construction Join OPand OQ.
Proof We know that, tangents drawn from an external point to a circle are equal in length.
AP = AQ
ㄥAPQ=ㄥAQP=x
In ∆APQ, ㄥPAQ = 180⁰-(ㄥAPQ+ㄥAQP)
= 180° - (x + x)
= 180° - 2x
∵ OP丄AP
∴ ㄥOPA=90⁰ ⇒ ㄥOPQ+ㄥAPQ=90⁰
⇒ ㄥOPQ + x = 90⁰
⇒ ㄥOPQ = 90⁰ - x
⇒ 2ㄥOPQ = 180⁰ - 2x
From Eqs. (i) and (ii), we have
ㄥPAQ = 2ㄥOPQ

Given, AC = 24 cm, BC = 10cm and AOB is a diameter.
Here, \(\angle ACB={ 90 }^{ o }\)  [since, angle in semi - circle is right angle]
In right angled delta ACB,
\(AB=\sqrt { { AC }^{ 2 }+{ BC }^{ 2 } } \)  [by Pythagoras theorem]
\(=\sqrt { { 24 }^{ 2 }+{ 10 }^{ 2 } } \\ =\sqrt { 576+100 } =\sqrt { 676 } =26\quad cm\)
Therefore, Radius of circle, \(r=\frac { AB }{ 2 } =\frac { 26 }{ 2 } =13cm\)
= Area of semi-circle - Area of delta ABC
\(=\frac { 1 }{ 2 } \times { \pi r }^{ 2 }-\frac { 1 }{ 2 } \times BC\times AC\\ =\frac { 1 }{ 2 } \times 3.14\times 13\times 13-\frac { 1 }{ 2 } \times 10\times 24\\ =\frac { 530.66 }{ 2 } -120=265.33-120=145.33{ cm }^{ 2 }\)
Hence, the area of the shaded region is 145.33 cm²

Given vertices of a quadrilateral are P(22, 5), Q(7, 10), R(12, 11) and S(3, 24).
Now, PQ = \(\sqrt { { (7-22) }^{ 2 }+{ (10-5) }^{ 2 } } \)   [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
=\(\sqrt { { (-15) }^{ 2 }+{ (5) }^{ 2 } } =\sqrt { 225+25 } \) 
=\(\sqrt { 250 } =5\sqrt { 10 } \quad \) units
QR = \(\sqrt { { (12-7) }^{ 2 }+{ (11-10) }^{ 2 } } =\sqrt { { \left( 5 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 25+1 } =\sqrt { 26 } \) units
RS =\(\sqrt { { (3-12) }^{ 2 }+{ (24-11) }^{ 2 } } =\sqrt { { \left( -9 \right) }^{ 2 }+{ \left( 13 \right) }^{ 2 } } \) 
= \(\sqrt { 81+169 } =\sqrt { 250 } =5\sqrt { 10 } \) units
and SP = \(\sqrt { { (22-3) }^{ 2 }+{ (5-24) }^{ 2 } } =\sqrt { { \left( 19 \right) }^{ 2 }+{ \left( -19 \right) }^{ 2 } } \) 
 = \(19\sqrt { 1+1 } =19\sqrt { 2 } \) units
Here, we see that opposite sides of a parallelogram are not equal i.e. \(QR\neq SP\)Hence, given vertices of a parallelogram are not formed a parallelogram.
Hence proved.

914.29 m, Rs. 12800.06

Given: ΔABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60^o. 

Required: ΔA'BC ∼ ΔABC with \(\frac { 3 }{ 4 } \) (reduced) scale-factor.
Steps of Construction : 
1. Draw line segment BC = 6 cm.
2. At B, construct an angle ∠CBY = 60^o.
3. With B as centre cut off BA = 5 cm and join AC.
4. Through B, construct an acute angle ∠CBX (<90^o).
5. Mark four points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
6. Join B₄C.
7. Through B₃, draw B₃C'||B₄C, intersecting BC in C'.
8. Through C', draw C'A'||CA, intersecting BA in A'. Hence, ΔA'BC' is the required triangle.

In rt. ΔABC, tan60^o = \(\frac { AB }{ BC } =\frac { h }{ x } \)
⇒ \(\sqrt { 3 } =\frac { h }{ x } \Rightarrow h=\sqrt { 3 } x\)
In rt. ∆ABD, tanፀ = \(\frac { AB }{ BD } \)
⇒  tanፀ = \(\frac { h }{ 3x } \)
⇒  tanፀ = \(\frac { \sqrt { 3 } x }{ 3x } =\frac { 1 }{ \sqrt { 3 } } \)⇒ ፀ = 30^o​​​

The arrangement of houses is
1,2,3, ..............x-I, x, x + 1,...... 49.
S₁= 1 + 2 + 3 + ...x-I
\(=\left( \frac { x-1 }{ 2 } \right) \left\lfloor 2\times 1+(x-2)1 \right\rfloor \)
\(=\left( \frac { x-1 }{ 2 } \right) [x]\)
\(=\frac { x(x-1) }{ 2 } \)
\(S_{ 2 }=x+1,x+2..49\)
\(= \frac { \left( 49-x \right) }{ 2 } [2(x+1)+(49-x-1])\)

We know that, sum of all angles of a triangle is equal to 180⁰ .
i .e .  A + B + C = 180⁰
\(\Rightarrow \frac { A }{ 2 } +\frac { B }{ 2 } +\frac { C }{ 2 } ={ 90 }^{ 0 }\)  [on dividing both sides by 2]
\(\Rightarrow \frac { B }{ 2 } +\frac { C }{ 2 } ={ 90 }^{ 0 }-\frac { A }{ 2 } \)
On taking sin both sides, we get
\(\sin { \left( \frac { B+C }{ 2 } \right) } =\sin { \left( { 90 }^{ 0 }-\frac { A }{ 2 } \right) } \)
\(\sin { \left( \frac { B+C }{ 2 } \right) } =\cos { \frac { A }{ 2 } } \left[ \because \sin { ({ 90 }^{ 0 }-\theta ) } =\cos { \theta } \right] \)
Hence proved.

Given, (a-b),a and (a+b) are the zeroes of the polynomial x³-3x²+x+1.
On comparing given polynomial with
Ax³+Bx²+cx²+D, we get
A=1, B=-3, C=1 and D=1
Now, sum of zeroes=(a-b)+a+(a+b)=-B/A
⇒ \(3a=\frac { -(-3) }{ 1 } =3\)              [∵ B=-3, A=1]
⇒ 3A=3⇒A=1
Sum of product of zeroes taken two at a time
=a(a-b)+a(a+b)+(a+b)(a-b)=\(\frac { C }{ A } \)
⇒ a²-ab+a²+ab+a²-b²=1/1  [∵ C=1, A=1]
⇒ 3a²-b²=1⇒3(1)²-b²=1      [from Eq.(i)]
⇒ 3-b²=1⇒b²=2⇒b=\( \pm \sqrt { 2 } \)
Hence, a=1 and b=\( \pm \sqrt { 2 } \)

Given, HCF (253, 440) = 11 and LCM (253, 440)
= 253 x R
\(\therefore LCM(253,440)=\frac { 253\times 440 }{ HCF(253,440) } \)
\(\Rightarrow 253\times R=\frac { 253\times 440 }{ 11 } \)
R = 40

d = 6, The next two terms are 24 and 30.

\(\left( 64\sqrt { 3 } -\frac { 231 }{ 4 } \right) { m }^{ 2 }\)

Let the coordinates of B and C be (a,b) and (x,y) respectively
Then \(\left( \frac { 1+a }{ 2 } ,\frac { -4+b }{ 2 } \right) \)  =(2.-1)

\(\Rightarrow \) \(\frac { 1+a }{ 2 } =2\quad and\quad \frac { -4+b }{ 2 } =-1\)
\(\Rightarrow \) 1+a=4 and -4+b=-2
\(\Rightarrow \) a=3 and b=2
Also  \(\left( \frac { 1+x }{ 2 } ,\frac { -4+y }{ 2 } \right) \)=(0,-1)
\(\Rightarrow \) \(\frac { 1+x }{ 2 } =0\quad and\quad \frac { -4+y }{ 2 } =-1\)
\(\Rightarrow \)  1+x=0 and -4+y=-2
\(\Rightarrow \)   x=-1 and y=2
Thus the coordinates of the vertices of ABC are A(1,-4),B(3,2) and C(-1,2)
Area of ABC =\(\frac { 1 }{ 2 } \) |1(2-2)+3(2+4)+(-1)(-4-2)|
= \(\frac { 1 }{ 2 } \)|0+18+6|
= \(\frac { 1 }{ 2 } \)(24)=12 sq.units 

\(6\frac { 2 }{ 3 } \)cm

(i)50m
(ii)2.13m