V = l \(\times\) b \(\times\) h = 1440 m³

height = 12 m

(i) l = 12 cm
b = 8 cm
h= 6 cm
Volume of the cuboid = lbh
= 12 \(\times\) 8 \(\times\) 6 cm³ = 576 cm³
(ii) l = 60 m 
b = 25 m 
h = 1.5 m
Volume of the cuboid = l b h = 60 \(\times\) 25\(\times\)1.5 m³ = 2250 m³

Total number of families S = 1500
n(S) = 1569
Let P be the event of selecting a family having part time maids and F be the event of selecting a family having full time maids.
(i) Both types of maids Let P \(\cap\) F be the event of selecting a family having both types of maids.
Let (P \(\cap\) F) = 259
\(\mathrm{P}(\mathrm{P} \cap \mathrm{F})=\frac{\mathrm{n}(\mathrm{P} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{250}{1500}=\frac{1}{6}\)
(ii) Part time maids Part time maids = only part time maid + both
= 860 + 250 = 1110 '
n(P) = 1110
\(\mathrm{p}(\mathrm{P})=\frac{\mathrm{n}(\mathrm{P})}{\mathrm{n}(\mathrm{S})}=\frac{1110}{1500}=\frac{111}{150}\)
(iii) No maids Let (P \(\cup\) P)'be the event of choosing a family not having maids and (P u F) be the event of choosing a family having part time or fuli time or both maids.
n(P \(\cup\) F ) = only n(P) + only n(F) + n(p \(\cap\) p) = 850 + 370 + 250
n(P\(\cup\)F) = 1480
n(P\(\cup\)F)' = n(S)-n(P\(\cup\)F)
= 1500 - 1480
n(P \(\cup\) F)' = 20
\(\mathrm{P}(\mathrm{P} \cup \mathrm{F})^{\prime}=\frac{\mathrm{n}(\mathrm{P} \cup \mathrm{F})^{\prime}}{\mathrm{n}(\mathrm{S})}=\frac{20}{1500}=\frac{1}{75}\)

(i) What is the probability that the spinner will not land on a multiple of 2?
(ii) What is the probability that the spinner will land on an odd number?

Number of cards n(S) = 52
No. of King cards n(A) = 4
No. of Queen cards n(B) = 4
No. of Jack cards n(C) = 4
Probability of drawing a King card
\(\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Queen card
=\(\frac { n(B) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Jack card
=\(\frac { n(C) }{ n(S) } =\frac { 4 }{ 52 } \)
∴ The Probability of drawing a King or a Queen or a Jack from a deck of cards
= p(A) + P(B)+ P(C) =\(\frac { 4 }{ 52 } +\frac { 4 }{ 52 } +\frac { 4 }{ 52 } =\frac { 4+4+4 }{ 52 } =\frac { 12 }{ 52 } =\frac { 3 }{ 13 } \).

l \(\times\) b \(\times\) h = 6 m \(\times\) 400 cm \(\times\) 1.5 m
l = 6m, b = 4 m, h = 1.5 m
Total surface area of the cuboid = Outer surface area
= 2 (lb + bh + hI)
= 2 ((6 \(\times\) 4) + (4 \(\times\) 1.5) + (1.5 \(\times\) 6))
= 2 (24 + 6 + 9) = 2 (39) m²
Cost of painting 1 m² = Rs. 22
Cost of painting 78 m²= 78 \(\times\) 22 = Rs. 1716

\(tan\ {42 }^{ 0 }=\cfrac { h }{ 60 } =0.9004\)
h = 0.9004 \(\times\) 60 = 54.024 m

(i) = 0.9111 + 0.9066 + 0.1793
= 1.9970
(ii) = 2.8982 + 0.9639 - 0.9962
= 3.8625 - 0.9962
= 2.8659

By the Pythagoras theorem
AB² = AD² + DB²
= 12²+16² = 144 + 256 = 400
AB = 20 cm
s = \(\frac { 34+20+42 }{ 2 } =\frac { 96 }{ 2 } \) = 48
∴ Area of the Δ ABC = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 48(48-34)(48-20)(48-42) } \)
= \(\sqrt { 48\times 14\times 28\times 6 } =\sqrt { 112896 } \)
= \(\\ \sqrt { 336\times 336 } \) = 336 sq.cm
Area of the triangle ABD

∴ Area of the unshaded region
= Area of Δ ABC - Area of Δ ABD
= 336 - 96 = 240 cm².

 
\({ \sin }^{ 2 }{ \tan }^{ 2 }+{ \cos }^{ 2 }{ 60 }^{ o }=\left( \cfrac { \sqrt { 3 } }{ 2 } \right) +\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }=\cfrac { 3 }{ 4 } +\cfrac { 1 }{ 4 } =\cfrac { 4 }{ 4 } =1\)

"The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle."
\(\therefore\) The mid points of the sides of the triangle are given as
  x₁y_{1 ,} x₂y, x₃y₃
\(\left( \frac { 3 }{ 2 } ,5 \right) \left( 7,\frac { -9 }{ 2 } \right) \left( \frac { 13 }{ 2 } ,\frac { -13 }{ 2 } \right) \)
\(\therefore\)Centroid \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 3 }{ 2 } +\frac { 7 }{ 1 } +\frac { 13 }{ 2 } }{ 3 } ,\frac { 5+\left( \frac { -9 }{ 2 } \right) +\frac { (-13) }{ 2 } }{ 3 } \right) =\left( \frac { \frac { 3+14+13 }{ 2 } }{ 3 } ,\frac { \frac { 10-9-13 }{ 2 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 30 }{ 2 } }{ 3 } ,\frac { \frac { -12 }{ 2 } }{ 3 } \right) =\left( \frac { 15 }{ 3 } ,\frac { -6 }{ 3 } \right) =(5,-2)\)

D (x, y) is the Mid point BC
\(\therefore \ D(x,y)=\left( \frac { 1+5 }{ 2 } ,\frac { -1+1 }{ 2 } \right) \)
\(=\left( \frac { 6 }{ 2 } ,\frac { 0 }{ 2 } \right) \)
= (3, 0)
AD is the median through A
Here A x₂ y₂ D x₃ y₃
(-1,3)     (3,0)
\( \therefore\) Length of AD \(=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-(-1)) }^{ 2 }+{ (0-3) }^{ 2 } } \)
\(=\sqrt { { (3+1) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5\) units 

In the figure,
\(\triangle\)OPM, \(\triangle\)OQN are similar triangles. In similar triangles the sides are in the same proportional.
\(\cfrac { QN }{ PM } =\cfrac { QO }{ PO } \)
\(\cfrac { h }{ 5 } =\cfrac { 35 }{ 25 } \)
\(h=\cfrac { 5\times 35 }{ 25 } \)

h = 7m

\(cos\theta :sin\theta =1:2\)
\(\cfrac { cos\theta }{ sin\theta } =\cfrac { 1 }{ 2 } \)
\(cos\theta =\cfrac { 1 }{ 2 } sin\theta \)
\(\sin\theta =2 \cos\theta \)

\(\therefore \cfrac { 8cos\theta -2sin\theta }{ 4cos\theta +2sin\theta } =\cfrac { 1 }{ 2 } \)

By the pythagoras theorem,
AB² = OA² + OB²
(1 + x²)² = (2x)² + OB²
OB² = (1 +x²)² - (2x)² = 1+ x⁴ + 2x² - 4x² = 1+x⁴ - 2x²
OB² = (1-x²)² B
OB = (1-x²)
\(\therefore sin\ A=\cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(tan\ A=\cfrac { 1-{ x }^{ 2 } }{ 2x } \)

x₁ y₁ x₂ y₂
A(-2, -3) B(2, 1)
m : n = 3 : 1
The point P divides AB in the ratio 3 : 1
\(P(x,y)=\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 3\times 2+1\times -2 }{ 3+1 } ,\frac { 3\times 1+1\times -3) }{ 3+1 } \right) \)
\(=\left( \frac { 6-2 }{ 4 } ,\frac { 3-3 }{ 4 } \right) =\left( \frac { 4 }{ 4 } ,\frac { 0 }{ 4 } \right) =(1,0)\)
P is at 25% distance from B on its left and C is at 25% distance from B on its right
\(\therefore\) B is the mid point of PC
Mid point of  \(\bar { PC } =\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((2,1)=\left( \frac { 1+{ x }_{ 2 } }{ 2 } ,\frac { 0+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { 1+{ x }_{ 2 } }{ 2 } =2\quad \frac { { y }_{ 2 } }{ 2 } =1\)
1 + x₂ = 4 y₂ = 2
⇒ x₂ = 3, y₂ = 2
∴ C(x₂, y₂) = (3, 2) is the solutions.

x₁ y₁      x₂ y₂
A(1, 2)   B (6, 7)
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 2\times 6+3\times 1 }{ 2+3 } ,\frac { 2\times 7+3\times 2) }{ 2+3 } \right) \)
\(=\left( \frac { 12-3 }{ 5 } ,\frac { 14+6 }{ 5 } \right) =\left( \frac { 15 }{ 5 } ,\frac { 20 }{ 5 } \right) =(3,4)\)

If \(2cos\theta =\sqrt { 3 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(x=\sqrt { { 2 }^{ 2 }-\sqrt { { 3 }^{ 2 } } } =\sqrt { 4-3 } =\sqrt { 1 } =1\)
\(\therefore sin\theta =\cfrac { 1 }{ 2 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(tan\theta =\cfrac { 1 }{ \sqrt { 3 } } \)
\(coseec\theta =2\)
\(sec\theta =\cfrac { 2 }{ \sqrt { 3 } } \)
\(cot\theta =\sqrt { 3 } \)

(i) \(sinB=\cfrac { 12 }{ 13 } \)
(ii) \(secB=\cfrac { 1 }{ cosB } \)
=  \(\cfrac { 1 }{ 5/3 } =\cfrac { 13 }{ 5 } \)
(iii) \(cotB=\cfrac { 1 }{ tanB } =\cfrac { 1 }{ 12/5 } =\cfrac { 5 }{ 2 } \)
(iv) 

(v) \(\tan C=\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } \)
(vi)  \(cosecC=\cfrac { 1 }{ sinC } =\cfrac { 1 }{ 12/20 } =\cfrac { 20 }{ 12 } =\cfrac { 5 }{ 3 } \)

By the pythagoras theorem,
\(AD=\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \)
= \(\sqrt { 144 } =12\)
AC =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\) 

By the Pythagoras theorem
In \(\triangle\)OAB, \(x=\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } } =\sqrt { 100-64 } =\sqrt { 36 } \)
\(sin\theta =\cfrac { 8 }{ 10 } =\cfrac { 4 }{ 5 } \)
\(cos\theta =\cfrac { 6 }{ 10 } =\cfrac { 3 }{ 5 } \)
\(tan\theta =\cfrac { 8 }{ 6 } =\cfrac { 4 }{ 3 } \)
\(cosec\theta =\cfrac { 10 }{ 8 } =\cfrac { 5 }{ 4 } \)
\(sec\theta =\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } \)
\(cot\theta =\cfrac { 6 }{ 8 } =\cfrac { 3 }{ 4 } \)
Study these thoroughly
\(sin\theta =\cfrac { Opp.side }{ Hypotenuse } \)
\(cos\theta =\cfrac { Adj.side }{ Hypotenuse } \)
\(tan\theta =\cfrac { Opp.side }{ Adj.side } \)
\(cosec\theta =\cfrac { Hypotenuse }{ Opp.side } \)
\(sec\theta =\cfrac { Hypotenuse }{ Opp.side } \)
\(cot\theta =\cfrac { Adj.side }{ Opp.side } \)

Mid point of Diagonal AC:
\(=\left( \frac { -a+a }{ 2 } ,\frac { 0+0 }{ 2 } \right) =\left( \frac { 0 }{ 2 } ,\frac { 0 }{ 2 } \right) =(0,0)\)
Mid point Diagonal of BD:
\(=\left( \frac { 0+0 }{ 2 } ,\frac { b+(-b) }{ 2 } \right) =\left( \frac { 0 }{ 2 } ,\frac { 0 }{ 2 } \right) =(0,0)\)
Mid point of AC = Mid point of BD
∴ Diagonals bisect each other

In the figure\(\angle \)POQ = 100⁰
\(\angle \)PQR = 30⁰
\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\times\)100 = 50⁰
In\(\triangle \)OPQ = \(\angle \)OQP =\(\angle \)POQ = 180⁰
2\(\angle \)OPQ = 180⁰
2\(\angle \) OPQ = 80⁰
\(\therefore \) \(\angle \)OPQ = 40⁰ 
In \(\triangle \)PRQ,
\(\angle \)R+\(\angle \)P+\(\angle \)Q = 180⁰
50⁰ + (40 + x) + 30⁰ = 180⁰
(40 + x)⁰ = 180⁰ - 80 = 100⁰
x⁰ = 100⁰ - 40⁰ = 60⁰
\(\therefore \) \(\angle \)RPO = x = 60⁰

If CD 2 cm and R is the radius, i.e. OC = OA = OB = R
\(\therefore\) OD = OC - DC = R - 2 cm
since AB = 6 cm
\(\Rightarrow \) AD = DB = \(\cfrac { 6 }{ 2 } \) = 3 cm
In \(\triangle \)OBD,

\(\Rightarrow \) OB² = OD² + BD² (By Pythagoras theorem)
\(\Rightarrow \) R² = (R - 2)² + 3²

\(\Rightarrow \) 4R = 13
\(\Rightarrow \)  R = \(\cfrac { 13 }{ 4 } \) = 3.25 cm

Straight line parellel to x-axis

(i) P is (-4, 4) [-4 is abscissa and 4 is ordinate]
(ii) Q is (3, 3) [3 is abscissa and 3 is ordinate]
(iii) R is (4, -2) [4 is abscissa and -2 is ordinate]
(iv) S is (-5, -3) [-5 is abscissa and -3 is ordinate]