Volume of wood = 12 \(\times\) 10\(\times\) 9 - 10 \(\times\) 8 \(\times\) 7
= 1080 - 560 = 520 cm³ \(\times\)2.00 = Rs. 1040

\(cot\theta =1\)
\(cot\theta =\cfrac { 1 }{ 3 } \)
\(\cfrac { adjacent }{ opposite } =\cfrac { 1 }{ 3 } \)
\(\sqrt { { 3 }^{ 2 }+1 } =\sqrt { 10 } \)
\(\cfrac { 3cos\theta -4sin\theta }{ 5sin\theta +4cos\theta } =\cfrac { 3\times \cfrac { 1 }{ \sqrt { 10 } } -4\times \cfrac { 3 }{ \sqrt { 10 } } }{ 5\times \cfrac { 3 }{ \sqrt { 10 } } +4\times \cfrac { 1 }{ \sqrt { 10 } } } =\cfrac { 3-12 }{ 15+4 } =\cfrac { -9 }{ 19 } \)

\((9,-3)=\left( \frac { m(12+n(7) }{ m+n } ,\frac { m(1)+n(-5) }{ m+n } \right) \)
\(\frac {13m+37n}{ m+n } =9\)
13m + 7n = 9m + 9n
4m = 2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\frac{m-5n}{m+nn}=-3\)
m -5n = -3m -3n
m + 3m= 5n-3n
4m=2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

Volume of a cuboid = 1800 cm³
l = 15 cm h = 12 cm b = ?
l \(\times\) b \(\times\) h = V
15 \(\times\) b \(\times\) 12 = 1800

Bredth = 10 cm

n(S) = 7000 
S - Total no. of lights.
n(A) = 25   
A - Defective ones.
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 25 }{ 7000 } =\frac { 1 }{ 280 } \).

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

\(n\left( A\cup B\cup C \right) \)= n(A) + n(B) + n(C) - \(n\left( A\cap B \right) -\left( B\cap C \right) -n\left( A\cap C \right) +\left( A\cap B\cap B \right) \)

27 = 3 \(\times\) 3 \(\times\) 3; 
therefore 27 = 3³

Distance of the chord from the centre
= \(\sqrt { { 15 }^{ 2 }-{ 12 }^{ 2 } } \)
= \(\sqrt { 225-144 } \)
= \(\sqrt { 89 } \)
= 9 cm

3.1, 3.2, 3.3, 2.1,1.3, 3.3, 3.1
In this given data 3.1, 3.3 occurs twice
\(\therefore\) mode = 3.1 and 3.3(bimodal)

In this example, three values 20, 21, 22 occur two times each. There are three modes for the given data!

m⁴-7m²+1 = (m³+3m+1)(m²-3m+1)

2y³+y²-2y-1 = 2y³-2y+y²-1
= 2y(y²-1)+(y²-1)
= (y²-1)(2y+1)
= (y+1)(y-1)(2y+1)

\(\sqrt{2}\) = 2^1/2

32 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 = 2⁵

7x +y

\({17\over 200}={17\over 2^3\times 5^2}\)
\(\therefore {17\over 200}\) has a terminating decimal expansion.

ㄥP = ㄥR= 168⁰

22

Here the set of all districts in tamilnadu is finite. 
So, infinite set (many lines can be drawn from a point)

Distance between the two points (3,- 9) and (-2, 3)
= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (-2-3)^{ 2 }+(3+9)^{ 2 } } =\sqrt { (-5)^{ 2 }+(12)^{ 2 } } \)
=\(\sqrt { 25+144 } =\sqrt { 169 } \) = 13 units

Let V be the set of people who had Vanilla ice cream and C be the set of people who had Chocolate ice cream.
Then n(V) = 35, n(C) = 30, n(VUC) = 60,
Let x be the number of people who had both ice creams.
From the Venn diagram

35 – x + x +30 – x = 60
65 – x = 60
x = 5
Hence 5 people had both ice creams.
(i) Number of people who had only Vanilla ice cream = 35 – x
= 35 – 5 = 30
(ii) Number of people who had only Chocolate ice cream = 30 – x
= 30 – 5 = 25.
We have learnt to solve problems involving two sets using the formula n(A U B) = n(A)+n(B)−n(A ∩ B). Suppose we have three sets, we can apply this formula to get a similar formula for three sets.

Complement of 27° =90° - 27°
= 63°

Sample space S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 4
E = {5, 6}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

Here, length (l) = 25 m, breadth (b) =15 m, height (h) = 5 m.
Area of four walls = LSA of cuboid
= 2(l + b) × h
= 2(25 +15) × 5
= 80 × 5 = 400 m²
Area of the floor = l × b

= 25 ×15
= 375 m²
Total renovating area of the hall = (Area of four walls + Area of the floor) = (400 + 375) m² = 775 m²
Therefore, cost of renovating at the rate of Rs.80 per m² = 80 × 775
= Rs. 62,000

Given x + 3y = 16 ... (1)
2x – y = 4 ... (2)

Let BC = 13 and AB = 5
sec θ = \(\frac { hypotenuse }{ adjacentside } =\frac { BC }{ AB } =\frac { 13 }{ 5 } \)
By the Pythagoras theorem,
\(AC=\sqrt { { BC }^{ 2 }-{ AB }^{ 2 } } \)
= \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \) = \(\sqrt { 144 } \) = 12
Therefore, \(sin\theta =\frac { AC }{ BC } =\frac { 12 }{ 13 } \) ; \(cos\theta =\frac { AB }{ BC } =\frac { 5 }{ 13 } \)
\(LHS=\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta } =\frac { 2\times \frac { 12 }{ 13 } =3\times \frac { 5 }{ 13 } }{ 4\times \frac { 12 }{ 13 } -9\times \frac { 5 }{ 13 } } =\frac { \frac { 24-15 }{ 13 } }{ \frac { 48-45 }{ 12 } } =\frac { 9 }{ 3 } =3\) = RHS

Let A(x, 3), B(6, y), C(8, 2) and D(9, 4) be the vertices of the parallelogram ABCD. By definition, diagonals AC and BD bisect each other.

Mid-point of AC = Mid-point of BD
\(\left( \frac { x+8 }{ 2 } ,\frac { 3+2 }{ 2 } \right) =\left( \frac { 6+9 }{ 2 } ,\frac { y+4 }{ 2 } \right) \)
equating the coordinates on both sides, we get
\(\frac { x+8 }{ 2 } =\frac { 15 }{ 2 } \)
x + 8 = 15
x = 7
\(\frac { 5 }{ 2 } =\frac { y+4 }{ 2 } \)
5 = y + 4
y = 1
Hence, x = 7 and y = 1.

Let R be the set of children forgotten their rulers.
Let P be the set of children forgotten their pencil.
n(R) = 20, n(P) = 17, n(R\(\cap \)P) = 5
By using venn-diagram

From the venn -diagram we get
(i) Number of children forgotten only pencil = 12
(ii) Number of children forgotten only ruler 15
(iii) Total number of children in the class 15 + 5 + 12 = 32

(4a+3b) (4a-3b) = (4a)² - (3b)² = 16a² - 9b²  [We have (a+b)(a-b) = a² - b²] Put [a = 4a, b = 3b]

(500000)⁵ \(\times\) (3000)³ 
\( =\left(5.0 \times 10^{5}\right)^{3} \times\left(3.0 \times 10^{3}\right)^{3} \\ =(5.0)^{2} \times\left(10^{5}\right)^{2} \times(3.0)^{3} \times\left(10^{3}\right)^{3} \\ =25 \times 10^{10} \times 27 \times 10^{9}=675 \times 10^{19} \\ =675.0 \times 1019=6.75 \times 102 \times 1019=6.75 \times 10^{21} \)

(i)\(\angle\)CAB = 25⁰
\(\therefore\) \(\angle\)BDC = 25⁰
(ii) \(\angle\)DBA = \(\angle\)DCA =180-(90+25⁰)
= 180⁰ - 115⁰
= 65
(iii)\(\angle\)COB = 2\(\angle\)CAB = 2\(\times\)25⁰ = 50⁰

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac { { 360 }^{ 0 } }{ 30 } \)
=12⁰
3x =3 \(\times\) 12 = 36°
5x = 5 \(\times\) 12 = 60°
9x = 9 \(\times\) 12 = 108°
13x = 13 \(\times\) 12 = 156°
The required angles of quadrilateral are 36°, 60° 108° and 156°.

(i) A = {1,3,5,7} ; B = {1,2,4,6,8}
A-B = {1,3,:5,7} - {1,2,4,6,8}
= {3,5,7}
B -A = {1,2,4,6,S} - {1,3,5,7}
= {2,4,6,8}
AΔB = (A - B)U(B-A)
= {3,5,7} U {2,4,6,8}
= {2,3,4,5,6,7,8}.
(ii) BM = (B-A) U (A-B)
= {2,4,6,8} u {3,5,7}
= {2,3,4,516,7,S}

(3, 5)

Let x = \(3.1\overline { 7 } \) = 3.1777 .......  (1)
Here period of decimal is 2, multiply equation (1) by 100
10x = 31.777 .........  (2)
(2)-(1)

\( =\frac{143}{45} \)

In the quadrilateral ABCD, join one of the diagonals, say AC.
Area of \(\triangle\)ABC = \(\frac{1}{2}\)\(\times\) base \(\times\) height
=\(\frac{1}{2}\)\(\times\)8\(\times\)15\(\times\) 60 cm²
By Pythagoras theorem, in right angled triangle ABC,
AC² = AB² + BC²
= 8² +15² = 64 + 225 = 289 cm
Therefore, AC =\(\sqrt{289}\) =17cm
Now, for\(\triangle\)ACD, let us consider a = 17 cm, b =12 cm, c =25 cm
then, s = \(\frac{a+b+c}{2}=\frac{17+12+25}{2}=\frac{54}{2}\) = 27cm

Area of \(\triangle\)ACD =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt{27(27-17)(27-12)(27-25)}\)
=\(\sqrt{27\times10\times15\times2}\)
=\(\sqrt{3\times3\times3\times2\times5\times5\times3\times2}\)
= 3 × 3 × 2 × 5 = 90cm²
Therefore, Area of quadrilateral ABCD
=Area of \(\triangle\)ABC + Area of \(\triangle\)ACD
= 60 + 90 = 150 cm²

Given,U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {-3, -1, 1, 3, 5, 7}
and B = {1, 2, 3, 4}
Law (i) \(\left( A\cup B \right) ^{ ' }\) =  \({ A }^{ ' }\cap { B }^{ ' }\)
\(\left( A\cup B \right) ^{ ' }\) = {-2, 0, 6, 8, 9} and                                ...(1)
Then,A' ={-2, 0, 2, 4, 6, 8, 9} and
B ' = {-3, -2, -1, 0, 5, 6, 7, 8, 9}
\({ A }^{ ' }\cap { B }^{ ' }\)  = {-2, 0, 6, 8, 9}
From (1) and (2) it is verified that                              ...(2)
\(\left( A\cup B \right) ^{ ' }\) =\({ A }^{ ' }\cap { B }^{ ' }\)
Law (ii) \(\left( A\cap B \right) ^{ ' }\) = \({ A }^{ ' }\cap { B }^{ ' }\)
Now,\(A\cap B\) = {1, 3}
\(\left( A\cap B \right) ^{ ' }\) = {-3, -2, -1, 0, 2, 4, 5, 6, 7, 8, 9}             ...(3)
Then,\({ A }^{ ' }\cup { B }^{ ' }\)  = {-3, -2, -1, 0,2 ,4, 5, 6, 7, 8, 9}      ...(4)
From (3) and (4) it is verified that
\(\left( A\cap B \right) ^{ ' }\) = \({ A }^{ ' }\cup { B }^{ ' }\)

Let p (x) 2x³ - x² - 12x - 9
Sum of the co-efficients = 2 - 1- 12- 9 = -20 \(\neq \) 0
Hence x-1 is not a factor
Sum of co-efficients of even powers with constant = -1 - 9 = -10
Sum of co-efficients of odd powers = 2 - 12= -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.

Then p (x) = (x + 1)(2x² - 3x - 9)
Now 2x² - 3x - 9 = 2x² - 6x + 3x - 9 = 2x (x - 3) + 3 (x - 3)
= (x - 3)(2x + 3)
Hence 2x³ - x² - 12x - 9 (x + 1) (x - 3) (2x + 3)

\(\angle\)In the cyclic quadrilateral \(\angle\) A+\(\angle\)C = 180^o
y + 4^o + 3y^o + 8^o = 180^o
4y^o + 12^o = 80^o
4y^o = 180^o -12^o = 168
y =\(\cfrac { 168 }{ 4 } \) = 42
\(\angle\)B + \(\angle\)D = 180^o
8x + 12 = 180^o
8x = 180^o - 12^o = 168
x = \(\cfrac { 168 }{ 8 } \) = 21^o
\(\therefore\) \(\angle\)A = y + 4^o = 42^o + 4^o = 46^o
\(\angle\)C = 3y + 8 = 3 \(\times\) 42 + 8 = 126 + 8 = 134^o
\(\angle\)B = 3x + 6 = 3 \(\times\) 21 + 6 = 63 + 6 = 69^o
\(\angle\)D = 5x + 6 = 5 \(\times\) 21 + 6
= 105 + 6 = 111^o

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
Now the rational numbers between \(-\frac { 202 }{ 24 } and\frac { 15 }{ 24 } are-\frac { 19 }{ 24 } ,-\frac { 18 }{ 24 } ,-\frac { 17 }{ 24 } ,....,\frac { 0 }{ 24 } ,\frac { 1 }{ 24 } ,\frac { 22 }{ 24 } ,...,\frac { 14 }{ 24 } \)
we can take any seven of them \(\frac { 1 }{ 24 } ,\frac { 2 }{ 24 } ,\frac { 3 }{ 24 } ,\frac { 4 }{ 24 } ,\frac { 5 }{ 24 } ,\frac { 6 }{ 24 } ,\frac { 7 }{ 24 } \)

Distance = \(\sqrt{(x_2+x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(3-3)^2+(2+2)^2}\)
=\(\sqrt{0+4^2}=\sqrt{16}=4\)
BC =\(\sqrt{(-1-3)^2+(2-2)^2}\)
\(\sqrt{(-4)^2+0}=\sqrt{16}=4\)
CD =\(\sqrt{(-1-1)^2+(2-2)^2}\)
\(\sqrt{0+(-4)^2}=\sqrt{16}=4\)
AD =\(\sqrt{(-1-3)^2+(-2+2)^2}\)
\(\sqrt{(-4)^2+0}=\sqrt{16}=4\)
AB = BC = CD = DA = 4. All the four sides are equal.

\(\therefore \)ABCD is a Rhombus .................(1)
Diagonal AC =\(\sqrt{(3+1)^2+(-2-2)^2}\)
\(=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}\)
Diagonal BD =\(\sqrt{(3+1)^2+(2+2)^2}\)
\(=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}\)
Diagonal AC = Diagonal BD =\(\sqrt{32}\) ................ (2)
From (1) and (2) we getABCD is a square.

Construction:
Step 1: Draw ΔABC with the given measurements AB = 8 cm, \(\angle\)A = 90^o and AC = 6 cm and construct the perpendicular bisector of any two sides (AB and AC) to find the mid points M and N of AB and BC respectively.
Step 2: Draw the medians (C and BN and let them meet at G. The point G is the centroid of the given ΔABC.

Construction :
Step 1: Draw \(\triangle \)ABC with AB = BC = CA = 6.5 cm
Step 2: Construct angle bisectors of any two angles (A and B) and let them meet at I. I is the incentre of \(\triangle \)ABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre, ID as radius draw the circle. This circle touches all the sides of triangle internally.
Step 5: Measure in radius. In radius = 1.9 cm.