Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of (AB and BC) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.


Circum radius = 4.3 cm.

Let ABCD be the rhombus.
Its perimeter = 4 × side = 400 m
Therefore, each side of the rhombus = 100 m
Given the length of the diagonal AC = 120 m
In \(\triangle\)ABC, let a =100 m, b =100 m, c = 120 m

s = \(\frac{a+b+c}{2}=\frac{100+100+120}{2}\) = 160 m
Area of \(\triangle\)ABC =\(\sqrt{160(160-100)(160-100)(160-120)}\)
= \(\sqrt{160 \times 60\times 60 \times40}\)
= \(\sqrt{40 \times 2 \times \times2\times60\times60\times40}\)
= 40 × 2 × 60 = 4800 m²
Therefore, Area of the field ABCD = 2 × Area of \(\triangle\)ABC = 2 × 4800 = 9600 m²

Let x be the digit at ten’s place and y be the digit at unit place.
Given that x + y = 5 …… (1)

Given, Original number − reversing number = 27
(10x + y) − (10y + x) = 27
10x − x + y −10y = 27
9x − 9y = 27
\(\Rightarrow\) x − y = 3 ... (2)
Also from (1), y = 5 – x ... (3)
Substitute (3) in (2) to get x − (5 − x) = 3
x − 5 + x = 3
2x = 8
x = 4
Substituting x = 4 in (3), we get y = 5 − x = 5 − 4
y = 1
Thus, 10x + y = 10 × 4 +1 = 40 +1 = 41.
Therefore, the given two-digit number is 41.
Verification :
sum of the digits = 5
x + y = 5
4 + 1 = 5
5 = 5 true
Original number – reversed number = 27
41 - 14 = 27
27 = 27 true

Let A(3,5), B(8,−10) be the given points and let the point P(x, y) divides the line segment AB internally in the ratio 3:2.
By section formula,
P(x, y) =P\(\left( \frac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n } ,\frac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } \right) \)
Here x₁ = 3, y₁ = 5, x₂ = 8, y₂ = −10 and m = 3, n = 2
Therefore P(x, y) = P\(\left( \frac { 3(8)+2(3) }{ 3+2 } ,\frac { 3(-10)+2(5) }{ 3+2 } \right) \) = P(6, -4)

(i) \({7\over 128}={7\over 2^7}\)

\(\therefore \frac{7}{128}\) has terminating decimal expansion.
(ii) \({21\over 15}={7\over 5}={7\over 5^1}\)
\(\therefore {21\over 15}\) has terminating decimal expansion.
(iii) \(4{9\over 35}={149\over 35}\)
  \(={149\over 5\times 7}\) (It is not in the form of \(\frac{P}{2^m\times 5^n}\))
\(\therefore 4\frac{9}{35}\) has a non-terminating recurring decimal expansion.
(iv) \({219\over 2200}={219\over 2^3\times 5^2\times 11}\) (It is not in the form of \({P\over 2^m\times 5^n}\))

\(\therefore {219\over 2200}\) has a non-terminating recurring decimal expansion.

Let the people who like tea be T. 
Let the people who like coffee be C 
By using formula 
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10 
The number of people who like both coffee and tea = 10.
(ii) The number of people who do not like tea 
n(T) = n(U) – n(T) = 45 – 35 = 10
(iii) The number of people who do not like coffee 
n(C) = n(U) – n(C) = 45 – 20 = 25.

\(\left( A\cup B\cup C \right) \) = {1,2,3,4,5,6,8}
\(\therefore \ n\left( A\cup B\cup C \right) \) =7
Also, n(A) = 5, n(B) = (C) = 4,
Further,\(A\cap B\) = {3,5,6} \(\Rightarrow \) \(n\left( A\cap B \right) \) = 3
\(B\cap C\) = {3,6} \(\Rightarrow \)  \(n\left( B\cap C \right) \) = 2
\(A\cap C\) = {3,5,6} \(\Rightarrow \) \(n\left( A\cap C \right) \) = 3
Also, \(A\cap B\cap C\) = {3,6} \(\Rightarrow \)n\(\left( A\cap B\cap C \right) \) = 2
now, \(n\left( A\cup B\cup C \right) \)= n(A) + n(B) + n(C) -\(n\left( A\cap B \right) -n\left( B\cap C \right) -n\left( A\cap C \right) -n\left( A\cap C \right) +\left( A\cap B\cap C \right) \)
7 = 5 + 4 + 4 - 3 - 2 - 3 + 2
7 = 13 - 8 + 2
7 = 7

(i)
  
\({ x }^{ 0 }=\cfrac { 1 }{ 2 } \angle BOC=\cfrac { 1 }{ 2 } \times \left( { 30 }^{ 0 }+{ 60 }^{ 0 } \right) =\cfrac { 1 }{ 2 } \times { 90 }^{ 0 }={ 45 }^{ 0 }\)
(ii)

\(\angle QPR=\cfrac { 1 }{ 2 } \angle QOR=\cfrac { 1 }{ 2 } \times { 80 }^{ 0 }={ 40 }^{ 0 }\)
In \(\triangle QPR\)
\(\angle R+\angle P+\angle Q={ 180 }^{ 0 }\)
\(\angle Q\) = 180^o-120^o= 60^o
\(\angle Q\) = x^o+ 50^o = 60^o
x^o = 60^o-50^o = 10^o
(iii)

\(\angle \) MPN = 90° (Angle subtended by the diameter is 90°)
\(\angle OMP+\angle OPM={ 180 }^{ 0 }-{ 110 }^{ 0 }={ 70 }^{ 0 }\)
\(\angle MPO=\cfrac { 70 }{ 2 } =35^{ 0 }\)
But \(\angle MPN={ 90 }^{ 0 }\)
\(\therefore\) \(\angle \)OPN = 90^o - 35^o = 55^o
(iv)

Central angle is twice that of angle subtended on the circumference
\(\angle \)YOZ = 120^o\(\times\)2 = 40^o
\(\therefore\) x+\(\angle\)YOZ = 360^o
x^o = 360^o- 240^o = 120^o
(v)

\(\angle\)BOC = 360^o- 240^o = 120^o
\(\therefore\) \(\angle\)BAC = x^o=\(\cfrac { { 120 }^{ 0 } }{ 2 } \) = 60^o

Let P(x) = ax²+5x+b
(x-2) is a factor of P(x), if P(2) = 0
P(2) = a(2)²+5(2)+b = 0
4a +10+b = 0
4a+b = -10...(1)
\((x-\frac{1}{2})\) is a factor of P(x), if P\((\frac{1}{2})\) = 0
P\((\frac{1}{2})\) = a\((\frac{1}{2})^2\) + 5\((\frac{1}{2})\)+ b = 0
\(\frac{a}{4}+\frac{5}{2}+b=0\)
\(\frac{a}{4}+b=\frac{-5}{2}\)
\(\frac{a+4b}{4}=\frac{-5}{2}\)
2a + 8b = -20
a + 4b = -10...(2)
From (1) and (2)
4a + b = -10...(1)
a + 4b = -10...(2)
(1) and (2) ⇒ 4a + b = a + 4b
3a = 3b
a = b
Hence it is proved.

(i) \(\sqrt{3}\times\sqrt{5}\times\sqrt{2}=\sqrt{3\times5\times2}=\sqrt{30}\)
(ii) \(\sqrt{35}\div\sqrt{7}=\sqrt{\frac{35}{7}}=\sqrt{5}\)
(iii) \(\sqrt[3]{27}\times\sqrt[3]{8}\times\sqrt[3]{125}=\sqrt[3]{27\times8\times125}=\sqrt[3]{3^3\times2^3\times5^3}=3\times2\times5=30\)
(iv) \((7\sqrt{a}-5\sqrt{b})(7\sqrt{a}+5\sqrt{b})=(7\sqrt{a})^2-(5\sqrt{b})^2=49a-25b\) = 21\(\sqrt 3\)
(v) \([\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}]\div\sqrt{\frac{16}{81}}\)

=\([\sqrt{\frac{15^2}{27^2}}-\sqrt{\frac{5^2}{12^2}}]\times\sqrt{\frac{9^2}{4^2}}\)
=\((\frac{15}{27}-\frac{5}{12})\times\frac{9}{4}=(\frac{5}{9}-\frac{5}{12})\times\frac{9}{4}\)
=\((\frac{20-15}{36})\times\frac{9}{4}=\frac{5}{ ̶3̶6̶}\times\frac{ ̶9̶}{4}=\frac{5}{16}\)

\(\sqrt[5]{100000}=(100000)^{\frac{1}{5}}=(10^{ ̶5̶})^{\frac{1}{ ̶5̶}}=10\)

n(C) = \(800\times\frac{1}{2}=400\)

n(A∩B) = 70
n(B∩C) = 55
n(C∩A) = 60

(i) Number of people who use exactly two branded soaps
= 50 + 35 + 40 = 125
(ii) Number of people who use at least one branded soaps
= 190 + 55 + 305 + 50 + 35 + 40 + 20 = 695
(iii) Number of people who do not use any of these brands
= 800 - [695] = 105

\(n\left( A\cup B\cup C \right) =n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\)
(i) A = {a,c,e,f,h}, B = {c,d,e,f}, C = {a,b,c,f}
n(A) = 5, n(B) = 4, n(C) = 4
n(A⋂B) = 3
n(B⋂C) = 2
n(A∩C) = 3
n(A∩B∩C) = 2
A∩B = {c,e,f}
B∩C = {c,f}
A∩C = {a,c,f}
A∩B∩C = {c,f}
AUBUC = {a,c,d,e,f,b,h}
ஃ n(AUBUC) = 7............(1)
n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A⋂C)+n(A∩B∩C).........(2)
= 5 + 4 + 4 - 3 - 2 - 3 + 2 = 15 - 8 = 7
(1) = (2)
⇒ n(AUBUC) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A⋂C) + n(A∩B∩C)
Hence it is verified.
(ii) A = {1,3,5}, B = {2,3,5,6}, C = {1,5,6,7}
n(A) = 3, n(B) = 4, n(C) = 4
n(A∩B) = 2
n(B∩C) = 2
n(C⋂A) = 2
n(A⋂B⋂C) = 1
n(AUBUC) = 6
n(AUBUC) = n(A) + n(B) + n(C) - n(A⋂B) - n(B∩C) - n(A⋂C) + n(A∩B∩C)
6 = 3 + 4 + 4 - 2 - 2 - 2 + 1 = 12 - 6 = 6
Hence it is verified.

We have a parallelogram ABCD in which diagonals AC bisect ㄥA.
ㄥDAC = ㄥBAC
(i) To prove that AC bisects ∠C
∵ ABCD is a parallelogram
∴ AB II DC and AC is a transversal
∴ ㄥ1 = ㄥ3 (Alternate interior angle)     .........(1)
.Also BC II AD and AC is a transversal.
∴  ㄥ2 = ㄥ4 (Alternate interior angle)  ...........(2)
But AC bisects ㄥA
∴ ㄥ1 = ㄥ2
From (1), (2) and (3) we get
ㄥ3 = ㄥ4
∴ AC bisects ㄥC.
(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ㄥ1 = ㄥ4 [ ∵ ㄥ1 = ㄥ2 = ㄥ4]
∴ BC = AB (side opposite to equal angles are equal) (4)
Similarly AD = DC            ....... (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram)        .......(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

Distance = \(\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(6-3)^2+(5-7)^2}\)
\(\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}\)
BC=\(\sqrt{(15-6)^2+(-1-5)^2}=\sqrt{(9)^2+(-6)^2}\)
\(\sqrt{81+36}=\sqrt{117}\)
\(\sqrt{9\times 13}-3\sqrt{13}\)
AC =\(\sqrt{(15-3)^2+(-1-7)^2}\)
\(\sqrt{12^2+(-8)^2}=\sqrt{144+64}\)
\(=\sqrt{208}=\sqrt{16\times 13}=4\sqrt{13}\)

AB + BC =AC\(\Rightarrow \sqrt{13}+3\sqrt{13}=4\sqrt{13}\) 
\(\therefore\)The points A,B,C are collinear.

Let f(x) = ax³ + 4x² + 3x –4 and g(x) = x³ – 4x + a, When f(x) is divided by (x–3), the remainder is f(3).
Now f(3) =a(3)³ + 4(3)² + 3(3) - 4
= 27a + 36 + 9 – 4
f(3) = 27a + 41 (1)
When g(x) is divided by (x–3), the remainder is g(3).
Now g(3) = 3³ – 4(3) + a
= 27 – 12 + a
= 15 + a (2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 + a
27a – a = 15 – 41
26a = –26
\(a={-26\over 26}=-1\)
Substituting a = –1,in f(3), we get
f(3) =
= – 27 + 41
f(3) = 14
∴ The remainder is 14.

\(\sqrt{2}=1.414\)
\(\sqrt{3}=1.732\)
The two irrational numbers between \(\sqrt{2}\) and \(\sqrt{3}\) are 1.514 and 1.632