Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (AB and BC) 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, Band C.


Circum radius = 4.3 cm.

(i) \(\sqrt { 10 } \) is an irrational number ( since 10 is not a perfect square number).
(ii) \(\sqrt { 49 } =7=\frac { 7 }{ 1 } \) a rational number(since 49 is a perfect square number).
(iii) 0.025 is a rational number (since it is a terminating decimal).
(iv) \(0.7\overline { 6 } \) = 0.7666…. is a rational number ( since it is a non – terminating and recurring decimal expansion).
(v) 2.505500555…. is an irrational number (since it is a non – terminating and non–recurring decimal).
(vi) \(\frac { \sqrt { 2 } }{ 2 } =\frac { \sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \) is an irrational number ( since 2 is not a perfect square number).

n(A - B) = 32 + x, n(B A) = 5x
n(A \(\cap\) B) = x
From the Venn diagram:
Given n(A) = n(B)
32 + x + x = x + 5x
32 + 2x = 6x
32 = 6x - 2x
32 = 4x
x = \({32\over 4}=8\)
The value of x = 8.

Let T be the set of families buy Tamil news Papers
Let E be the set of families buy English news Papers
n(TUE) = 100
n(T) = 65 and n(E) = 55
Let "X" be the number families buy both newspapers.
By using venn-diagram
From the venn-diagrarn we get
65 - x + x + 55 - x = 100
120 - x = 100
120 -100 = x
20 = x
(i) Families buy both newspapers = 20
(ii) Families buy Tamil newspapers only = 65 - x
= 65 - 20
= 45
(iii) Families buy English newspapers only = 55 - x
= 55 - 20
= 35.

\(1\over 13\) = 0.07692307 .........
= \(0.\overline{076923}\)

Length of the period of decimal is 6.

Length of the rectangle = 3x + 2 units
Breadth of the rectangle = 3x - 2 units
Area of the rectangle = (3x + 2 ) (3x - 2)
= 9x² - 6x + 6x - 4
= 9x²- 4
When x = 20
Area of the rectangle = 9(20)2 - 4
= 9(400) - 4
= 3600 - 4 = 3596 sq.units.

ABCD is a parallelogram AP and BP are bisectors of consecutive angles ∠A and ∠B.
Since the consecutive angles of a parallelogram are supplementary
∠A + ∠B = 180⁰
\(\frac{1}{2}\angle{A} +\frac{1}{2}\angle{B}=\frac{180^{0}}{2}\)
 ⇒ ∠PAB + ∠PBA = 90^O
In ΔAPB,
∠PAB + ∠APB + ∠PBA = 180^O (angle sum property of triangle)
∠APB = 180^O – [∠PAB + ∠PBA]
= 180^O – 90^O = 90^O
Hence Proved.

\({6\over 7}=0.\overline {857142}\)
\({12\over 13}=0.\overline{923076}\)

The two irrational numbers are  0.866142 ...and 0.903076.......

Here A = (7, 10), B = (-2, 5), C = (3,-4)
\(AB=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 5-10 \right) ^{ 2 } } \)
\(=\sqrt { \left( -9 \right) ^{ 2 }+\left( -5 \right) ^{ 2 } } \)
\(=\sqrt { \left( 81+25 \right) } \)
\(=\sqrt { 106 } \)
\(A{ B }^{ 2 }=106\)............1
\(BC=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 3-\left( -2 \right) \right) ^{ 2 }+\left( -4-5 \right) ^{ 2 } } \)
\(=\sqrt { \left( 5 \right) ^{ 2 }+\left( -9 \right) ^{ 2 } } \)
\(=\sqrt { \left( 81+25 \right) } \)
\(=\sqrt { 106 } \)
\({ BC }^{ 2 }=106\)...............2
\(AC=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 3-7 \right) ^{ 2 }+\left( -4-10 \right) ^{ 2 } } \)
\(=\sqrt { \left( -4 \right) ^{ 2 }+\left( -14 \right) ^{ 2 } } \)
\(=\sqrt { \left( 16+196 \right) } \)
\(=\sqrt { 212 } \)
\({ AC }^{ 2 }=212\)..........3
From (1), (2) & (3) we get
AB²+BC² = 106 +106 = 212 = AC²
AB² + BC² = AC² 
ΔABC is a right angled triangle, right angled at B.