x⁸-y⁸=(x⁴)²-(y⁴)²
=(x⁴+y⁴)(x⁴-y⁴)
=(x⁴+y⁴)[(x²)²-(y²)²]
=(x⁴+y⁴)(x²+y²)(x²-y²)
=(x⁴+y⁴)(x²+y²)(x+y)(x-y)

Construction: Join BD and produce upto E

Proof \(\angle ADE=\angle A+\angle ABD\)
| An exterior angle of a traingle is equal to the sum of its two interior opposite angles
\(\angle CDE=\angle C+\angle ABD\)
|An exterior angle of a traingle is equal to the sum of its two interior  opposite angles
Adding (1) and (2) we get
\(\angle ADE=\angle CDE=\angle A+\angle ABD+\angle C+\angle CBD\)
\(\Rightarrow \angle ADC=\angle A+(\angle ABD+\angle CBD)+\angle C\)
\(\Rightarrow \angle ADC+\angle A+\angle B+\angle C\) 

Given PQ || RS and T is any point
To prove \(\angle PQT+\angle QTS+\angle RST=360^{ 0 }\)
Construction Through T, draw TU || PQ || RS

PQ || UT  |By construction and a transversal QT intersects then
\(\therefore \angle PQT+\angle QTU=180^{ 0 }\)
The Sum of consecutive interior angles on the same sides of a transversal is \(180^{ 0 }\)
UT || RS
| By construction and a transversal TS intersect them
\(\therefore \angle UTS+\angle RST=180^{ 0 }\)
The Sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)
Adding (1) and (2) ,we get
\(\angle PQT+(\angle QTU+\angle UTS)+\angle RST=360^{ 0 }\) 
\(\Rightarrow \angle PQT+\angle QTS+\angle RST=360^{ 0 }\) 

Let us represent time (in hour) by x and distance (in km) by y. Then, we have y =80x.
Table of solution

We plot the points (1, 80), (2, 160), (3, 240) and (4, 320) on a graph paper and join these points by a ruler to get the line which is the graph of the equation y = 80x.

In right triangle AOB
OA² + OB² = AB² [By Pythagoras Theorem]
\(\Rightarrow \) 4² + OB² = 5²
\(\Rightarrow \) 16 + OB² = 25
\(\Rightarrow \) OB² = 9
\(\Rightarrow \) OB = 3
\(\Rightarrow \) B = (0, 3)

Let a = 0 and b=\(\sqrt { 2 } \)
Then, ab = 0\(\times \sqrt { 2 } \) which is rational.
Hence, 'ab' is not necessarily an irrational.

\((x+40)^0+(x-20)^o+\left(\frac{x}{2}-10\right)^o=180^o\)
\(\Rightarrow \frac{5x}{2}-70=180^o\)
\(\Rightarrow \frac{5x}{2}=250^o\)
x=100^o
So, angles are 60^o, 80^o and 40^o

0.5101001000100001...and
0.502002000200002...

84 cm²

384 cm²

(a) (-3,-2)
(b) 0
(c) 0
(d) R, Q, S, T, O

\((6x-5y)(36x^2+30xy+25y^2)\)

\({ (\sqrt { 2 } +\sqrt { 3 } ) }^{ 2 }+(\sqrt { 5 } -\sqrt { 2 } { ) }^{ 2 }\)=\({ (\sqrt { 2 } ) }^{ 2 }+{ (\sqrt { 3 } ) }^{ 2 }+2\times \sqrt { 2 } \times \sqrt { 3 } +{ (\sqrt { 5 } ) }^{ 2 }+{ (\sqrt { 2 } ) }^{ 2 }-2\times \sqrt { 5 } \times \sqrt { 2 } \)
=2+3+2\(\sqrt { 6 } \)+5+2-2\(\sqrt { 10 } \)
= 12+2\(\sqrt { 6 } \)-2\(\sqrt { 10 } \)
=2(6+\(\sqrt { 6 } \)-\(\sqrt { 10 } \))

\({ \left( 729 \right) }^{ \frac { -1 }{ 6 } }={ \left( { 3 }^{ 6 } \right) }^{ \frac { -4 }{ 6 } }={ 3 }^{ -4 }\)
=\(\frac{1}{3}\)

For one triangular piece = 20 cm, b = 50 cm, c = 50 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+50+50 }{ 2 } =\frac { 120 }{ 2 } =60\) cm
\(\therefore \)  Area of one triangle \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-50)(60-50) } \)
\(=\sqrt { (60)(40)(10)(10) } =200\sqrt { 6 } \) cm²
\(\therefore \)  Area of 5 triangles of one colour \(=5(200\sqrt { 6 } )\) cm² = 1000\(\sqrt { 6 } \) cm²
Hence, 1000\(\sqrt { 6 } \) cm² cloth of each colour is required for the umbrella.

Rs.5376

(a)(4, 2)
(b)\(\left({36\over13},{24\over13}\right)\)

The given equation is
y=3
(i)In one variable
The representation of y = 3 on the number line is as shown below:

(ii)In two variables

It is a linear equation in two variables x and y .This is represented by a line. All the values of x are permissible because O.x is always O.However, y must satisfy the relation y = 3. Hence, two solutions of the given equation are x = 0, y = 3 and x = 2, y = 3.Thus the graph AB is a line parallel to the x-axis at a distance of 3 units above it

II

(i) \(B\rightarrow (-5,2)\)
(ii) \(C\rightarrow \left( 5,-5 \right) \)
(iii) E
(iv) G
(v) 6
(vi) -3
(vii) \(L\rightarrow (0,5)\)
(viii) \(M\rightarrow (-3,0)\)