we have \([{5^6\over 5^3}]\times 5^2\)  =[5^6-3]x5²      \((\because {a^m\over a^n}=a^{m-n})\)
= 5³ X 5² 
= 5^{3 + 2}                                         \((\because a^m \times a^n =a^{m+n})\)
= 5⁵
Thus,\([{5^6\over 5^3}]\times 5^2=5^5\)

Steps of construction:
I. Draw a line segment BC = 4.5 cm. 
II. Construct \(\angle\)CBX = 60° at B.
III. From BX, cut off line segment BA = 5.5 cm,
IV. Join AC
Thus, ABC is the required triangle.

Given, perimeter of a rectangular sheet = 100 cm and length of a rectangular sheet, I = 35 cm
Let b be the breadth of a rectangular sheet .
We know that, perimeter of a rectangular sheet = 2 (/ + b)
\(\Rightarrow\)  2(/ + b) = 100  \(\Rightarrow\) 2(35 + b) = 100
\(\Rightarrow\) \(35+b=\frac{100}{2}\)    [dividing by 2 on both sides]
\(\Rightarrow\) 35 + b = 50  \(\Rightarrow\) b = 50 - 35  \(\Rightarrow\) b = 15 cm
\(\therefore\) Area of a rectangular sheet = I X b = 35 x 15 = 525 cm²
Hence, the breadth and area of a rectangular sheet are 15 cm and 525 cm², respectively.

Given, two sides and an angle of ΔPQR are QR = 8 cm, PR = 10 cm and mㄥQ = 90°.
To construct a triangle with these two sides and one right angle, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch with measures marked on it.

Step II Draw a line segment QR = 8 cm.

Step III At point Q, draw QX 丄 QR.

Step IV With R as centre and radius 10 cm, draw an arc which intersects ray QX at P.

Step V Join PR.

Thus, ΔPQR is the required triangle.

Let us consider a square ABCD.

Obviously, each of the above four times, the figure fits on-to-it self.
∴ It has a rotational symmetry of order 4.

In 5n² + 5n - 2, we have
For n = - 2, 5n - 2 = - 12 and,5n² = 5 x (- 2)² = 5 x 4 = 20
Combining,
5n² + 5n - 2 = 20 - 12 = 8

Length of the field = 240 m,
Breadth of the field = 110 m
\(\therefore\) Area of the field = (240 x 110) m²
= 26400 m²
= hectare = 2.64 hectare
= \(\frac { 26400 }{ 10000 } \) hectare = 2.64 hectare
[\(\therefore\) 10000 m² = 1 hectare]

\([(5^2)\times 5^4]\div 5^7=\frac{5^6\times 5^4}{5^7}\)
\(=\frac{5^{6+4}}{5^7}=\frac{5^{10}}{5^7}\)
= 5^10-7
= 5³.

We have, \(\frac{-1}{3}\) and \(\frac{1}{3}\)
\(\therefore\) Half of \(\frac{-1}{3}\) and \(\frac{1}{3}\)=\(​​\frac { \left( \frac { -1 }{ 3 } +\frac { 1 }{ 3 } \right) }{ 2 } =\frac { \left( \frac { -1+1 }{ 3 } \right) }{ 2 } \)
=\(\frac { \left( \frac { 0 }{ 3 } \right) }{ 2 } \)=\(\frac{0}{2}\)=0
(b) We have, \(\frac{1}{6}\) and \(\frac{1}{9}\)

LCM of 6 and 9 = 3 x 2 = 18
\(\therefore\) Half of \(\frac{1}{6}\) and \(\frac{1}{9}\)=\(​​\frac { \left( \frac { 1 }{ 6 } +\frac { 1 }{ 9 } \right) }{ 2 } =\frac { \left( \frac { 3 }{ 18 }+\frac { 2 }{ 18 } \right) }{ 2 } \)
\(\left[ \frac { 1 }{ 6 } \times \frac { 3 }{ 3 } =\frac { 3 }{ 18 } ,\frac { 1 }{ 9 } \times \frac { 2 }{ 2 } =\frac { 2 }{ 18 } \right] \)
=\(\frac { \left( \frac { 5 }{ 18 } \right) }{ 2 } \)=\(\frac{5}{18}\times\frac{1}{2}\)=\(\frac{5}{36}\)

(a) \(\frac { 6 }{ 7 } \times \left( -\frac { 3 }{ 5 } \right) =\frac { 6\times (-3) }{ 7\times 5 } \)
=\(\frac{18}{35}\)
(b) \(\left( -\frac { 11 }{ 4 } \right) \times \frac { 5 }{ 7 } =\frac { (-11)\times 5 }{ 4\times 7 } \)
=-\(\frac{55}{28}\)

Here, Length of the rectangular park = 70 m
Breadth of the rectangular park = 45 m
The cross paths are shown by EFGH and PQRS in the figure.
Now, PQ =3 m and PS = 45 m
EH = 3 m and EF = 70 m
KL = 3 m and KN = 3 m
Now, area of the path = [Area of rectangle PQRS] + [Area of rectangle EFGH]- [Area of square KLMN]
= [PS x PQ] + [EF x EH] - [KL x KN]
= [45 x 3] m² + [70 x 3] m² - [3 x 3] m²
= 135 m² + 210 m² - 9 m² = 336 m²
Now, cost of constructing the path = Rs. 110 x 336 = Rs. 36960.
Note: While finding the area of cross roads, the area of the middle square (here, KLMN) is taken twice
which is to be subtracted once.

(a) (6^-1 - 8^-1)^-1 + (2^-1 - 3^-1)^-1
\(=(\frac{1}{6}-\frac{1}{8})^{-1}+(\frac{1}{2}-\frac{1}{3})^{-1}\)
\(=(\frac{4-3}{24})^{-1}+(\frac{3-2}{6})^{-1}\)
\((\frac{1}{24})^{-1}+(\frac{1}{6})^{-1}\)
= 24 + 6 = 30
(b) \({ \left\{ { 6 }^{ -1 }+{ \left( \frac { 3 }{ 2 } \right) }^{ -1 } \right\} }^{ -1 }\)
\(=(\frac{1}{6}+\frac{2}{3})^{-1}\)
\(=(\frac{1+4}{6})^{-1}=(\frac{5}{6})^{-1}\)
\(=\frac{6}{5}\)