We can write this as \(\sqrt { 42.25 } =\sqrt { \cfrac { 4225 }{ 100 } } =\cfrac { \sqrt { 4225 } }{ \sqrt { 100 } } \) 
Now, it is easy to compute the square root of the whole number 4225, without any botheration of decimal symbol.
\(\sqrt { 4225 } =65\) and so, we now get \(\sqrt { 42.25 } =\sqrt { \cfrac { 42.25 }{ 100 } } =\cfrac { \sqrt { 4225 } }{ \sqrt { 100 } } =\cfrac { 65 }{ 10 } \) 
Th is is another way of tackling problems of square root of decimal numbers instead of long division method.

(i)

725 = 5 x 5 x 29
= 52 x 29
The second prime number 29 does not have a pair. Hence 725 is not a perfect square number.
(ii)

190 = 2 x 5 x 19
All the three prime numbers do not have pairs.
Hence 190 is not a perfect square number.
(iii) 841

841 = 29 x 29 = 29²
\(\sqrt { 841 } =\sqrt { 29^{ 2 } } =29\) 
\(\sqrt { 841 } =29\) 
Hence 841 is a perfect square number.
(iv) 1089
 
1089 = 3 x 3 x 11 x 11
= 3²X11²
\(\sqrt { 1089 } =\sqrt { { 3 }^{ 2 }\times { 11 }^{ 2 } } \) 
= \(\sqrt { (3\times 11)^{ 2 } } \) 
= 3 x 11
= 33
\(\sqrt { 1089 } =33\) 
Hence 1089 is a perfect square number.

Given:
NI = 4.5 cm, IC = 4.3 cm,
NE = 3.5 cm and two diagonals,
NC = 5.5 cm and IE = 5 cm


Steps:
1. Draw a line segment NI = 4.5 cm.
2. With N and I as centres, draw arcs of radii 5.5 cm and 4.3 cm respectively and let them cut at C.
3. Join NC and IC.
4. With N and I as centres, draw arcs of radii 3.5 cm and 5 cm respectively and let them cut at E.
5. Join NE, IE and CE .
6. NICE is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral NICE = \(\frac12\)× d × (h₁+ h₂) sq. units
= \(\frac12\) x 5 x (2.4 + 3.1)
= 2.5 x 5.5 = 13.75 cm²

Given: DE = 6 cm, EA = 5 cm, AR = 5.5 cm,
RD = 5.2 cm and a diagonal DA = 10 cm


Steps:
1. Draw a line segment DE = 6 cm.
2. With D and E as centres, draw arcs of radii 10 cm and 5 cm respectively and let them cut at A.
3. Join DA and EA.
4. With D and A as centres, draw arcs of radii 5.2 cm and 5.5 cm respectively and let them cut at R.
5. Join DR and AR.
6. DEAR is the required quadrilateral
Calculation of Area:
Area of the quadrilateral DEAR = \(\frac12\) x d x (h₁+ h₂) sq. units
= \(\frac12\) x 10 x (1.9+ 2.3)
= 5 x 4.2 = 21 cm²

We have, (5y³ − 25y² + 8y) ÷ 5y
= \(\cfrac { { 5y }^{ 3 }-25{ y }^{ 2 }+8y }{ 5y } \)

 = \({ y }^{ 3-1 }-{ 5y }^{ 2-1 }+\cfrac { 8 }{ 5 } \)
= \({ y }^{ 2 }-5y+\cfrac { 8 }{ 5 } \)

The given field has four triangles (I, III, IV & V) and a trapezium (II).
Area of the triangle (I) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 5\times 6=15{ m }^{ 2 }\)
Area of the trapezium (II) \(=\frac { 1 }{ 2 } h(a+b)=\frac { 1 }{ 2 } \times 13\times (6+4)=65{ m }^{ 2 }\)
Area of the triangle (III) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 8\times 4=\frac { 32 }{ 2 } =16{ m }^{ 2 }\)
Area of the triangle (IV) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
Area of the triangle (V) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
∴ The total area of the field = 15 + 65 + 16 + 65 + 65 = 226 m²

(i) 17²  = 17 x 17 = 289
(ii) 203² = 203 x 203 = 41209
(iii) 1098² = 1098 x 1098 = 1205604

LCM of 5 , 7 = 35
\(\text { So, } \frac{7}{5}+\frac{5}{7} =\frac{7 \times 7}{5 \times 7}+\frac{5 \times 5}{7 \times 5} \)
\(=\frac{49}{35}+\frac{25}{35}=\frac{49+25}{35}=\frac{74}{35}\)

\(\frac{1}{3}=0.3333 \ldots\) (by actual division and it is recurring and non-terminating)

There are two triplets. We need one more 3.
There fore 1944 is not a perfect cube.

(10x - 7y + 5z) by 6xyz
(10x - 7y + 5z) x 6xyz
Multiplying each term ofthe polynomial by the monomial
= (10x) (6xyz) - (7y) (6xyz) + 5z (6xyz)
= 60x²yz - 42xy²z + 30xyz²

\(\frac{-21}{5} \div \frac{-7}{-10} =\frac{-21}{5} \times \frac{10}{7} \)
\(=\frac{-3}{1} \times \frac{2}{1}=-6 \)

\(\frac{9}{132} \times \frac{-11}{3}=\frac{3}{12} \times \frac{-1}{1}=\frac{-1}{4} \)

The rational number for the point marked on the number line is  -3\(\frac{2}{3}\) = \(\frac{-11}{3}\)

\(\frac { 15 }{ -4 } =-3\frac { 3 }{ 4 } \)
\(\frac { 15 }{ -4 } \) lies between -3 and - 4.

∴ \(\frac{9}{4}\) lies between 2 and 3

\(\cfrac { { 3 }m^{ 2 } }{ m } +\cfrac { { 2m }^{ 4 } }{ { m }^{ 3 } } \) = 3m + 2m = (3 + 2)m = 5m

Cuboid

Cube

3x²y −3xy³, x²y²  = -9x⁵y⁶