Represent the pocket money along x- axis and number of students along the y–axis.
Draw a histogram for the given data. Now, mark the midpoints of the upper sides of the consecutive rectangles. Also mark the midpoints of two imagined class intervals 0-10 and 90-100 whose frequency is 0 on x- axis. Now, join all the midpoints with the help of ruler. We get a frequency polygon imposed on the histogram.

The given distribution is discontinuous. If we represent the given data as it is by a graph we shall get a bar graph, as there will be gaps in between the classes. So, convert this into a continuous distribution using the adjustment factor 0.5.
The first class interval can be written as 9.5-15.5 and the remaining class intervals are changed in the same way. There are no changes in frequencies.
The new continuous frequency table is

The histogram is constructed as below

Given: diagonal = 8 cm


(i) Draw a line segment RM = 8 cm.
(ii) Draw the perpendicular bisector XY to RM. Let it bisect RM at O.
(iii) With O as centre, draw arcs of radius 4 cm on either side of O which cut OX at P and OY at A.
(iv) Join RA, AM, MP and PR.
(v) RAMP is the required square.
Calculation of area:
Area of square RAMP = a² sq.units
= 5.7 x 5.7 = 32.49sq.cm

Given: LI = 6 cm and IE = 7 cm

Steps:
(i) Draw a line segment LI = 6 cm.
(ii) At L, construct LX ⊥LI .
(iii) With I as centre, draw an arc of radius 7 cm and let it cut LX at E.
(iv) With I as centre and LE as radius draw an arc. Also, with E as centre and LI as radius draw an another arc. Let them cut at M.
(v) Join IM and EM.
(vi) LIME is the required rectangle.
Calculation of area:
Area of rectangle LIME = l x b sq.units
= 6 x 3.6 = 21.6 sq.cm

Given: FR = 7 cm and ∠F = 80°


Steps:
(i) Draw a line segment FR = 7 cm.
(ii) At F, make ∠RFX = ∠RFY = 40° on either side of FR.
(iii) At R, make ∠FRP = ∠FRQ = 40° on either side of FR.
(iv) Let FX and RP cut at M and FY and RQ cut at A.
(v) FARM is the required rhombus
Calculation of area:
Area of rhombus FARM =\(\cfrac { 1 }{ 2 } \times { d }_{ 1 }\times { d }_{ 2 }sq.units\) 
= \(\cfrac { 1 }{ 2 } \times 7\times 5.9=20.65sq.cm\)

Given: KE = 6 cm and ∠L = 65°


(i) Draw a line segment LE = 6 cm.
(ii) At L on LE, make ∠ELX = 65° 
(iii) With L as centre draw an arc of radius 6 cm. Let it cut LX at F.
(iv) With E and F as centres, draw arcs of radius 6 cm each and let them cut at A.
(v) Join EA and AF.
(vi) LEAF is the required rhombus
Calculation of area:
Area of rhombus LEAF = \(\cfrac { 1 }{ 2 } \times { d }_{ 1 }\times { d }_{ 2 }sq.units\) 
= \(\cfrac { 1 }{ 2 } \times 6.4\times 10.2=32.64\ sq.cm\)

1642.398 =1.642398 × 10³

Since 4 = 2², we write 4⁻⁵ = (2² )⁻⁵ = 2^2×(−5)
= 2⁻¹⁰ [Here, we use (a^m)ⁿ = a^mn ]

Given that the area of the square field = 3136 m².

ஃ The side of square field =\(\\ \sqrt { 3136 } \ m=\) 56 m
ஃ The perimeter of the square field = 4 x side
= 4 x 56
= 224 m

\(\sqrt { 2\cfrac { 7 }{ 9 } } =\sqrt { \cfrac { 25 }{ 9 } } =\sqrt { \cfrac { { 5 }^{ 2 } }{ { 3 }^{ 2 } } } =\cfrac { 5 }{ 3 } =1\cfrac { 2 }{ 3 } \)

Remembering the rule,\(\sqrt { a } \times \sqrt { b } =\sqrt { a } \times \sqrt { b } \) 
\(\sqrt { 12 } \times \sqrt { 3 } =\sqrt { 12\times 3 } =\sqrt { 36 } =\sqrt { { 6 }^{ 2 } } =6\)

(3.7 x 10^-5)(2 x 10^-3) = 7.4 x 10^-8

\(\left( -3 \right) ^{ 4 }\times \left( \cfrac { 5 }{ 3 } \right) ^{ 4 }={ 3 }^{ 4 }\times \cfrac { { 5 }^{ 4 } }{ { 3 }^{ 4 } } ={ 5 }^{ 4 }=625\)

(- 2)⁵´(- 2)^-3 = (-2) ^{5 - 3} = ( - 2) ² = - 2 x - 2 = 4 .

\({ 4 }^{ -3 }=\cfrac { 1 }{ { 4 }^{ 3 } } =\cfrac { 1 }{ 4\times 4\times 4 } =\cfrac { 1 }{ 64 } \)