We have,  Area of ∆ABC = 27 cm2 Or, 12 ⨉ AB ⨉ BC = 27cm2 Or, 12⨉AB ⨉12cm = 27cm2 Or, AB = 4.5 cm. ∴    AB = 4.5 .....

Here, ∡ECB = 100° - 10° = 90°  In right angled ∆ECB, Area = 12 ⨉EB ⨉ EC          = 12 ⨉10cm ⨉ 6cm          = 30 cm2 Now, area of para. ABCD = 2 ⨉ area of ∆ECB [ .....

Given, CA = b = 10 cm    AB = c = 12 cm  Area of ∆ABC = 60 cm2 We know, Area of ∆ABC = ½ bc SinA  or, 60              = ½ ⨉ 10 ⨉ 12 SinA  or, 60     .....

Here,        b = 32 cm       c = 41 cm       ∠A = 130o . We know,        Area of triangle ABC = ½ bc SinA                 .....

Let, AB = Original height of the pole CD = length of the broken part of the pole Let D be the point on the ground at which the top of the pole meets after broken. BC = CD Here, AD = 8 m and ∠ADC = 30o In right angled triangle ACD, tan60o = .....

Here, Join AC. Let, length of rhombus be a cm. Now, Area of ∆ACD = 12 ⨉ Area of rhombus ABCD                          = 12⨉18√3= 9√3 cm2 [ Diagonal bisect rhombus .....

Let, AB = height of the house CD = height of the tower AE = BD = distance between the house and the tower. Here, In right angled triangle CAE, Tan30o  = 12CE/AE or,1√3 = CE60m or, CE   = 20√3 m or, CE   = 34.64 .....

Here, Join AC. We know,         Area of ∆ACD = 12 ⨉ area of //gm ABCD      [Both are standing on the same base and same parallels ] Or, 12⨉ AD⨉DC⨉sin(∡D) = 12⨉ 96√2 Or, 12⨉16 ⨉ .....

Here,  AC = 8 cm and BC = 6 cm Area of ∆ABC = 12√2 cm2 Area of ∆ABC     = ½ ⨉ BC ⨉ CA SinC      or, 12√2          = ½ ⨉ 6 ⨉ 8 ⨉ .....

Here,        b = 2.7 cm       c = 4 cm       ∠A = 100o . We know,          Area of ▲ABC = ½ . b.c SinA               .....

Here, BC = 9 cm   ∡C = 60o Area of ∆ABC = 27 cm2. We know, Area of ∆ABC = ½ ⨉ BC ⨉ AC SinC           or, 27     = ½ ⨉ 9 ⨉ AC Sin60o           or, 27 .....

Let, AB = height of the hill = 90 m  BC = distance of the object from the foot of the hill. In right angled triangle ABC,      Tan45o = ABBC or, 1 = 90mBC or, BC = 90 m. So, the distance of the object from the foot of the hill .....

Here, i.    CE = CD – ED = CD – AB = 25.3 – 1.3 = 24 cm  ii.    In right angled ∆ ACE, Tan60° = CEAE or, AE  = CEtan60              = .....

We know,        Area of ∆SMR = 12 ⨉ area of para. PQRS     [  Both are standing on the same base and same parallels ] Or,12⨉ SM⨉ SR⨉ sin(∡MSR) = ½ ⨉ 60 Or, 12 ⨉ 5√3 ⨉ 8 ⨉ .....

We have given data,          10, 16, 22, 28, 38, 40, 46         So, n = 7. The mean of the given series is given by,     .....

Here,

Marks (C.I)

.....

Here, Calculation of A.M.

Wages (Rs.) (C.I)

Calculation of mean:

Height (cm) (C.I)

Let the missing frequency be “a”. Then,

C.I

Here,

Rainfall (mm) (C.I)

.....

Here,

Marks

.....

Here,

Marks

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        Calculation of  Median. Class interval No. of student (f) Cumulative frequency .....

Marks

                                               Frequency Distribution Table:

We have given,          N= 240,          c.f = 56,          f = 85 and         Median class= 50 – 70.         Median = .....

Here, median = 24, so median class = 20 – 30.                Cumulative frequency table of given data is given by:

            Cumulative frequency table of given data is given by:

.....

            The given data can be constructed in class interval of 10 as follow:  

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Calculation of Median: Marks No. of students (f) Cumulative frequency (cf) 35-45 7 7 45-55 8 15 55-65 10 25 65-75 9 34 75-85 6 40   N = .....

Here,         N = 30,         c.f = 8 ,         f = 14 and         median class = 30 – 40  . The median is given by,  M = .....

Here,           Third quartile (Q3) = 12, so Q3 class = 10 – 12.               Cumulative frequency table of given data is given by:

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Here, the series given are in ascending order.         No. of items  = 7. Now,        Median = (N+12)th term                  .....

Here, the given series are in ascending order.  No. of terms = N = 7. Now, First quartile = (N+14)th term                      = (7+14)th .....

Here, the given series are in ascending order. No. of items = N = 7. Now, Third quartile = 3(N+1)4th term                        = 3(7+1)4th .....

Given,        ∑fx=40 and N=200. We have,  = ΣfxN       = 40200       = 0.2. So, mean is .....

  Solution:

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Calculation of Average production:

Production (m. tons) (C.I)

Solution:                                                        Frequency table

.....

Solution:

Class interval

The given data can be constructed in class interval of 10 as follow:

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The given data can be represented as:

Marks obtained

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                                                 Calculation of mean:

.....

Given,         ∑fx = 60 + 45a and ,          ∑f = 12 + 9a.         Mean () = ? We have,  Mean () = ∑fxN             .....

Here,   X F Cumulative frequency (cf) 5-10 4 4 10-15 10 14 15-20 x .....

Here,           N = 60,           c.f = 15,            f = 16 and            median class = 70 – 80.   The median is .....

Given,         Mean= 10, ∑fx=a and N=15, We have,       = ΣfxN Or, 10 = a​​​​​15 Or,   a = 15×10           .....

Table:   Marks No. of plants (f) Cumulative frequency (cf Less than .....

Here,         Third quartile (Q3) = 60, so Q3 class = 60 – 70 .                           Cumulative frequency table of given data is given .....