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SA1 Answers4

Probability Introduction :

SA1 Answers4

19)

Given : \(\Delta ABC\)  ~ \(\Delta DEF\)
           ar(\(\Delta ABC\)) = ar(\(\Delta DEF\))

RTP   : \(\Delta ABC\) \(\cong\) \(\Delta DEF\)

Proof :

           \(\large \frac{ar(\Delta ABC}{ar \Delta DEF} = \frac{AB²}{DE²}\)

           1    = \(\large \frac{AB²}{DE²}\)

           AB²= DE²

           AB = DE

           Similarly, we can prove BC = EF, AC = DF.

           In \(\Delta ABC\)  and  \(\Delta DEF\) are congruent by SSS.

                            AB      =      DE
                            BC      =      EF
                            CA      =      FD                            

            Hence proved.

              (OR)                   

Given : AB \(\perp\)BC, DE\(\perp\)AC, GF\(\perp\)BC.

RTP   : \(\triangle AED \)~\( \triangle GCf\)

Proof : In \(\triangle ABC\)  and \(\triangle AED\)
               \(\angle ABC\)   =   \(\angle DEA\) = 90   
               \(\angle BAC\)   =   \(\angle EAD\)          

           Therefore, \(\triangle ABC\)  ~ \(\triangle AED\) by AA similarity
           criterion.     

           Similarly we can prove that \(\triangle ABC\) ~ \(\triangle GFC\) by 
           AA similarity criterion. 

           \(\triangle AED\) ~ \(\triangle ABC\) ~ \(\triangle GFC\)

           So, \(\triangle AED\) ~ \(\triangle GFC\)

           Hence proved.


20)

Given  : ABCD is a rhombus, AC and BD are diagonals and E is the meeting
            point of these diagonals.

RTP    : AC² + BD² = 4AB²

Proof  : In triangle AEB,
           AB² = AE² + BE²

           AB² = \(\large \frac{AC²}{4} + \frac{BD²}{4}\)

           AB² = \(\large \frac{AC² + BD²}{4}\)

           4AB² = AC² + BD²
           Hence Proved.


21)

LHS

= \(\large \frac{cos(90-A) sin(90-A)}{tan(90-A)}\)

= \(\large \frac{(sinA cosA)}{\frac{cosA}{sinA}}\)

= \(\large \frac{(sinA cosA)sinA}{cosA}\)

= sin²A

= RHS

Hence Proved.



22)

cosΘ = \(\large \frac{adj}{hyp}\) = \(\large \frac{3}{5}

adj = 3 , hyp = 5

opp = √(hyp² - adj²)

      = √(5² - 3²)

      = √(25 - 9)

      = √16

      = 4.

cosecΘ = \(\large \frac{hyp}{opp}\)

           = \(\large \frac{5}{4}

cosΘ + cosecΘ = \(\large \frac{3}{5} + \frac{5}{4}\)

                      = \(\large \frac{12+25}{20}\)

                      = \(\large \frac{35}{20}\)

                      = \(\large \frac{7}{4}\)

(OR)

= cos²20° + cos²70° + sin48° sec42° + cos40° cosec50°

= cos²20° + cos²(90-20)° + sin(90-42)° sec42° + cos(90-50)° cosec50°

= 1 - sin²20° + sin²20° + cos42° sec42° + sin50° cosec50°

= 1+1+1

= 3



23)

mode = l + \(\large \left [ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right]\) h

Here, l = 20, \(f_1 = 22, f_0 = 12, f_1 = 8\), h = 10

mode = 20 + \(\large \left [ \frac{22 - 12}{2(22) - 12 - 8} \right]\) 10

          = 20 + \(\large \frac{100}{22 } \)

          = \(\large \frac{220 + 50}{11}\)

          = 24.5.

Therefore mode is 24.5.



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