Assignment 6

WEEKLY ASSIGNMENT 6

1.    In the given figure alongside, ABCD is a rectangle and M is the point on DC. If area of rectangle is 82cm², find the area of ABM.
Solution :       Area of ABM = ½ Area of rectangleABCD
                                                  = ½  82     =  41cm²

2.    Find the LCM : 3(a³+b³), a(a²–ab+b²)
Solution :       First expression     = 3(a³+b³)
                                                              = 3(a + b)(a²­ab+b²)
                  Second expression      = a(a² – ab + b²)
                                          LCM           = CF   RF
                                                        = (a² – ab + b²)  3(a+b)  a
                                                              = 3a(a³+b³)

3.    Find the HCF : x²+5x–6 , x³–1
Solution :        First expression   = x² – 5x +6
                                                              = x² – (3+2)x + 6
                                                              = x² – 3x – 2x +6
                                                              = x(x – 3 ) – 2(x – 3)
                                                              = (x – 3)(x – 2)
               Second expression       = x³ – 1
                                                              = (x)³ – (1)³
                                                              = (x – 1)(x²+x.1 + 1²)
                                                              = (x – 1)(x²+x + 1)
                              HCF                        = 1

4.    In the given figure, ABCD is a square whose length of side is 8cm. If AE=10cm, find the length of BF
Solution :                       Area of square ABCD = 2  Area of AEB
                                  or,            (side)2              = 2  ½ AEBF
                                      or,              (8)²                   = 10  BF
                                  or,   64                              = 10  BF
                                      or,       6.4 cm                   = BF

5.   In the given figure, ABCD is a parallelogram and M is the midpoint of BC. If AD=2MD=20cm, DMC =30^o, find the area of  ABM
Solution :
We are given,
          AD =BC = 20cm, 2MD = 20cm so, MD = 10cm
          Since M is the midpoint of BC, 2MC = BC
                                                          i.e, 2MC = 20 cm so, MC = 10cm
          Now,
                          Area of ∆DCM = ½ DM ×MC × Sin ∡DMC
                                                      = ½ ×10 ×10 ×Sin30^o
                                                      = ½  ×10 ×10 × ½
                                                      = 25cm²
 Since, ∆DCM and ABM lie on equal base and same parallels , Area of ABM = 25cm²

6.   In a continuous series, if mean(= 20, If fm = 25a+80 and f= a+10, find the value of a.
Solution :
          We are given,
                          mean(= 20,  = 25a+80 and f= a+10
           Now,
                          Mean( =
          or,            20          =
          or,  20(a + 10)      = 25a + 80
          or, 20a + 200        = 25a + 80
          or, 200 – 80          = 25a – 20a
          or, 120                    = 5a
          or, 24                       = a
          Hence the value of a is 24.

7.    Two children were given birth by a couple at the interval of 4 years. Illustrate the
probabilities of having son or daughter by drawing a tree diagram and calculate the
probability of both are daughters.
Solution :
          Let S and D denote the set of Son and Daughter respectively.
          n(S)= 1, n(D) = 1
















                         

                          P(DD) = ½ × ½ = ¼
          Hence the probability of both are daughter is ¼. 

8.    In a group of 65 players, 11 play basketball only and 33 play Cricket only. If the number of players who play Cricket is twice the number of students who play Basketball, by using Venn diagram find the number of players who play both games and who don’t play any game.
Solution :
          Let B and C denote the set of players who play Basketball and Cricket respectively.
          We are given,
                          n(U)       = 65
                          n_o(B)      =11
                          n_o(C)      = 33
                          n(B∩C) = ? x (say)
          n()= ?y (say)
          According to question,
                          n(C)       = 2 × n(B)
          or,          33+x      = 2(11+x)
          or,          33+x      = 22+2x
          or,          33–22    = 2x–x
          or,          11           = x
          From above Venn-diagram,
                          n(U)       = 11+x+33+y
          or,          65           = 11+11+33+y = 55+y
          or,          65–55    = y
          or,          10           = y
          Hence, the number of players who play both is 11 and who play none is 10.

9.    An umbrella is made by stitching 8 triangular pieces of cloths of two different colors each piece measuring 30cm, 113cm, 113cm. How much cloth is required for the umbrella? Find the cost of cloth at the rate of Rs 2 per cm².
Solution:
          Here, measure of sides of triangle are 113cm , 113cm , 30cm
          Now,
                          Area of isosceles triangle              =  
                                                                                          =  
                                                                                          = 1680cm²
          So, Area of 8 triangular pieces                    = 8 × 1680 = 13,440cm²
          Hence, 13,440cm² of cloth is required.
          Total cost             = 13,440 × 2 = Rs 26,880

10.   Simplify :      +    + 

Solution:              +    + 
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