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2014-12-28T18:48:09+05:30

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5.

It is not clear if the diagram is a parallelogram.

Sum of angles = 180 – b + x + 180 – a + y = 360 deg

      X + y = a + b

If it is a parallelogram, as the lines seem to be parallel,

X = y = b = a.

 

6.

  Join OB.  OA = OB = radius.  So OAB is an isosceles triangle. So perpendicular OM on to AB will meet at the center of AB.

AM = AB/2  = 12cm.    OA = 13 cm

Applying Pythagoras theorem,  OM = √(OA² – AM²) = 5 cm

 

12.

Mass of a solid sphere = volume * density

       = 4/3 π R³ * d = 4/3 π * 4.2³ * 10 grams


13. 

side of the cube = a = 8 cm

surface area of one face of a cube = a² = 8² cm

when two cubes are joined side by side, only 5 faces of each cube are visible.  So surface area will be = 5 * a² + 5 a².  When a third cube is joined, one surface gets hidden on the third cube and one of the first two cubes.  Hence

total surface area =  5 a²  + (10 - 1) a² = 14 a²  (totally 14 faces are only visible.


15.

You have to do this on a graph paper.

3 x - 2 y = 4  and  x + y - 3 = 0

the point of intersection will be,  x = 2 and y = 1.  

drawing graph is easy.  Join points (0, -2) , (2, 1), (4, 4)  to get the first line.

       Join points (0, 3),  (2,1),  (3, 0) to get the second line.

Have the scale : 1 division (square) on graph = 1 or 1/2.


16.

C = 5/9 * (F - 32)

Take F on the x- axis and  C on the  y axis.  Keep the origin at C = 0, F = 32. 

Mark  the point   C = 0, F = 32.  Mark 1 division on x axis as 9 deg F and on y-axis  scale is 1 division (square) = 5 deg C.

Now mark a point for  F = 41, C = 5 deg.    and  F = 50 deg,  C = 10 deg.

Join the marked points..  Draw a horizontal line from  y axis at  y = 5 divisions (ie., C = 25 deg), then from the intersection of the line, draw a vertical line on to x - axis.  You should find F = 77 deg. (between 8 and 9 divisions on x  axis.


17. 

Please draw the diagram: marking ABCD, E , and F.  Draw perpendiculars from C and D on to AB. Perpendicular from C intersects Ef at H and AB at J.  Perpendicular from D intersects EF at G and AB at I. 

triangles DEG and DAI are similar, as sides are all parallel in both. 

As DE = 1/2 DA, so EG = 1/2 AI.

triangles CHF and CJB are similar as their sides are all parallel.

as CF = 1/2 CB, then HF = 1/2 JB.

NOW,  EF = EG + GH + HF = 1/2 AI + CD + 1/2 JB 

         = 1/2 (AI + 2 CD + JB ) = 1/2 ( AI + IJ + JB + CD ) = 1/2 (AB+CD)

 we used the fact that ,  CD = GH = IJ,  they are parallel and cut by perpendicular lines.


19.

Please take a paper or a graph and do as follows.

   1. Draw BC horizontally and of length 4.5 cm.

   2. Draw a line at 45 deg at B using instruments you have. Let it be 9 cm or so (long enough).

   3.  mark 2.5 cm on this line from B.  Call this point D.

   4. Draw a perpendicular bisector of CD to intersect the inclined line.  This point will be A. 

   5.  Join  ABC.

 Justification :  AB = AD + 2.5 cm.  AC = AD, as ADC is isosceles.  I hope you know to draw perpendicular bisector of CD.  at mid point of CD, draw a perpendicular.  Or there is a better way by drawing circular arcs from D and C.


20.

curved surface area = 2 π R H    total surface area = 2 π R H + 2 π R ² 

ratio = H / (H + R ) = 2/3    =>  3 H = 2 H + 2 R      =>  H/R = 2


21.

So after deducting the waste of 2 m², the lateral surface area of Conical tent = 550 m².

  Let radius of base = R = 7 m and Height = H at the center, and lateral length (height) be L.

   π R L = 550 m²    => L = 25 m.    

   H² = L² - R²  => H = 24 m 

 Volume = 1/3 π R² H = 1/2 π * 7² * 24  m³


22.  

Take a graph sheet and draw vertical bars of  width 5 mm each.  The height of bars in millimeters will be amount of manure produced.  1 mm = 1 thousand tonne.

Draw x axis (year) and y axis (amount of manure).  Have a distance of 1 cm between one bar and another on x axis.

Draw first bar  of height 18 mm, next 35 mm, 45 mm, 30 mm, 40 mm, 20 mm.  The highest decrease occurred during  the last year 1996-97, as its 20 mm and durin g 1994-95, the decrease is 15 mm.  During other years, there is an increase.


23 .

Probability (age >= 40 yrs) = number of persons with age >= 40 / total number of persons

               = (95+55+15) / (50+35+95+55+15)  = 165 / 250    - simplify

Probability (30 <= age <=39) = 35 / 250  - simplify

P (39 < age < 60) = (95 + 55) / 250


24.

total number of cards  = 24.

numbers which are multiples of 2 and 3 are multiples of 6:  6, 12, 18, 24.

Probability =  4 / 24    ,  as there are four such  cards, multiples of 2 and 3.


25.

(2 x - 3) / 5 + ( x + 3)/4 = (2 x + 3)/ 4

multiply by 20:    4(2 x - 3 ) + 5 ( x + 3 ) = 5 ( 2 x + 3)

            8 x - 12 = 5 x      => x = 4


28.

Let centers of circles be AB. Let them intersect at P and Q.  Let PQ intersect AB at C.  PQ is perpendicular to AB.  PC = CQ.    APC and BPC are right angle triangles.

PC² = AP² - AC² = 10² - AC²

PC² = BP² - BC² = 17² - (21 - AC)² = - 152 - AC² + 42 AC

So,  AC = 252 / 42 = 6 cm

Then PC = √(10²-6²) = 8 cm 

Chord = 2 * PC = PQ = 16 cm


29.

volume of water in cone = V = 4/3 π R₁² H₁ = 1/3 π 20² * 30  cm³

volume of water in cylinder = V = π R₂² H₂ = π 8² H₂ 

     H₂ = π/3 * 400 * 30 / π 64 = 62.50 cm

30.






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2014-12-29T00:50:38+05:30

This Is a Certified Answer

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
27. 
Draw a square ABCD.  Draw the diagonals AC and BD, intersecting at O.

Triangles ABO and CDO are similar, as OD || OB, OA || OC,  included angles COD = angle AOB,  and AB || CD.

AB || CD as angle B = 90 and angle  A = 90 and C = 90 and hence exterior angle at C is also 90.

    AB/CD =  OB/OD =  OC/ OA    = 1,  as  AB = CD in a square.

  So O is the midpoint of AC and BD.  Hence proved.


30.
See diagrams enclosed.  histogram and frequency polygon both.

31.
Origin is  O (0,0)
  Line is  X + Y = 2
Let x = 0,  y =2,  means the Line intersects  y axis at A (0,2)
Let y = 0,  x = 2,    so line intersects  x axis at B (2, 0)

Area of RIGHT angle triangle  AOB,  1/2 * AB * AO
           = 1/2 * 2 * 2 = 2 units square

31 part (ii)
     Draw a parallelogram  ABCD with base AB horizontal. See diagram enclosed.
     Area of both is same = b * h
   Now perimeter of parallelogram P1 = 2 ( b + L ) = 2 ( b + √(h²+CE²)
       perimeter of rectangle = 2 ( b + h) = P2  <  P1
   proved.

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