Math, asked by rppilania, 1 year ago

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Answers

Answered by kvnmurty
4

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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Answered by BrainlyGood
4
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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