its my holiday home work plz dont delete it plz
Answers
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.
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.
