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Answer:
Step-by-step explanation:
Q.22 is not visible properly
Q23
ans=
Lateral height of cone = \sqrt{7^2 + 24^2 } =25cm
Surface Area of Article = CSA of Cylinder + CSA of Cone + CSA of Cone
= 2\pi (7)(40)+ \pi (7)(25) + 2 \pi (7)^2
=2618 sq cm
Volume of article = Volume of cylinder - [Volume of cone + volume of hemisphere]
= \pi (7)^2 (40) - [ (1/3) \pi (7)^2 (24) + (2/3) \pi (7)^3 ]
= 12628/3 cubic cm
Q24 also not visible properly
Q25
ans
given in photos
Q26
i can telll u method only
1st draw base of 8cm
then perpenticular bisect it
mark 4cm on altitude
then join other side os base to mark point on altitude
2nd find sides of other triangle
which is similar to 1st triangle then draw the new triangle
by normal triangle method
Q26 part(2)
method
draw triangle abc such that given
then perpendicular bisect the ac
now ad is altitude
therefore triancle adc is right angle triangle
angle d is 90
now draw perpendicular bisector for bc and draw circle such that point a,d and c should be on the circle
then answer is over
Q27
ans
Let the Tower be AC = 100m
Distance between cars is BD
BD=BC+CD
Let the BC be 'x' and CD be 'y'
In Traingle ACD
Cot45°= Base/Perpendicular
Cot45°=y/100
1=y/100
y=100m
In Traingle ACB
Cot30°=x/100
1.732=x/100
1.732×100=x
x=173.2m
Distance between cars = BC+CD
=x+y
=173.2+100
=273.2m
Q28
ans
Let the time taken by man be x days
time taken by boys be y days
work done by 1 man in one day=1/x
work done by 1 boy in one day=1/y
8/x+12/y = 1/10............eq 1
6/x+ 8/y = 1/14..........eq2
Let u=1/x & v=1/y
8u+ 12y=1/10.........eq3
6u+8v= 1/14...........eq4
Multiplying eq3 by 2 & eq4 by 3
16u+ 24y=2/10.........eq5
18u+24v= 3/14.........eq6
subtract eq 5 & 6
-2u= -3/14+2/10
-2u = (-3×10 +2×14)/140
-2u=( -30+28)/140
-2u=-2/140
u=1/140
put this value in eq 5
16u+ 24y=2/10
16×1/140 +24y =1/5
4/35+24y= 1/5
24y= 1/5-4/35
24y= (1×7 -4)/35
24y = (7-4)/35
24y= 3/35
y= 3/35×24
y= 1/280
u=1/x
1/140 = 1/x
x=140
v=1/y
1/280 = 1/y
y=280
A man complete the work in 140 days
A boy can complete the work in 280 days
Q29
ansLet A (x\(_{1}\), y\(_{1}\)), B (x\(_{2}\), y\(_{2}\)) and C (x\(_{3}\), y\(_{3}\)) are the three vertices of the ∆ABC .
Let D be the midpoint of side BC.
Since, the coordinates of B (x\(_{2}\), y\(_{2}\)) and C (x\(_{3}\), y\(_{3}\)), the coordinate of the point D are (\(\frac{x_{2} + x_{3}}{2}\), \(\frac{y_{2} + y_{3}}{2}\)).
Let G(x, y) be the centroid of the triangle ABC.
Then, from the geometry, G is on the median AD and it divides AD in the ratio 2 : 1, that is AG : GD = 2 : 1.
Therefore, x = \(\left \{\frac{2\cdot \frac{(x_{2} + x_{3})}{2} + 1 \cdot x_{1}}{2 + 1}\right \}\) = \(\frac{x_{1} + x _{2} + x_{3}}{3}\)
y = \(\left \{\frac{2\cdot \frac{(y_{2} + y_{3})}{2} + 1 \cdot y_{1}}{2 + 1}\right \}\) = \(\frac{y_{1} + y _{2} + y_{3}}{3}\)
Therefore, the coordinate of the G are (\(\frac{x_{1} + x _{2} + x_{3}}{3}\), \(\frac{y_{1} + y _{2} + y_{3}}{3}\))
Hence, the centroid of a triangle whose vertices are (x\(_{1}\), y\(_{1}\)), (x\(_{2}\), y\(_{2}\)) and (x\(_{3}\), y\(_{3}\)) has the coordinates (\(\frac{x_{1} + x _{2} + x_{3}}{3}\), \(\frac{y_{1} + y _{2} + y_{3}}{3}\)).
Note: The centroid of a triangle divides each median in the ratio 2 : 1 (vertex to base).
Solved examples to find the centroid of a triangle:
1. Find the co-ordinates of the point of intersection of the medians of trangle ABC; given A = (-2, 3), B = (6, 7) and C = (4, 1).
Solution:
Here, (x\(_{1}\) = -2, y\(_{1}\) = 3), (x\(_{2}\) = 6, y\(_{2}\) = 7) and (x\(_{3}\) = 4, y\(_{3}\) = 1),
Let G (x, y) be the centroid of the triangle ABC. Then,
x = \(\frac{x_{1} + x _{2} + x_{3}}{3}\) = \(\frac{(-2) + 6 + 4}{3}\) = \(\frac{8}{3}\)
y = \(\frac{y_{1} + y _{2} + y_{3}}{3}\) = \(\frac{3 + 7 + 1}{3}\) = \(\frac{11}{3}\)
Therefore, the coordinates of the centroid G of the triangle ABC are (\(\frac{8}{3}\), \(\frac{11}{3}\))
Thus, the coordinates of the point of intersection of the medians of triangle are (\(\frac{8}{3}\), \(\frac{11}{3}\)).
Q30
if you know formula find normaly and graphical method
method is
1st find cf(cumilative fre)(opp from median cf that is less than previous class)(c.m.)and other method also in same graph paper
draw polygon graph of cf
two differnt polygon line where intersect each other that poin is median of frequencies
Q30part 2
ans
sum of frequencies is 150
sigma fi=150
12+21+x+52+y+11=150
x+y=150-96
x+y=54 equ1 (it is in use for after )
multiplying by 5
5x+5y=270 equ2 (it is important dont leave this)
mean=sigma fixi/sigmafi
(i have did it directly you dont do directly)
91*150=180+945+5460+1815+75x+135y
91*150=8400+75x+135y
dividing by 15 on b.s.
910=560+5x+9Y
350=5x+9y equ3
subtacting equ3by2
4y=80
y=2
putting y=2 in equ1
x+y=54
x=54-20
x=34
all answers are over please mark it as brainlist
