Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

Let assumed mean A = 30
C = 5

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

D= b²-4ac
⇒ 4(ac + bd)² - 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² - (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd - a²c²b²c² - a²d² - b²d²]
⇒ 4[2acbd - a²d² - b²c²]
⇒ 4[a²d² + b²c² - 2adbc]
⇒-4[ ad - bc]²
We have ad≠ bc
∴ ad- be of 0
⇒ (ad - bc)² > 0
⇒ 4(ad - bc)² < 0 ⇒ D < 0
Hence the given equation has no real roots.

We have 6 = 2¹ x 3¹ and
20 = 2 x 2 x 5 = 2² x 5¹
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 x 2 x 3 x 5 = 60.
As done in your earlier classes. Note that HCF (6, 20) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 2² x 3¹ x 5¹ = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given paints. One way at showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonals should be equal.
Now,
AB = \(\sqrt { (1-4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
BC = \(\\ \sqrt { (4+1)^{ 2 }+(2+1)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
CD = \(\sqrt { (-1+4)^{ 2 }+(-1-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
DA =\(\sqrt { (1+4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
AC = \(\sqrt { (1+1)^{ 2 }+(7+1)^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
BD = \(\\ \sqrt { (4+4)^{ 2 }+(2-4)^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
Since, AB = BC = CD = DA and AC = BD, all the four sides at the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

Given points (1, - 5) and (4, 2)
Mid-point of the line joining the points (1, - 5), (4, 2)
\(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{1+4}{2}, \frac{-5+2}{2}\right) \)
\(=\left(\frac{5}{2},-\frac{3}{2}\right) \)
Equation of a straight line passing through
\(\left(\frac{5}{2},-\frac{3}{2}\right) \) and parallel to Y-axis is x = a
\(\text { i.e., } x=\frac{5}{2} \Rightarrow 2 x-5=0\)

\(\frac { { x }^{ 2 }-16 }{ { x }^{ 2 }+8x+16 } =\frac { \left( x+4 \right) \left( x-4 \right) }{ { \left( x+4 \right) }^{ 2 } } =\frac { x-4 }{ x+4 } \)

A = B = {(p, q)
A x B = {p, q} x {p, q}
= {(p, p), (p, q), (q, p), (q, q)}
A x A = {p, q} x {p, q}
= {(p, p),(p, q),(q, p),(q, q)}
B x A = {p, q} x {p, q}
= {(p, p),(p, q),(q, p),(q, q)}
A x B = A x A = B x A
Since A = B

If the range = 36.8 and
the smallest value =13.4
Range R = L - S
36.8 = L-13.4
= 36.8 + 13.4 = 50.2
The largest value L = 50.2

Radius of conical flask = 'r' units
Height of conical flask = 'h' units
Volume of conical flask = Volume of water
\(=\frac{1}{3} \pi r^{2} h \text { cu. units }\)
Since, water is poured into the cylindrical flask
Volume of cylinder = Volume of water
\(\pi(\mathrm{xr})^{2} H=\frac{1}{3} \pi r^{2} h\)
[xr - radius of cylinder, H - height]
\(\mathrm{X}^{2} \mathrm{r}^{2} \mathrm{H}=\frac{r^{2}}{3} h\)
Height of the water in cylinder flask
\(\mathrm{H}=\frac{h}{3 x^{2}}\)

in the right triangle ABC [see figure. (a)]
tan \(\theta \) =\(\frac { opposite\ side\ }{ adjacent\ side\ } =\frac { 4 }{ 5 } \)
= tan^-1(0.8)
\(\theta \) = \(38.7°\)(since tan \(38.7°\) = 0.8011)
\(\angle \)BAC = \(38.7°\)

3, 7, 11, ... upto 40 terms.
We have sum of n terms of an A.P.
S_n = \(\frac{n}{2}\) (2a + (n - 1)d)
Here d = 7 -3 = 11 - 7 = 4
t₂- t₁ = t₃ - t₂ and it forms an A.P.
where a = 3,d = 4.
S₄₀ = \(\frac{40}{2}\) (2 X 3 + 39d)
= 20 [6 + (39 x 4)] - 20 [6 + 156]
= 20 [162] = 3240
Sum upto 40 terms = 3240

In \(\triangle\)PST and \(\triangle\)PQR,
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } ,\frac { PT }{ PR } =\frac { 4 }{ 4+2 } =\frac { 2 }{ 3 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is common
Therefore, by SAS similarity,
\(\triangle\) PST~\(\triangle\)PQR

From \(\Delta ADC\) we have
AC² = AD² + CD²       ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD²     ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

Σ(x -10)² = 79 = Σx²- 20x + 100 = 79
= Σx²- 20Σx + 100 x 9 = 79
= Σx²- 20 x 99 + 900 = 79
Σx² = 79 + 1980 - 900 = 1159
Σ(x - \(\bar { x } \))² = Σ(x - 11)² = Σ(x² - 22x + 121)
= Σx² - 22Σx + 121 x 9
= 1159 - 22 x 99 + 1089 = 70
∴ Σx² = 1159, Σ(x - \(\bar { x } \))² = 70

1,1,1,2,2,2,3,3,3
t₂ - t₁ = 1-1 = 0
t₃ - t₂ = 1-1 = 0
t₄ - t₃ = 2-1 = 1
Here t₂ - t₁ ≠ t₃ - t₂
ஃ It is not an A.P.

The area of the triangle formed by the given points is equal to
= \(\frac { 1 }{ 2 } \) [-1.5 (-2 - 4) + 6 (4 - 3) + (-3) (3 + 2)]
= \(\frac { 1 }{ 2 } \) [9 + 6 - 15] = 0
We can have a triangle at area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 x 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

84,90 and 120
First we will find the H.C.F. of 84 and 90.
H.C.F. (90, 84)
Applying Euclid's Division Algorithm until we get remainder zero.
90 = 84 (1) + 6
84 = 6 (14) + 0
Remainder = 0.
H.C.F. (90, 84) = 6
Now finding H.C.F. (120 , 6) we have
120 - 6(20) + 0
Remainder = 0
H.C.F. is 6.
So H.C.F. of 84, 90 and 120 is 6.

Let x₁ are the mid values of the given set
Assumed mean A = 11

Standard deviation
\(\sigma =\sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{N}-\left(\frac{\Sigma f_{i} d_{i}}{N}\right)^{2}} \)
\(=\sqrt{\frac{78}{50}-\left(\frac{8}{50}\right)^{2}}\)
\(=\sqrt{\frac{78}{50}-\frac{64}{2500}} \)
\(=\sqrt{\frac{3900-64}{2500}}=\sqrt{\frac{3836}{2500}} \)
\(=\sqrt{1.5344} \simeq 1.238 \simeq 1.24 \)
Standard deviation \( \sigma \simeq 1.24 \)

Let AB be the building of height 60 m.
DC be the lamp post.
DC = BE
o
o
In the right triangle ADE
\(\tan 38^{\circ}=\frac{A E}{D E}\)
\(0.7813 =\frac{A E}{C B} \)
\(C B =\frac{A E}{0.7813} \)
From the right triangle ACB
\(\tan 60^{\circ} =\frac{A B}{B C} \)
\(\sqrt{3} =\frac{60}{B C} \)
\(B C =\frac{60}{\sqrt{3}}=\frac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\(=\frac{60 \sqrt{3}}{3} C B =20 \sqrt{3} \)
From (1) and (2)
\(\frac{A E}{0.7813}=20 \sqrt{3}\)
AE = 20 x 1.732 x 0.7813
= 34.64 x 0.7813 = 27.064232 = 27.06 m
Now height of the lamp post
= DC = EB = AB - AE = 60 - 22.06 = 32.93 m
Height of the lamp post = 32.93 m

The relationship between selling price and demand is linear. Taking selling price along x- axis and demand along y- axis. We have two points from the data. (14,980) and (16,1220)
Equation of a straight line joining the points (x₁, y₁) and (x₂, y₂) is
\(\frac{y-y_{1}}{y_{2}-y_{1}} =\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-980}{1220-980} =\frac{x-14}{16-14} \)
\(\frac{y-980}{240} =\frac{x-14}{2} \)
y - 980 = 120x - 1680
120x - y - 700 = 0
(or) y = 120x - 700
when x = 17, y = 120(17) - 700
= 2040 - 700 = 1340
Hence the owner could sell 1340 litres of milk weekly at Rs. 17/ litre.

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)

A = \(\frac { x }{ x+1 } \), B = \(\frac { 1 }{ x+1 } \)
\(\frac { { \left( A+B \right) }^{ 2 }+{ \left( A-B \right) }^{ 2 } }{ A+B } \)
\(\frac { { A }^{ 2 }+2AB+{ B }^{ 2 }+{ A }^{ 2 }-2AB+{ B }^{ 2 } }{ (A\div B) } \)
\(=\frac { 2{ A }^{ 2 }+{ 2B }^{ 2 } }{ A\div B } =\frac { 2({ A }^{ 2 }+{ B }^{ 2 }) }{ (A\div B) } \)
\(=\frac { 2\left( { \left( \frac { x }{ x+1 } \right) }^{ 2 }+{ \left( \frac { 1 }{ x+1 } \right) }^{ 2 } \right) }{ \frac { x }{ x+1 } \div \frac { 1 }{ x+1 } } \)

Let h, l, R and r be the height, slant height, outer radius and inner radius of the frustum.
Given that, diameter of the top = 10 m; radius of the top R = 5 m.
diameter of the bottom = 4 m; radius of the bottom r = 2 m, height h = 4 m
Now, \(l=\sqrt { { h }^{ 2 }+\left( R-{ r } \right) ^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+(5-2)^{ 2 } } \)
\(l=\sqrt { 16+9 } =\sqrt { 25 } =5m\)
Here, C.S.A. = \(\pi\)(R + r)l sq. units
\(\frac { 22 }{ 7 } (5+2)\times 5={ 110m }^{ 2 }\)
T.S.A. = \(\pi\)(R + r)l + \(\pi\)R² + \(\pi\)r² sq. units
\(\frac { 22 }{ 7 } \left[ (5+2)5+25+4 \right] =\frac { 1408 }{ 7 } =201.14\)
Therefore, C.S.A. = 110 m² and T.S.A. = 201.14 m²

x² - 8x + 16 = 0
Step 1 Prepare the table of values for the equation y = x² - 8x + 16

Step 2: Plot the points for the above ordered pairs (x, y) on the graph using suitable scale.

Step 3: Draw the parabola and mark the coordinates of the parabola which intersect with the X axis.
Step 4: The roots of the equation are the x coordinates of the intersecting points of the parabola with the X axis (4,0) which is 4.
Since there is only one point of intersection with X axis, the quadratic equation x² - 8x + 16 = 0 has real and equal roots.

y=x²+3x-4

Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4,24).

To solve: X² + 3x - 4 = 0 subtract X² + 3x - 4 = 0 from y = X² + 3x - 4

The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x - co-ordinates (-4, 1) is the solution, set for the equation X2 + 3x - 4 = 0.

Given radius r = 3.6 cm
Length of the tangents PA = PB = 6.2cm

Construction:
Steps
(1) with centre at o, drawn a circle of radius 3.6 cm.
(2) Draw a line OP = 7.2 cm,
(3) Draw a perpendicular bisector of OP, which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 6.2 cm.

Given, diameter (d) = 6 cm, we find radius \((r)=\cfrac { 6 }{ 2 } =3cm\)

Construction
Step 1: With centre at O, draw a circle of radius 3 cm.
Step 2: Draw a line OP of length 8 cm.
Step 3: Draw a perpendicular bisector of OP, which cuts OP at M.
Step 4: With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
Step5: Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 7.4 cm.
Verification : In the right angle triangle OAP,PA² = OP² - OA² = 64 -9 = 55
\(PA=\sqrt { 55= } 7.4\ cm\) (approximately) .