Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

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 P(x, y) be equidistant from the points A (7, 1) and B (3, 5).
We are given that AP = BP. So, AP² = BP²
(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
x²- 14x + 49 + y - 2y + 1 = x²- 6x + 9 + y -10y + 25
x - y = 2
Which is the required relation.

in right triangle ABC [see fig.(b)]
tan\(\theta \) =\(\frac { 8 }{ 3 } \)
= tan^-1(2.66)
\(\theta \) = \(69.4°\)(since tan \(69.4°\)=2.6604)
\(\angle \)BAC = \(69.4°\)

Here each term is decreased by 3. So the next three terms are -7, -10, -13.

P(A) = 0.37, P(B) = 0.42, P(A∩B) = 0.09
P(AUB) = P(A) + P(B) - P(A∩B)
P(AUB) = 0.37 + 0.42 - 0.09 = 0.7

Compare x² + 2x - 2 with the standard form ax² + bx + c = 0
a = 1, b = 2, c = -2
x = \({-b \pm \sqrt{b^2-4ac} \over 2a}\)
substituting the values of a, b and c in the formula we get,
x = \(\frac { -2\pm \sqrt { { \left( 2 \right) }^{ 2 }-4\left( 1 \right) \left( -2 \right) } }{ 2\left( 1 \right) } =\frac { -2\pm \sqrt { 12 } }{ 2 } =-1\pm \sqrt { 3 } \)
Therefore, x = \(-1+\sqrt { 3 } , -1-\sqrt { 3 } \)

Given point (- 1,2), Slope m = \(-\frac{5}{4}\)
Equation of the line passing through (x₁ ,y₁) and having slope 'm' is
y - y₁  = m(x - x₁)
\(y-2=-\frac{5}{4}(x+1)\)
4y - 8 = -5x - 5
5x + 4y - 3 = 0

\( \cot \theta+\tan \theta=\sec \theta \operatorname{cosec} \theta \)
\(\text { LHS } =\cot \theta+\tan \theta \)
\(=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta} \)
\(=\frac{1}{\sin \theta} \times \frac{1}{\cos \theta} \)
\(=\operatorname{cosec} \theta \sec \theta \)
\(=\sec \theta \operatorname{cosec} \theta \)
= RHS
LHS = RHS

The equation of any straight line parallel to X axis is y = b.
Since it passes through (5, 7), b = 7 .
Therefore, the required equation of the line is y = 7.

Let the radii of two cones be r₁ and r₂ and heights be h₁ and h₂
Given ratio of their radii = \(\frac{r_{1}}{r_{2}}=\frac{1}{3}\)
\(r_{1}=\frac{r_{2}}{3} \)
\(h_{1}=3 r_{1}, h_{2}=3 r_{1} \)
[ r₁ is the radius of smaller cone]
Slant heights \(l_{1} =\sqrt{h_{1}^{2}+r_{1}^{2}} \)
\(=\sqrt{9 r_{1}^{2}+r_{1}^{2}}=\sqrt{10} r_{1} \)
\(l_{2} =\sqrt{h_{2}^{2}+r_{2}^{2}} \)
\(=\sqrt{9 r_{1}^{2}+9 r_{1}^{2}}=\sqrt{18 r_{1}^{2}}=3 \sqrt{2} r_{1}\)
Ratio of curved surface areas
\(=\frac{\text { CSA of I cone }}{\text { CSA of II cone }} \)
\(=\frac{\pi r_{1} l_{1}}{\pi r_{2} l_{2}} =\frac{r_{1}\left(\sqrt{10} r_{1}\right)}{\left(3 r_{1}\right)\left(3 \sqrt{2} r_{1}\right)} \)
\(=\frac{\sqrt{10}}{9 \sqrt{2}}= \frac{\sqrt{5} \sqrt{2}}{9 \sqrt{2}}=\frac{\sqrt{5}}{9} \)
Ratio of C.S.A = \(\sqrt{5}: 9\)

It is not a function. Since, a vertical line intersects the curve at more than one points.

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

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

Let BC be the height of the tower and CD be the height of the pole
Let ‘A’ be the point of observation.
Let BC = x and AB = y.
From the diagram,
ㄥBAD = 60° and ㄥXCA = 45° = ㄥBAC
In right triangle ABC, tan 45^o = \(\frac{BC}{AB}\)
gives 1 = \(\frac{x}{y}\) so, x = y         ...(1)
In right triangle ABD, tan60° = \(\frac{BC}{AB}\) = \(\frac{BC+CD}{AB}\)
gives \(\sqrt3\)  = \(\frac{x+5}{y}\) so, \(\sqrt3\)y = x + 5
we get \(\sqrt3\) x = x + 5 [From (1)]
so, \(\frac { 5 }{ \sqrt { 3 } -1 } =\frac { 5 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 5(1.732+1) }{ 2 } \) = 6.83
Hence, height of the tower is 6.83 m.

We are given that sin 3A = cos (A - 26°) ...(1)
Since sin 3A = cos(90° - 3A) we can write (1) as
cos (90° - 3A) = cos (A - 26°)
Since 90° - 3A and A - 26° are both acute angles
90° - 3A = A - 26°
which gives A = 29°

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr² = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

Let P and Q be the points of trisection at AB.
i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1:2. Therefore, the coordinates at P, by applying the section formula, are
\(\left[ \frac { 1(-7)+2(2) }{ 1+2 } ,\frac { 1(7)+2(-2) }{ 1+2 } \right] \) i.e., (-1,10)
Now, Q also divides AB internally in the ratio 2:1, so, the coordinates at Q are
\(\left[ \frac { 2(-7)+1(2) }{ 2+1 } ,\frac { 2(4)+(-2) }{ 2+1 } \right] \) i.e., (-4,2)
Therefore, the coordinates at the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).

Area of quadrilateral BCEG
Vertices are B (- 4, - 2), C (2, - 1), E (1.5, 1) and G (- 4.5, 0.5)
Area of quadrilateral \(=\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)\right. \left.-\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \text { sq. units } \)
\(=\frac{1}{2}[(-4-1.5)(-1-0.5)-(2+4.5)(-2-1)] \)
\(=\frac{1}{2}[(-5.5)(-1.5)-(6.5)(-3)] \)
\(=\frac{1}{2}[8.25+19.5]=\frac{1}{2} \cdot[27.75] \)
= 13.875 = 13.88 sq. units

(A U B) x  C = (A x C) U (B x C)
A = {0,1} , B = {2,3,4} , C = {3,5}
\(A\cup B\) = {0,1,2,3,4}
\((A\cup B)\times C\) = {0,1,2,3,4} x {3,5}
= {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}             ...(1)
A x C = {0,1} x {3,5}
= {(0,3),(0,5),(1,3),(1,5)}
B x C = {2,3,4} x {3,5}
= {(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}
\((A\times C)\cup (B\times C)\) = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}                .....(2)
From (1) and (2), it is clear that
(A U B) x  C = (A x C) U (B x C)
Hence verified.

F^o = (c^o x 1.8) + 32
σ_c = 5°C
σ_F = (1.8 x 5°C) . 9°F
Adding value to data doesn't affect standard deviation.
New variance = σ²_F = 81°F.

Given the number 107,
It is a positive odd integer.
Let a = 107 and b = 4
Applying division algorithm we have,
107 = 4q + r where 0 < r < 4
The possible r = 0, 1, 2, 3.
But 107 is odd, the remainders cannot be 0 or 2.
i.e. 4q or 4q + 2 is not possible to express 107.
The other possibilities are 4q + 1 or 4q + 3
Suppose 4q + 1 = 107
4q = 107- 1
= 105
\(q=\frac{106}{4}\) not a natural numbers
Only possibility is 107 = 4q + 3.
107 - 4q + 3
4q = 107 - 3 = 104
\(q=\frac{104}{4}=26\)
q = 26

Area of the flower bed = a²
Area of the gravel path = 100 - a²
Area of total garden = 100
given cost of flower bed + gravelling = Rs.364
3a² + 4 (100 - a²) = Rs.364
3a² + 400 - 4a² = 364
∴ a² = 400-364
= 36 ⇒ a = 6
∴ width of gravel path \(=\frac { 10-6 }{ 2 } =\frac { 4 }{ 2 } =2cm\)

Coin is in the form of a cylinder
Diameter of the coin = 1.5 cm
Radius of the coin = \(\frac{1.5}{2}\)
Thickness = height = 2 mm = \(\frac{2}{10}=0.2 \mathrm{~cm}\)
Volume of coin (cylinder) = \(\pi r^{2} h\)
\(=\pi\left(\frac{1.5}{2}\right)^{2}(0.2) \)
\(=0.1125 \pi \mathrm{cm}^{3} \)
Diameter of cylinder = 4.5 cm
radius = \(\frac{4.5}{2}=2.25 \mathrm{~cm}\)
height = 10 cm
volume = \(\pi r^{2} h\ sq. units \)
= \(\pi(2.25)^{2}(10) \)
= \(50.625 \pi \)
No.of coins \(=\frac{\text { Volume of cylinder }}{\text { Volume of Coin }} \)
\(=\frac{50.625 \pi}{0.1125 \pi}=450 \text { coins. } \)

Given \(\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a} \)
\(\therefore a=\frac{1+\sin \theta}{\cos \theta} \)
\(\mathrm{LHS}=\frac{a^{2}-1}{a^{2}+1} \)
\(=\frac{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}-1}{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}+1}\)
\(=\frac{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}-1}{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}+1}\)
\(=\frac{\frac{1+\sin ^{2} \theta+2 \sin \theta-\cos ^{2} \theta}{\cos ^{2} \theta}}{\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{\cos ^{2} \theta}}\)
\(=\frac{\left(1-\cos ^{2} \theta\right)+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{1+\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+2 \sin \theta}\)
\(=\frac{\sin ^{2} \theta+\sin ^{2} \theta+2 \sin \theta}{1+1+2 \sin \theta}\)
\(=\frac{2 \sin ^{2} \theta+2 \sin \theta}{2+2 \sin \theta} \)
\(=\frac{2 \sin \theta(\sin \theta+1)}{2(1+\sin \theta)} \)
= sin \(\theta \) = RHS

Let r₁ and r₂ be the radii of sphere and hemisphere respectively.
Given TS.A of sphere = T.S.A of hemisphere
\(4 \pi r_{1}^{2} =3 \pi r_{2}^{2} \)
\(\frac{r_{1}^{2}}{r_{2}^{2}} =\frac{3}{4} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{\sqrt{3}}{2} \)
Ratio of their volumes : \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{2}{3} \pi r_{2}^{3}}=2\left(\frac{r_{1}}{r_{2}}\right)^{3}\)
\(=2\left(\frac{\sqrt{3}}{2}\right)^{3} \)
\(=\frac{3 \sqrt{3}}{4}=3 \sqrt{3}: 4 \)
Ratio of their volumes = \(3 \sqrt{3}: 4\)

Clearly the ladder AC make a right triangle with the wall AB force at a distance BC. \(\angle\)B - 90^o

By Pythagoras theorem
AC² = AB² + BC²
5² = 4² + BC²
BC² = 25 - 16
BC² = 9
BC = 3m
If C moves 1.6 m towards the wall BC becomes
3m - 1.6m = 1.4m
Now in \(\triangle\)ABC
AC² = AB² + BC²
5² = AB2 +(1.4)²
25 - 1.96 = AB²
AB² = 23.04
AB = 4.8m
The new height of the wall = 4.8 m
Difference =4.8 - 4 = -0.8m
The ladder would be placed 0.8 m upward the wall.

Let the first aeroplane starts from O and goes upto A towards north, (Distance=Speed × time)
where \(OA=\left( 100\times \frac { 3 }{ 2 } \right) km=1500km\)
Let the second aeroplane starts from O at the same time and goes upto B towards west,
where \(OB=(1200\times \frac { 3 }{ 2 } )=1800km\)
The required distance to be found is BA.
In right angled triangle AOB, AOB, AB² = OA² + OB²
AB² = (1500)² + (1800)² = 100² (15² +18²)
= 100² x 549 = 100² x 9 x 61
\(AB=100\times 3\times \sqrt { 61 } =300\sqrt { 61 } kms.\)

y = ax + b; x = forehand length; y = height

For all the x-values, there is an image which is 'y
Moreover, the difference between two consecutive 'y' values is constant
In y = ax + b,
(i) The Relation 
R = { (35, 56), (45, 65), (50, 69.5), (55, 74) } is a function
(ii) In y = ax + b
when x = 35,y = 56
56 = 35a + b                 ..............(1)
when x = 45,y = 65
65 = 45a + b                ..............(2)
Solving (1) and (2), we get a = 0.90 and b = 24.5
(iii) Given, forehand length is 40 cm
i.e., when x = 40,y = ax + b
So, y = (0.90) (40) + 24.5 = 60.5
Height of person is 60.5 inches.
(iv) Given height is 53.3 inches
i.e. when y = 53.3, x = ?
53.3 = 0.9x + 24.5
53.3 - 24.5 = 0.9x
x = \(\frac{28.8}{0.9}\)
x = 32 cm
Length of forehand is 32 cm.

(2x-3)(x+2)=0
2x² - 3x + 4x - 6 = 0
2x² + 1x-6 = 0
Let y = 2x² +X - 6= 0

Step 2:
The points to be plotted: (-4,22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2,4), (3,15), (4, 30)
Step 3:
Draw. the parabola and mark the co-ordinates of the intersecting point of the parabola with the x-axis.

Step 4:
The points of intersection of the parabola with the x-axis are (-2, 0) and (1.5,0).
Since the parabola intersects the x-axis at two points, the equation has real and unequal roots
∴ Solution {-2, 1.5}

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.

Given, radius r = 3 cm
Construction
Step 1: Draw a circle with centre at O of radius 3 cm.
Step 2: Take a point P on the circle. Join OP.
Step 3: Draw perpendicular line to OP which passes through P.
Step 4: TT'  is the required tangent.

Construction:
Step (1) Draw a line segment QR = 5 cm.
Step (2) At Q, draw QE such that \(\angle RQE\) = 40°.
Step (3) At Q, draw QF such that \(\angle EQF\) = 90^o
Step (4)Drawn a perpendicular bisector to QR, which intersects QF at 'O' and QR at G.
Step (5) With O as centre and OQ as radius, draw a circle
Step (6) From G marked arcs of radius 4.4 cm on the circle. Marked them as P and S.
Step (7) Joined QP and PR. Now \(\triangle\)PQR is the required triangle
Step (8) From P draw a line PN which is \(\bot \) to LR. LR meets PN at M.
Step (9) The length of the altitude is PM = 2.1cm