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

Total number of cards = 52
Event of selecting a spade card = A
Event of selecting a heart card = B
n(A) = 13, n(B) = 13
P(A) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \), P(B) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \).
A ⋂ B refers to a card is both spade and heart. It is not possible

A, B are exclusive events.
P(A,B) = P(A) + P(B) =\(\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \).

No. of coins required \(=\frac{Volume\ of\ the\ cylinder}{Volume\ of\ 1\ coin}\)
\(=\frac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { \pi \times \frac { 42 }{ 20 } \times \frac { 45 }{ 20 } \times 10 }{ \pi \times \frac { 15 }{ 20 } \times \frac { 15 }{ 20 } \times \frac { 2 }{ 10 } } \)

= 450

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.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

Given vertices are A (- 5, - 4), B (1, 6) and C (7, - 4)
Area of triangle \(=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\right. \left.\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \text { sq. units } \)
Area of triangle ABC \(=\frac{1}{2}[-5(6+4)+ 1(-4+4)+7(-4-6)] \)
\(=\frac{1}{2}[-50+0-70]=\frac{-120}{2}=-60\)
[Area cannot be negative].
Area = 60 sq. units.
Given that one bucket of paint can be applied for 6 sq. feet
No. of buckets \(=\frac{60}{6}=10\)

LHS = A(B + C)
B + C = \(\left[ \begin{matrix} 1 & 2 \\ -4 & 2 \end{matrix} \right] +\left[ \begin{matrix} -7 & 6 \\ 3 & 2 \end{matrix} \right] =\left[ \begin{matrix} -6 & 8 \\ -1 & 4 \end{matrix} \right] \)
A(B + C) = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} -6 & 8 \\ -1 & 4 \end{matrix} \right] =\left[ \begin{matrix} -6-1 & 8+4 \\ 6-3 & -8+12 \end{matrix} \right] =\left[ \begin{matrix} -7 & 12 \\ 3 & 4 \end{matrix} \right] \) ......(1)
RHS = AB + AC
AB = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 2 \\ -4 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1-4 & 2+2 \\ -1-12 & -2+6 \end{matrix} \right] =\left[ \begin{matrix} -3 & 4 \\ -13 & 4 \end{matrix} \right] \)
AC = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} -7 & 6 \\ 3 & 2 \end{matrix} \right] =\left[ \begin{matrix} -7+3 & 6+2 \\ 7+9 & -6+6 \end{matrix} \right] =\left[ \begin{matrix} -4 & 8 \\ 16 & 0 \end{matrix} \right] \)
Therefore, AB + AC = \(\left[ \begin{matrix} -3 & 4 \\ -13 & 4 \end{matrix} \right] +\left[ \begin{matrix} -4 & 8 \\ 16 & 0 \end{matrix} \right] =\left[ \begin{matrix} -7 & 12 \\ 3 & 4 \end{matrix} \right] \) ....(2)
From (1) and (2), A(B + C) = AB + AC. Hence proved.

Number of green balls is n(G) = 6
Let number of red balls is n(R) = x
Therefore, number of black balls is n(B) = 2x
Total number of balls n(S) = 6 + x + 2x = 6 + 3x
It is given that, P(G) = 3 x P(R)
\(\frac { 6 }{ 6+3x } =3\times \frac { x }{ 6+3x } \)
3x = 6 gives, x = 2
(i) Number of black balls = 2 x 2 = 4
(ii) Total number of balls = 6 + (3 x 2) = 12

Let AC be the pole and let point 'B' divide it in the ratio 
\(\not x: 9 \not x=1: 9\)
Let 'D' be the point 25 m.
\(tan\alpha =\frac { x }{ 25 } \) \(tan2\alpha =\frac { 10x }{ 25 } \)
\(tan2\alpha =\frac { 2tan\alpha }{ 1-{ tan }^{ 2 }\alpha } \)
\(\frac{\not 10 x^{5}}{\not 25}=\frac{2 \times \frac{x}{25}}{1-\frac{x^{2}}{625}}\)

Height of pole = 10x
= \(100\sqrt { 5 } \)
x = \(10\sqrt { 5 } \) m

i) f(0) = 4
ii) \(f(3)=\sqrt { 3-1 } =\sqrt { 2 } \)
iii)  \(f(a+1)=\sqrt { a+1-1 } =\sqrt { a } \)

Let r be the radius of the hemisphere.
Given that, volume of the hemisphere = 29106 cm³
Now, volume of new hemisphere = \(\frac{2}{3}\)(Volume of original sphere)
= \(\frac{2}{3}\) x 29106
Volume of new hemisphere = 19404 cm³
\(\frac { 2 }{ 3 } \pi { r }^{ 3 }=19404\)
\({ r }^{ 3 }=\frac { 19404\times 3\times 7 }{ 2\times 22 } =9261\)
\(r=\sqrt [ 3 ]{ 9261 } =21cm\)
Therefore, r = 21 cm

Let a be the first term and d be the common difference. It is given that
t₁ = x, t_m = y, t_n = z
Using the general term formula
a + (l - 1)d = x .............(1)
a + (m - 1)d = y ............(2)
a + (n -1)d = z ................(3)
We have, x (m- n) +y (n -l) + z(l -m)
= a[(m-n) + (n - l) + (l - m)]  + d [(m - n)(l - 1) + (n - l) (m -1) + (l - m) (n - 1)]
= a[0] + d[lm - ln - m + n + mn - lm - n + l + ln - mn - l + m]
= a(0) + d(0) = 0

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),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

σ =\(\sqrt { \frac { \Sigma x^{ 2 } }{ n } -\left( \frac { \Sigma x }{ n } \right) ^{ 2 } } \)
=\(\\ \sqrt { \frac { 20300 }{ 7 } -\left( \frac { 350 }{ 7 } \right) ^{ 2 } } \)
=\(\sqrt { 400 } \) = 20

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.

4x + 3y + 12 = 0
4x + 3y = -12
Dividing by - 12
\(\Rightarrow \ \frac{4 x}{-12}+\frac{3 y}{-12} =1 \)
\(\Rightarrow \ \frac{x}{(-3)}+\frac{y}{(-4)} =1 \)
\(\frac{x}{a}+\frac{y}{b} =1 \) 
x intercept = a = -3
y intercept = b = -4.

In \(\triangle\)PST and ​​\(\triangle\)PQR
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+3 } =\frac { 2 }{ 5 } ,\frac { PT }{ PR } =\frac { 2 }{ 2+3 } =\frac { 2 }{ 5 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is is common
Therefore, by SAS similarity,
\(\triangle\)PST~\(\triangle\)PQR

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

 f(x) = 4x² - 1, g(x) = 1 + x
f o g = f(g(x)) = f(1 + x) = 4(1 + x)² - 1
= 4(1 + 2x + x²) - 1
= 4 + 8x + 4x² - 1
= 4x² + 8x + 3
g o f = g(f(x)) = 1 + 4x² - 1
= 4x²
f o g ≠  g o f

Sample space
S = {1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T};
n(S) = 12
Let A be the event of getting an odd number and a head.
A = {1H, 3H, 5H}; n(A) = 3
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 12 } =\frac { 1 }{ 4 } \)

\( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} =\sec \theta+\tan \theta \)
\(\mathbf{L H S} =\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}\)
[Multiplying the Numerator and denominator by \(\sqrt{1-\sin \theta}\)]
\( =\sqrt{\frac{1^{2}-\sin ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\sqrt{\frac{\cos ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{\cos \theta}{1-\sin \theta} \)
\(=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \)
[Multiplying Numerator and denominator by \(1+\sin \theta\)]
\( =\frac{\cos \theta(1+\sin \theta)}{1^{2}-\sin ^{2} \theta}=\frac{\cos \theta(1+\sin \theta)}{\cos ^{2} \theta} \)
\({\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]} \)
\(=\frac{1+\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta+\tan \theta=\text { RHS }\)

The 17^th term can be obtained by subtracting the sum of first 16 terms from the sum of first 17 terms
\({ S }_{ 17 }=\frac { 5\times \left( 17 \right) ^{ 2 } }{ 2 } +\frac { 3\times 17 }{ 2 } =\frac { 1445 }{ 2 } +\frac { 51 }{ 2 } =748\)
\({ s }_{ 16 }=\frac { 5\times \left( 16 \right) ^{ 2 } }{ 2 } +\frac { 3\times 16 }{ 2 } =\frac { 1280 }{ 2 } +\frac { 48 }{ 2 } =664\)
Now, t₁₇ = S₁₇ - S₁₆ = 748 - 664 = 84

Let r₁ and r₂ be the radii of the balloons.
Given that, \(\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 12 }{ 16 } =\frac { 3 }{ 4 } \)
Now, ratio of C.S.A. of balloons \(=\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } ={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }={ \left( \frac { 3 }{ 4 } \right) }^{ 2 }=\frac { 9 }{ 16 } \)
Therefore, ratio of C.S.A. of balloons is 9:16.

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.

Draw the parabola using the points (-4, 39), (-3, 22), (-2, 9), (-1, 10), (0, -5), (1, -6), (2, -3), (3, 4), (4, 15).

To solve 2x²- 4x - 6 = 0, subtract it from y = 2x²- 3x - 5
 is a straight line

Draw a straight line using the points (-2, -1), (0, 1), (2, 3). The points of intersection of the parabola and the straight line forms the roots of the equation.
The x-coordinates of the points of intersection forms the solution set.
∴ Solution {-1, 3}

Construction:
Step (1) Draw \(\bar { AB } =5.5cm\)
Step (2) Draw \(\angle BAE={ 25 }^{ 0 }\)
Step (3) Draw \(\angle FAE={ 90 }^{ 0 }\)
Step (4) Drawn the perpendicular bisector XY to AB which intersects AF at O and AB at G.
Step (5) With O as center and OA as radius drawn a circle
Step (6) XY intersects AB at G. On XY from G marked an arc at M such that GM = 4 cm
Step (7) Drawn PQ through M which is parallel to AB.
Step (8) PQ meets the circle at C and S.
Step (9) Joined AC and BC. Now \(\triangle\)ABC is the required triangle

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.