\(\angle QOR+\angle QPR={ 180 }^{ ° }\)
(Supplementary angle)
\(\Rightarrow \quad \angle QOR+{ 46 }^{ ° }={ 180 }^{ ° }\)
\(\Rightarrow \quad \angle QOR={ 180 }^{ ° }-{ 46 }^{ ° }={ 134 }^{ ° }\)

Distance of the point,
\(a=\sqrt { { \left( 5-3 \right) }^{ 2 }+{ \left( 8-4 \right) }^{ 2 } } \)
\(=\sqrt { 4+16 } =\sqrt { 20 } =2\sqrt { 5 } \)
\(\because 2\sqrt { 5 } \) is less than 7
\(\therefore\)  The point lies inside the circle.

Let the first term of an AP. be a, common difference,
d =-6
a₁₆ = a + (16- 1)(- 6)
= a-90
a₁₂ = a + (12-1)(-6)
= a-66
a₁₆ - a₁₂ = (a - 90) - (a - 66)
=a-90-a+66
=-24

According to the question,
Sum of zeroes = \(\frac{21}{8}\)
and Product of zeroes = \(\frac{5}{16}\)
So, quadratic polynomial = x² - (Sum of zeroes)x + Product of zeroes
= x² - \(\left( \frac { 21 }{ 8 } \right) \)x + \(\left( \frac { 5 }{ 16 } \right) \)
= \(\frac { 1 }{ 16}\left( 16{ x }^{ 2 }-42x+5 \right) \)
 = \(\left( 16{ x }^{ 2 }-42x+5 \right) \frac { 1 }{ 16}\)

\(\frac { \sqrt { 2+\sqrt { 3 } } }{ 2 } \) 

\(\frac { \sqrt { 5 } +\sqrt { 21 } }{ 4 } \) and \(\frac { \sqrt { 5 } -\sqrt { 21 } }{ 4 } \)

Total number of outcomes, n(S)=100
Let E=Event of getting a prime number
={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,71,73,79,83,89,97}
∴ n(E)=25
Hence, P (getting a prime number)
\(=\frac { n(E) }{ n(S) } =\frac { 25 }{ 100 } =\frac { 1 }{ 4 } \)

We know that, tangents drawn from an external point to a circle are of equal length. 
AR = AP
BR= BQ  ....(I)
Now,  perimeter of ATB = TA + AB - TB
= TA + AR + BR + TB 
= TA + AP + BQ + TB    [using (i)]
= TP + TQ = TP + TP
= 2TP     [∵ TQ = TP]
Semi-perimeter \(\triangle\)TAB = \(1\over2\) x 2TP = TP
= 13 cm

Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere
= 2\(\pi\)rh + 2\(\pi\)r² = 2\(\pi\)r(h + r)
\(=2 \times \frac{22}{7} \times 30(145+30) \mathrm{cm}^2\)
= 33000 cm² = 3.3 m²

90^o

2.049km

Since distance between the store and first flag towards right side in 2 m.

\(\therefore\)  Distance covered by Ruchi to place the flag first towards right side of the store = 4 m
Distance covered by Ruchi to place the second flag towards right side of the store = 8 m
Distance covered by Ruchi to place the third flag towards right side of the store = 12 m
Similarly, distance covered by Ruchi to place the thirteenth flag towards right side of the store = a + 12d = 4 + 12 x 4 = 25 m
Now,  total distance covered in completing the job = 2S13
\(=2\{ \frac{13}{2}(2\times4+12\times4) \}=13(8+48)\)
= 13 x 56 = 728 m
Maximum distance covered by Ruchi for carrying a flag is 52 m.
Armed forces flag day is celebrated on 7th December, every year.

Let (x,y) be the coordinates of third vertex of the triangle.
\(\therefore \frac { x-3+0 }{ 3 } =0\) and \(\frac { y+1-2 }{ 3 } =0\) 
\(\Rightarrow \ x=3\ \) and x=1
Hence, the coordinates of the third vertex is (3,1).

Coordinates of the centroid of \(\Delta \) are
\(({3-4+4\over3},{-4+2+8\over3}) \) i.e., (1,2)

126 cm²

a_n=9-5n
a₁=9-5x1=4
a₂=9-5x2=2=-1
a₃=9-5x3=-6
sequence is 4, -1, -6,....

Steps of Construction: 
1. Draw a line segment BC = 6 cm and at point B draw a ㄥABC = 60^o. 
2. Cut AB 5 cm. Join AC. We obtain ABC is triangle. 
3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A. 
4. Locate 4 points A₁,A₂,A₃ and A₄ on the ray BX so that BA₁=A₁A₂=A₂A₃=A₃A₄.
5. Join A₄ to C. 
6. At A₃ draw A₃C' || A₄C. Where C' is a point on the line segment BC.
7. At C' draw C'A' || CA, where A' is a point on the line segment BA.

Δ A'BC' is the required triangle.
Justification:
In Δ A'BC' and ΔABC A'C' || AC
∴ By BPT \(\frac { A'B }{ AB } =\frac { BC' }{ BC } \)    ...(i)
From (i) and (ii),
\(\frac { A'B' }{ AB } =\frac { 3 }{ 4 } \Rightarrow A'B=\frac { 3 }{ 4 } AB\)
In ΔBA₃C' and ΔBA₄C
\(\frac { BC' }{ BC } =\frac { { BA }_{ 3 } }{ { BA }_{ 4 } } =\frac { 3 }{ 4 } \)      ...(ii) 
∴ Sides of new triangle formed are \(\frac { 3 }{ 4 } \)  times the corresponding sides of first triangle.

Total number of outcomes = 2³ = 8
(i) P(three heads) = \(\frac{1}{8}\)
(ii) P(atleast two tails) = \(\frac{4}{8}\)=\(\frac{1}{2}\)

In triangles LMK and PNK,
\(\angle\)M = \(\angle\)N = 50°   (Given) ​​​
\(\angle\)K =\(\angle\)K    (Common)
\(\triangle\)LMK~\(\triangle\)PNK   (AA similarity)
\(\Rightarrow \frac { LM }{ PN } =\frac { KM }{ KN } \)
\(\Rightarrow \frac { a }{ x } =\frac { b+c }{ c } \)
\(\therefore x=\frac { ac }{ b+c } \)

Here given, a₃ = 9 ⇒ a + 2d = 9 ....(i)
a₈ - a₅ = 6
⇒ (a + 7d) - (a + 4d) =6
⇒ 3d =6
⇒ d = 2 .......(ii) 
Substituting (ii) in (i), we get
⇒ a + 2(2) = 9
⇒ a = 5  
So, AP. is 5, 7, 9, 11, .....

10.41 m

The area of rectangular sheet ABCD = 40 x 28 
= 1120 cm² 
The diameter of semicircle = 28 cm 
ஃ Radius of semicircle = 14 cm 
Area of semicircle = \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 14\times 14\quad { cm }^{ 2 }\)
= 308 cm² 
Thus, area of remaining paper = Area of paper - Area of semicircle 
= 1120 - 308 = 812 
∴ Required area = 812 cm² 

Let coordinates of B and C are (x₂,y₂) and (x₃,y₃) respectively.
D is mid-point of AB.
∴ 1= \(\frac { 0+{ x }_{ 2 } }{ 2 } \) ⇒ x₂=2
and 0=\(\frac { -1+{ y }_{ 2 } }{ 2 } \)⇒ y₂=1
∴ Coordinates of B are (2,1).
E is mid-point of AC
∴ 0=\(\frac { 0+{ x }_{ 2 } }{ 2 } \) ⇒ x₃=0
and 1=\(\frac { -1+{ y }_{ 2 } }{ 2 } \)⇒ y₃=3
∴ Coordinates of C are (0,3).
Area △ABC
= \(\frac{1}{2}\)|0(1-3)+2(3+1)+0(-1-1)|
=\(\frac{1}{2}\)x8 =4 sq.units
F is mid point of BC
∴ Coordinates F are \((\frac{2+0}{2},\frac{1+3}{2})\), i.e., (1,2).
Area △DEF ,
=\(\frac{1}{2}\)|1(2-1)+1(1-0)+0(0-2)|
=\(\frac{1}{2}\)|1+1|=1 sq.units. 

Given: T BP and TCQ are tangents to the circle whose centre is O. 
Also,  \(\angle \)PBA = 60°
\(\angle \)ACQ = 70°

To determine: \(\angle \)BAC and \(\angle \)BTC
Sol. Join OB and OC
\(\angle \)OBP = 90°
[Tangent makes 90° angle with the radius at the point of contact]
⇒ \(\angle \)OBA + \(\angle \)ABP = 90°
⇒ \(\angle \)1 + 60° = 90° [Given ABP = 60°]
⇒ \(\angle \)1 = 30°       ......(i)
Also  \(\angle \)OCQ = 90°
⇒ OCA + 70° = 90°  ......(i)
⇒ \(\angle \)OCA = 20°   .....(ii)
In OBA ,   OB = OA = radii
⇒ \(\angle \)1 = \(\angle \)4 = \(\angle \)30°     .....(iii)
[ Angles opposite to equal sides of a triangle are equal]
Similarly, In \(\triangle\)OCA
OC = OA
\(\angle \) 5 = 20°​​​​​​​      .....(iv)
From (iii) and (iv)
\(\angle \)BAC = 20°​​​​​​​ + 30°​​​​​​​ = 50°​​​​​​​
⇒  \(\angle \)BOC = 2x50°​​​​​​​ = 100°​​​​​​​
\(\angle \)BOC + BTC = 180°​​​​​​​
100°​​​​​​​ + \(\angle \)BTC = 180°​​​​​​​
⇒ \(\angle \)BTC = 80°​​​​​​​

Let the distance of Pole PR = Y and PQ = 80 - Y
From \(\Delta BPR\),
tan 60°=\(\frac{x}{y}\)
x=y\(\sqrt{3}\) ...(i)
From \(\Delta APQ\),
tan 30° =\(\frac{x}{80-y}\)
\(\sqrt{3}\)x=80-y ...(ii)
Solving (i) and (ii),
y = 20,x = 20\(\sqrt{3}\)m
Height of pole = 20\(\sqrt{3}\)m
Distance of Pole, PR = 20 m.
QP = 80 - 20 = 60 m.

Since \(\alpha\) and \(\beta\) are the zeroes of the cubic polynomial x²  +4x + 3
then, \(\alpha+\beta=-4 \)
and \(\alpha\beta=3\)
Sum of zeroes  \(=1+\frac { \beta }{ \alpha } +1+\frac { \alpha }{ \beta } \)
\(=\frac { \alpha \beta +{ \beta }^{ 2 }+\alpha \beta +{ \alpha }^{ 2 } }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta }{ \alpha \beta } \)
\(=\frac { \left( \alpha +\beta \right) ^{ 2 } }{ \alpha \beta } =\frac { \left( -4 \right) ^{ 2 } }{ 3 } =\frac { 16 }{ 3 } \)
Product of zeroes = \(\left( 1+\frac { \beta }{ \alpha } \right) \left( 1+\frac { \alpha }{ \beta } \right) \)
\(=1+\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } +\frac { \alpha \beta }{ \alpha \beta } \)
\(\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta }{ \alpha \beta } =\frac { \left( \alpha +\beta \right) ^{ 2 } }{ \alpha \beta } \)
\(=\frac { \left( -4 \right) ^{ 2 } }{ 3 } =\frac { 16 }{ 3 } \)
But required polynomial = x² - (Sum of the zeroes) x + Product of the zeroes
\(={ x }^{ 2 }-\left( \frac { 16 }{ 3 } \right) x+\frac { 16 }{ 3 } \)
or \(=\left( { x }^{ 2 }-\frac { 16 }{ 3 } x+\frac { 16 }{ 3 } \right) \)
\(=\left( 3{ x }^{ 2 }-16x+16 \right) \frac { 1 }{ 3 } \)

Convert the given distribution into continuous grouped frequency distribution and find the median.
60.85 kg

Let present age of Jacob and his son be x and y, respectively.
According to the question,
Condition I. (x-5)=7(y-5) \(\Rightarrow\)x-7y+30=0 ...(i)
Condition II. (x+5)=3(y+5) \(\Rightarrow\)x-3y+10=0 ...(ii)
Hence, algebraic representation is
x-7y+30=0  \(\Rightarrow\)x-3y+10=0
For graphical representation, draw graphs Eq. (i) and Eq. (ii).

Time taken by Sonia to drive one round of the field = 18 min
Time taken by Ravi  to drive one round of the field = 12 min
The LCM of 18 and 12 gives the exact number of minutes after which they will meet at the starting point again.
Now,  18 = 2 x 3 x 3 = 2 x 3²
and   12 = 2 x 2 x 3  = 2² x 3
\(\therefore \)  LCM of 18 and 12 = 2² x 3² = 2 x 2 x 3 x 3 = 36
Hence, Sonia and Ravi will meet again at the starting point after 36 min.

(i) Number of all possible outcomes = 16
(a) Let E₁ be the event of getting blood group O.
Then, number of outcomes favourable to E₁ = 7
\(\therefore \)  Required probability = P (E₁) = \(\frac{7}{16}\)
(b) 1/4
(ii) (a) Habit of donation   (b) Social works

\(\frac { 4 }{ 3 } h\)

7200km/h

Radii and height of the frustum i.e., bucket are
R= 20 cm ,r = 8cm , h= 16 cm 
Volume of bucket = \(\frac { 1 }{ 3 } \pi h(R^{ 2 }+r^{ 2 }+Rr)\)
= \(\frac { 1 }{ 3 } \times 27\times 16(8^{ 2 }+20^{ 2 }+8\times 20)\)

\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16(64+100+160)\)
\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16\times 624cm^{ 3 }=10459.42cm^{ 3 }\)
Area of sheet required = area of the bucket 
\(=\pi (R+r)+\pi r^{ 2 }\)
Where l =  r\(\sqrt { h^{ 2 }+(R-r)^{ 2 } } \)
\(l=\sqrt { (16)^{ 2 }+(20-8)^{ 2 } } =\sqrt { 256+(12)^{ 2 } } =\sqrt { 256+144 } =\sqrt { 400 } =20\)
Area of sheet required =  \(\pi (R+r)+\pi r^{ 2 }\)
= \(\frac { 22 }{ 7 } \left[ 20(20+8)+64 \right] =\frac { 22 }{ 7 } (20\times 28+64)\)
\(\frac { 22 }{ 7 } \left( 560+64 \right) =\frac { 22 }{ 7 } \times 624=1961.142cm^{ 2 }\)
Cost of 100 cm² of metal sheet = Rs 15
Cost of 1 cm² of metal sheet = Rs \(\frac { 15 }{ 100 } \)
Cost of 1961.142 cm² = Rs \(\frac { 15 }{ 100 } \times 1961.142=Rs\quad 294.17\)​​​​​​​

Diameter of front wheel=80cm
∴ Radius=40cm
Distance covered in 1 revolution=\(2\pi r={2\times22\over 7}\times40={1760\over 7}cm\)
∴ Distance covered in 1400 revolution
\(={1400\times1760\over 7}=352000cm=3520m\)
Diameter of rear wheel=2m
∴ Radius of rear wheel=1m
Distance covered in 1 revolution=\(2\pi r={2\times\over 7}\times1m={44\over 7}m\)
∴ No. of revolutions to cover 3520m=\({3520\times7\over44}=560\)