\(\frac { 2\sqrt { 3 } +3-\sqrt { 21 } }{ 3 } \)

(a) (0,5)
(b) (0,-3)
(c) (5,0)
(d) (-2,0)

(i) \(\Delta \)PDC and quadrilateral ABCD lie on the same base DC and between the same parallels DC and AB.
(ii) \(\Delta \) TRQ and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.
(iii) Quadrilaterals APCD and ABQD lie on the same base AD and between the same parallels AD and BQ.

For one tile a = 9 cm, b = 28 cm, c = 35 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 9+28+35 }{ 2 } =36\) cm
\(\therefore \) Area of one tile  \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 36(36-9)(36-28)(36-35) } \)
\(\\ =\sqrt { 36(27)(8)(1) } =\sqrt { 36\left( 9\times 3 \right) \left( 4\times 2 \right) } \)
 \(=6\times 3\times 2\sqrt { 6 } =36\sqrt { 6 } \) cm²
\(\therefore \)  Area of 16 tiles \(=36\sqrt { 6 } \times 16=576\sqrt { 6 } \)cm²
 \(\therefore \) Cost of polishing the tiles at the rate of 50 p per cm².
\(=576\sqrt { 6 } \times 50\) p = Rs. \(\frac { 576\sqrt { 6 } \times 50 }{ 100 } \) 
= Rs. \(288\sqrt { 6 } \) = Rs. 705.60

(i) Diameter = 14 cm
Radius (r) = \(\frac { 14 }{ 2 } cm=7cm\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 7 \right) }^{ 2 }=616{ cm }^{ 2 }.\)
(ii) Diameter = 21 cm
Radius (r)   = \(\frac { 21 }{ 2 } cm\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
 \(=4\times \frac { 22 }{ 7 } \times { \left( \frac { 21 }{ 2 } \right) }^{ 2 }=1386{ cm }^{ 2 }.\)
(iii) Diameter = 3.5 m
Radius (r)   = \(\frac { 3.5 }{ 2 } m=1.75m\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 1.75 \right) }^{ 2 }\)
\(\\ =38.5{ m }^{ 2 }.\)

Let, p(x)=x⁶-ax⁵+x⁴-ax³+3x-a+2
(x-a) is a factor of the polynomial p(x), then
p(a)=0
\(\Rightarrow\) a⁶-a\(\times\) a⁵+a⁴-a\(\times\)a³+3\(\times\)a-a+2
= 0
\(\Rightarrow\) a⁶-a⁶+a⁴-a⁴+3a-a+2
= 0
\(\Rightarrow\) 2a = -2 \(\Rightarrow\) a=-1.

In \(\triangle\)PQR, we have
PR > PQ (Given)
\(\angle\).PQR > \(\angle\)PRQ
(\(\therefore\) angle opp. to larger side is greater)
\(\angle\)PQR + \(\angle\)1 > \(\angle\)PRQ + \(\angle\)1
(Adding L.l on both sides)
\(\angle\)PQR + \(\angle\)1 > \(\angle\)PRQ + \(\angle\)2 ...(i)
(\(\therefore\)PS is the bisector at \(\angle\)P \(\therefore\) \(\angle\)1 = \(\angle\)2)
Now, in MQS and MSR, we have
\(\angle\)PQR + \(\angle\)1 + \(\angle\)PSQ = 180⁰
and
\(\angle\)PRQ + \(\angle\)2 + \(\angle\)PSR = 180⁰
\(\angle\)PQR + \(\angle\)1 = 180⁰- \(\angle\)PSQ       ...(ii)
and \(\angle\)PRQ + \(\angle\)2 = 180⁰-\(\angle\)PSQ     ....(iii)
Substitude eq. (ii) & (ill) in eq. (i)
\(\therefore\) 1800 - \(\angle\)PSQ > 180⁰-\(\angle\)PSR
\(\Rightarrow\) \(\angle\)PSQ > - \(\angle\)PSR
\(\Rightarrow\) \(\angle\)PSQ < \(\angle\)PSR i.e.,\(\angle\)PSR > \(\angle\)PSQ

Vol. of cuboid = lbh
=24 \(\times\) 18 \(\times\) 4
= 1728 cu.cm.
Edge of a cube=\(\sqrt [ 3 ]{ 1728 } \)
= 12 cm
LSA = 4a²
=4 \(\times\)12 \(\times\)12
= 576 cm².

Total number of balls faced by Sachin = 40
No. of balls on which he hit a six = 12
Let E₁, be the event of hitting a six.
\(\therefore \)No. of outcomes = 12
\(\therefore P(E_1)=\frac{12}{40}=\frac{3}{10}\)
Now, total No. of balls faced by Saurav = 30
Let E₂ be the event of Saurav did not hit the boundary
No. of outcomes = 30 - 9 = 21
\(P(E_2)=\frac{21}{30}=\frac{7}{10}\)

2

1000x = 237.\(\overline{237}\)
x = \(\frac{237}{999}\)

\(\sqrt[4]{16}=\sqrt[4]{2\times2\times2\times2}=2\)
\(\sqrt[3]{343}=\sqrt[3]{7\times7\times7}=7\)
\(\sqrt[5]{243}=\sqrt[5]{3\times3\times3\times3\times3}=3\)
\(\sqrt{196}\) = 14
ஃ \(\sqrt[4]{16}-6\sqrt[3]{343}+18\sqrt[5]{243}-\sqrt{196}\)
= 2- 6 x 7 + 18 x 3 - 14
= 2 - 42 + 54 - 14
= 56 - 56 = 0

Things which are equal to the same thing are equal to one another.

Median = average of 5th and 6th observations
24 = \(\frac{x+1+2x-13}{2}\)
⇒ 48 = 3x-12
⇒ 60 = 3x
⇒ x = 20

\(px+q+r=0\)
\(\Rightarrow px=-(q+r) \)
\(\Rightarrow -x=-(\frac{q+r}{p})\)
 \(\therefore\)  \(-(\frac{q+r}{p})\) is the required zero.

3x + y= 8
⇒ y = 8 - 3x
Table of solution

We plot the points (0, 8) and (3, - 1) on the graph paper and join the same by a ruler to get the line which is the graph of the equation 3x + y = 8

∵ Point (2, 2) lies on the graph
∴ x = 2, y = 2 is a solution of the given equation.

Given PQ || RS and T is any point
To prove \(\angle PQT+\angle QTS+\angle RST=360^{ 0 }\)
Construction Through T, draw TU || PQ || RS

PQ || UT  |By construction and a transversal QT intersects then
\(\therefore \angle PQT+\angle QTU=180^{ 0 }\)
The Sum of consecutive interior angles on the same sides of a transversal is \(180^{ 0 }\)
UT || RS
| By construction and a transversal TS intersect them
\(\therefore \angle UTS+\angle RST=180^{ 0 }\)
The Sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)
Adding (1) and (2) ,we get
\(\angle PQT+(\angle QTU+\angle UTS)+\angle RST=360^{ 0 }\) 
\(\Rightarrow \angle PQT+\angle QTS+\angle RST=360^{ 0 }\) 

Given: \(\Delta \)ABC, lines are drawn through A, B and C parallel respectively to the sides BC, CA and AB forming \(\Delta \)QPR.
To Prove: BC = \(1\over2\) QR

Proof:    \(\because\)  AQ IICB and AC IIQB
\(\therefore\) AQBC is a parallelogram
\(\therefore\)  BC  = QA     .....(1)    I Opposite sides of a II gm
\(\because\) AR II BC and AB II RC
\(\therefore\) ARCB is a parallelogram
\(\therefore\) BC = AR     I Opposite sides of a II gm
From (1) and (2),
QA= AR = \(1\over2\) QR.  ...(2)
From (1) and (3), BC = \(1\over2\) QR.

Given: In figure, AB and CD are equal chords of a circle whose centre is O. OM 丄 AB  and ON 丄 CD.
To Prove: \(\angle OMN=\angle ONM\).
Proof: ∵ Chord AB = Chord CD
 ∴ OM=ON     .........(1)
 | Equal chords of a circle are equidistant from the centre of the circle
In \(\Delta OMN\),
OM=ON     I From (1)
∴ \(\angle OMN=\angle ONM\) | Angles opposite to equal sides of a triangle are equal.

Steps of Construction
1. Draw a line segment MN = 5 cm.
2. With M as centre and 5 cm as radius, draw an arc on one side of MN.
3. With N as centre and 5 cm as radius, draw another arc on the same side of MN to intersect the former arc at L.
4. Join LM and LN. Then, \(\Delta \) LMN is the required equilateral triangle.

5. Taking M as centre and any radius, draw an arc to intersect the line segments MN and ML at P and Q respectively.
6. Next, taking P and Q as centres and with 1 the radius more than \(\frac { 1 }{ 2 } \) PQ, draw arcs to intersect each other, say at R.
7. Draw the ray MR. This ray MR is the required bisector of the \(\angle M\).

Let the sides forming the right angle be a cm and b cm. Then,
      
a + b + 10 = 24
\(\Rightarrow \)  a + b = 14               ....(1)
Also,  a² + b² = (10)²     |By Pythagoras Theorem
\(\Rightarrow \)   a² + b² = 100      ...(2)
We know that   (a + b)² = a² + b² + 2ab
\(\Rightarrow \)  (14)² = 100+2ab
\(\Rightarrow \) 2ab = 96
\(\Rightarrow \)  ab=48                   ...(3)
Also,    (a - b)² = a² + b²- 2ab
= 100 - 2\(\times \)48
= 100 - 96 = 4       | if a>b  
\(\Rightarrow \)  a-  b = 2
Solving (1) and (4), we get   a=8cm, b=6cm
\(\therefore \)   Area = \(\frac { 1 }{ 2 } \)ab = \(\frac { 1 }{ 2 } \).8.6 = 24 cm²

\(\therefore\)   Mean \(=\frac {{\sum fx}}{\sum f}= \frac {750}{30}=25\)