2\(\pi\)r = 22
\(2\times \frac { 22 }{ 7 } \times r=22\)
\(r=\frac { 7 }{ 2 } \)
V = \(\pi\)r²h
\(=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times 10\)
= 385 cm²

 We know that each angle of equilateral triangle is 60^o,So have to draw an angle 60^oand bisect it.
Construction:
i) Draw any line OP.
ii) With 0 as centre and any suitable radius, draw an arc to meet OP at R.
iii) With R as centre and same radius draw an arc to meet the previous arc at S.
iv) Join OS and Produce it to Q, then ㄥPOQ=60^o
v) With R as centre and any suitable radius (not necessarily)equal to radius of step 1 (but >\(\frac { 1 }{ 2 } \)RS),draw an arc. Also, with 5 as centre and radius draw another arc to meet the previous arc at Y
vi) Join OY and produced it, the OY is the required bisector of ㄥPOQ(i.e,ㄥPOY=30^o)

Let AB be perpendicular to a line l and AP is any other line segment.
In right \(\triangle ABP,\angle B>\angle P,(\therefore \angle B=90^o)\)
\(\ \Rightarrow AP>AB\ or\ AB\)

\((i)\frac { 53 }{ 100 }\) 
\((ii)\frac { 3 }{ 25 } \)

58

6 cm

12 cm²

\(130^{ 0 }\)

2x+0y-3=0

6 Units.

\(-\frac{1}{2}\)

(i) False, because zero is a whole number but not a natural number.
(ii) True, because every integer m can be expressed in the form m/1, and so it is a rational number.
(iii) False because 3/5 is a rational number but not an integer.

Steps of construction:
i) Draw a line segment BC = 4.7 cm. and at point B construct an angle of 45^o i.e ㄥXBC=45^o
ii) Cut the line segment BD = 2 cm (equal to AB-AC)on ray BX
iii) Join DC and draw the perpendicular bisector PQ of DC
iv) The perpendicular bisector intersects BXat point A. Jon AC ΔABC is the required triangle.

AD = BD
\(\Rightarrow\) \(\angle\)ABD = \(\angle\)DAB = 59⁰
(Angles opp. to equal sides are equal) 
In \(\triangle\)ABD,
59⁰ + 59⁰ + \(\angle\)ADB = 180⁰
\(\Rightarrow\) \(\angle\)ADB = 180⁰- 118⁰ = 62⁰
and \(\angle\)ACD = 62⁰- 32⁰ = 30⁰
(Exterior angle is equal to the sum of interior opposite angles)
In \(\triangle\)ABD,
(Side opp. to greatest angle is the longest) 1
Also in \(\triangle\)ABC, AB < AC
BD < AC

In figure, AD is the median of \(\Delta\)ABC.

\(\therefore \ ar(\triangle ABD)=\frac { 1 }{ 2 } ar(\triangle ABC)\)      ...........(i)
Now, BE is the median of \(\Delta\)ABD
\(\therefore \ ar(\triangle BED)=\frac { 1 }{ 2 } ar(\Delta ABD)\)       ..............(ii)
From (i) and (ii), we get
ar(\(\Delta\)BED) = \(\frac { 1 }{ 2 } ar(\triangle ABC)\)

So, 2y-2 must be subtracted.

We shall first make the class intervals continuous. Then, the modified table is as follows:

For tank
\(r=\frac { 3.8 }{ 2 } m=1.9m\)
\(\\ h=4.9m\)
 Curved surface area
\(=2\pi r\left( h+r \right) \)
\(\\ =2.\frac { 22 }{ 7 } .\frac { 19 }{ 10 } \left( 4.9+1.9 \right) { m }^{ 2 }\)
\(\\ =2.\frac { 22 }{ 7 } .\frac { 19 }{ 10 } .\frac { 68 }{ 10 } { m }^{ 2 }\)
\(\\ =81.21{ m }^{ 2 }\)

Area of \(\Delta \)ABC = \(=\frac { AB\times BC }{ 2 } =\frac { 24\times 10 }{ 2 } \) = 120 m².

Given: In figure, EF is a line passing through the centre 0 of a circle. EF bisects chords AB and CD of the circle.
To Prove: AB || CD.
Proof: ∵ EF bisects chord AB
∴ OL bisects chord AB
∴ \(\angle OLB=\angle OLA=90°\)                        ...(1)
| ∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord
∵ From (1) and (2),
  \(\angle OLB=\angle OMC=90°\)
But these angles form a pair of equal alternate interior angles
∴ AB || CD.

Given ABCD is a quadrilateral
To prove \(\angle A+\angle B+\angle C+\angle D=360^{ 0 }\)
Construction Join AC
Proof In \(\triangle \) ABC

\(\angle 1+\angle B+\angle 3=180^{ 0 }\)
|Angle sum property of a traingleIn 
In \(\triangle \) ADC
\(\angle 2+\angle D+\angle 4=180^{ 0 }\)
|Angle sum property of a traingleAdding (1) and (2) we get
\((\angle 1+\angle 2)+\angle B+(\angle 3+\angle 4)+\angle D=360^{ 0 }\)
\(\Rightarrow \angle A+\angle B+\angle C+\angle D=360^{ 0 }\) 

\(\angle \)2 = \(\angle \)4          ,,,,,,,(1) | Given
\(\angle \)1= \(\angle \)4          ........(2) | Given
\(\angle \)3 =\(\angle \)2         .......(3) | Given
From (1), \(\angle \)4 = \(\angle \)2           ......(4)
From (3) and (4),
\(\angle \)3 = \(\angle \)4               .....(5)
I Things which are equal to the same thing are equal to one another
From (2) and (4),
\(\angle \)1 = \(\angle \)3
I Things which are equal to the same thing are equal to one another

The given linear equation
\(\Rightarrow\ \ y={2\over3}x+{1\over3}\ \ \ \ . . . .(1)\)
Table of solution

We plot the points (1, 1) and (4, 3) on a graph paper and join the same by a ruler to get the line which is the graph of the equation \(y={2\over3}x+{1\over3}\)

From graph, we see that the point (7, 5) lies on the graph, so it is a solution of the linear equation.

(i) (3,2)
(ii) (-3,2)
(iii) (-3,-2)
(iv)(3,-2)

Let x = \(0.15\overline { 9 } \)
x = 0.159999...
100x = 15.9999...      ...(1)
1000x = 159.9999...    ...(2)
Subtracting (1) from (2), we get
900x = 144
\(x=\frac { 144 }{ 900 } \)
\(\\ x=\frac { 4 }{ 25 } \)
Here, p = 4, q = 25(\(\neq 0\))

Required probability=\(\frac { 25 }{ 200 } =\frac { 1 }{ 8 } \)

v = 5 \(\times\) (6 \(\times\) 2 \(\times\) 1.5)

\(\angle BDA=\angle BCA=40°\) [Angles in the same segment]
Now, Since AD || BC
\(\angle\)DBC=\(\angle\)BDA   [Alternate interior angles]
\(\therefore\)  \(\angle\)DBC=40^o

\(x=90^{ 0 }-x)+30^{ 0 }\Rightarrow x=60^{ 0 }\)

For each value of x, we get the corresponding value of y.

Use factor theorem

AE = BE
\(\angle\)A = \(\angle\)B      (angle opposite to equal sides of a \(\Delta\)are equal)
\(\angle\)EDC = \(\angle\)B;
\(\angle\)ECD = \(\angle\)A (exterior angle of cyclic quad.)
\(\therefore\)   \(\angle\)EDC = \(\angle\)A (Corresponding Angles)
\(\therefore\)    AB II DC Proved.

Let r and h be the radius and height of the cylinder given r : h = 5 : 7
\(\therefore\) The radius of the cylinder(r) = 5x
and The height of the cylinder (h) = 7x
Volume of the cylinder = \(\pi\)r²h
\(\therefore\) 4400=\(\frac { 22 }{ 7 } \times { (5x) }^{ 2 }\times 7x\)
\(\left( \because \pi =\frac { 22 }{ 7 } \right) \)
\(\Rightarrow \ 4400=\frac { 22 }{ 7 } \times 5x\times 5x\times 7x\)
\(\Rightarrow \ { x }^{ 3 }=\frac { 4400\times 7 }{ 22\times 5\times 5\times 7 } \)
\(\Rightarrow\) x³ = 8 = 2³
\(\therefore\) x = 2
Hence, the radius of the cylinder = 5x
= 5(2)
= 10 cm.

Given: PQR is a triangle and A and B are points on QR such that QA = AB = BR
To prove that: ar(\(\Delta \)PQB) = 2ar(\(\Delta \)PBR)
Let h be the height of triangle PQR.
Now, \(ar\left( \Delta PQA \right) =\frac { 1 }{ 2 } \times QA\times h\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } QR\times h\)           ..............(i)
\(ar(\Delta PAB)=\frac { 1 }{ 2 } \times AB\times h\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } QR\times h\)           ..............(ii)
\(ar(\Delta PBR)=\frac { 1 }{ 2 } \times BR\times h\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } QR\times h\)           ..................(iii)
\(ar(\Delta PQB)=\frac { 1 }{ 2 } \times QB\times h\)
\(=\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } QR\times h\)              ..................(iv)
From (i), (ii) and (iii), we conclude
ar(\(\Delta \)PQA) = ar(\(\Delta \)PAB) = ar(\(\Delta \)PBR)   ......(v)
Now, ar(\(\Delta \)PQB) = ar(\(\Delta \)PQA) + ar(\(\Delta \)PAB)
\(\therefore\) ar(\(\Delta \)PQB) = ar(\(\Delta \)PBR) + ar(\(\Delta \)PBR)
[from (iv)]
\(\therefore\) ar(\(\Delta \)PQB) = ar(\(\Delta \)PBR)     Hence Proved.
From (iii) and (iv), we get
\(ar\left( \Delta PQB \right) =\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } QR\times h\)
\(=2\left( \frac { 1 }{ 2 } \times \frac { 1 }{ 3 } QR\times h \right) \)
= 2 ar(\(\Delta \)PBR) Hence Proved.

y=180^o-(25^o+125^o)(Linear pair)
\(\Rightarrow\) y=30^o
y+z=110^o (Exterior angle)
z=110^o-30^o=80^o
x+25^o=z (Exterior angle)
x=80^o-25^o=55^o

A parallelogram PQRS in which the bisectors of LP and LQ meet SR at O.
To Prove: ㄥPOQ=90°
Now, since PQRS is a parallelogram. Therefore,
PSIIQR
Now, PS II QR and transversal PQ intersects them.
ㄥP+ㄥQ=180°
(∵ Sum of consecutive interior angles is 180°)
\(\frac{1}{2}\)ㄥP+\(\frac{1}{2}\)ㄥQ=90°
⇒ ㄥ1+ㄥ2=90° ( OP is bisector of ㄥP...(i) and OQ is bisector of ㄥQ.
ㄥ1=\(\frac{1}{2}\) ㄥP and ㄥ2=\(\frac{1}{2}\) ㄥQ)
Now, in ΔPOQ
ㄥ1+ㄥPOQ+ㄥ2=180°
⇒ 90°+ㄥPOQ=180°
⇒ ㄥPOQ=90°

250x³-432y³=2[125x³-216y³]
=2[(5x)³-(6y)3]
=2(5x-6y)[(5x)2+(6y)2+5x\(\times\)6y]
\(\because\) a³-b³=(a-b)(a²+b²+ab)
=2(5x-6y)(25x²+36y²+30xy).

2^xx(2²)^x = (2³)^1/3x(2⁵)^1/5
2^3x = 2²
x =\(\frac{2}{3}\)

3x=y+3; three solutions are x=1, y=0; x=2, y=3 and x=0, y=-3

From graph it is clear that line meets x-axis at(1,0) and y-axis at (0,-3)

\((i)\frac { 3 }{ 25 }\) 
\((ii)\frac { 7 }{ 25 }\) 
\((iii)\frac { 9 }{ 25 }\) 
\((iv)\frac { 6 }{ 25 } \)

\(1500\sqrt { 3 } \) cm²

Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \(\angle PQR\)
To prove: (i) \(\triangle ABM\cong \triangle PQN\) (ii) \(\triangle ABC\cong \triangle PQR\)
Proof: In \(\triangle ABM\) and \(\triangle PQN\)
AB = PQ
AM = PN
BC = QR
2BM = 2QN | M and Nare the mid-points of BC and QR respectively
BM = QN
In view of (1), (2) and (3)
\(\triangle ABM\cong \triangle PQN\) | SSS rule
(ii) \(\triangle ABM\cong \triangle PQN\)
\(\angle ABM=\angle PQN\) | C.P.C.T
\(\angle ABC=\angle PQR\)
In \(\triangle ABC\) and \(\triangle PQR\)
AB = PQ
BC = QR
\(\angle ABC=\angle PQR\)
\(\triangle ABC=\triangle PQR\) | SAS rule

Height of the cone(h) = 24 cm
Let r cm be the radius of the base and l cm an can be the slant height of the cone, then
\(l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+{ 24 }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+576 } \)
Now, curved surface area=\(\pi\)rl
\(\Rightarrow \ \frac { 22 }{ 7 } \times r\times \sqrt { { r }^{ 2 }+576 } =550\)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =550\times \frac { 7 }{ 22 } \)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =175\)
Squaring both the sides we get
r² (r² + 576) = 30625
(r²)² + 576r²-30625 = 0
Let r² = x
x² + 576x - 30625 = 0
\(\Rightarrow\) x² + 625x - 49x - 30625 = 0
\(\Rightarrow\) x(x + 625)-  49(x + 625) = 0
\(\Rightarrow\) (x + 625)(x - 49) = 0
\(\Rightarrow\) x + 625=0 or x - 49 = 0
\(\Rightarrow\) x = 625 or x = 49
not possible x = 49
\(\therefore\) r²= 49
\(\Rightarrow\) r = 7 cm
Volume & the cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+24\)
= 1232 cm³.

(i) Let us assume that A, Band C are the position of Priyanka, Sania and David respectively on the boundary of circular park with centre O.
Draw AD\(\bot\) BC
Since the centre of the circle coincides with the centroid of the equilateral \(\triangle\) ABC
\(\therefore\) Radius of circumscribed circle = \(\frac{2}{3}\) AD
\(\Rightarrow 20=\frac { 2 }{ 3 } AD\)
\(\Rightarrow AD=20\times \frac { 2 }{ 3 } \)
\(\Rightarrow AD=30m\)
Now, AD\(\bot\)BC, and let AB=BC=CA=x
\(\Rightarrow BD=CD=\frac { 1 }{ 2 } BC=\frac { x }{ 2 } \)

In rt. \(\triangle\)BDA, D=90⁰
By Pythagoras Theorem, we have
AB²=BD²+AD²
\(\Rightarrow { x }^{ 2 }={ \left( \frac { x }{ 2 } \right) }^{ 2 }+{ (30) }^{ 2 }\)
\(\Rightarrow{ x }^{ 2 }-{ \frac { { x }^{ 2 } }{ 4 } }=90\)
\(\Rightarrow \frac { 3 }{ 4 } { x }^{ 2 }=90\)
\(\Rightarrow { x }^{ 2 }=900\times \frac { 4 }{ 3 } \)
\(\Rightarrow { x }^{ 2 }=1200\)
\(\therefore x=\sqrt { 1200 } =20\sqrt { 3 } \)
Hence, the distance between each of them is \(20\sqrt { 3 } .\)
(ii) Properties of the circle, equilateral triangle and Pythagoras theorem.
(iii) Live and let live.

Proof: In \(\triangle\) PQR,
PQ =PR
\(\angle\) PQR = \(\angle\)PRQ
(Angles opp. to equal sides are equal) ...(i) 
In \(\triangle\)PQS, \(\angle\)PQR > \(\angle\)PSQ 
(Ext. angle of a 6 is greater than each of interior opp. angle)
\(\angle\)PRQ > \(\angle\)PSQ, using (i)
\(\Rightarrow\) \(\angle\)PRS >\(\angle\)PSR \(\Rightarrow\) PS > PR 
PS>PQ (\(\because\) PR = PQ)
(Side opp. to greater angle is larger)

(i) Required number of students = 8 + 32 = 40
(ii) Here, We notice that classes are continuous but class-size is not the same for all the classes. We notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of rectangle (adjusted frequency) for each class.
Length of rectangle (adjusted frequency) =\(\frac { Frequency\ of\ Class }{ Width\ of\ class } \times Minimum\ class-size\)

Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies as heights.
Histogram representing marks obtained by students in unit test of Mathematics.

(iii) Hardwork and Dilligence.

(i) Since sum of adjacent angles of a parallelogram is 180°
ஃ We have ㄥA+ㄥB=180
⇒ 5x + 7 + 3x - 3 = 180
⇒ 8x + 4 = 180
⇒ 8x = 176
⇒ x=\(\frac{176}{8}\)=22

ㄥA=(5x+7)°=(5x22+7)=117°
ㄥB=(3x-3)°=(3x22-3)=63°
ㄥC=ㄥA=117°
and ㄥD=ㄥA=63°
(ii) Properties of parallelogram.
(iii) Energy conservation is necessary for a happy and prosperous future.

\(\frac { { 3 }^{ 30 }+{ 3 }^{ 29 }+{ 3 }^{ 28 } }{ { 3 }^{ 31 }+{ 3 }^{ 30 }-{ 3 }^{ 29 } } =\frac { { 3 }^{ 38 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }+1 \right) }{ { 3 }^{ 29 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }-1 \right) } \)
\(=\frac { \left( 9+3+1 \right) }{ 3\left( 9+3-1 \right) } \)
\(=\frac { 13 }{ 3\times 11 } =\frac { 13 }{ 33 } \)