Steps of construction:
i) Draw ㄥDEF=72^o ,using protractor
ii) Bisect it. Let the bisected angle be ㄥDEK
iii) Again bisect ㄥDEK
iv) Now ㄥGEF = \(\frac { 3 }{ 4 } \)ㄥDEF

Here, \(OX=\frac{1}{2}XY, PX=\frac{1}{2}XZ\)
XY = 2(OX), XZ = 2(PX)
Also. OX = PX(Given)
XY = XZ
(because things which are double of the same things are equal to one another)

(i) 0.21  
(ii) 0.465  
(iii)0.79  (iv)0 (v) 1

60.6

84 cm²

Given 7x-3y=15
\(\Rightarrow x=\frac{15+3y}{7}\)
At y-axis x=0
\(\therefore\) 7(0)-3y=15
\(\Rightarrow\) 0-3y=15
\(\therefore \ y=\frac{15}{-3}=-5\)
Given line intersects the y-axis at y=-5

(i) (5,2) (ii) (-5,2) (iii) (-5,-2) (iv) (5,-2)

8

\(54\sqrt { 3 } \)

Required probability=\(\frac { 58+122 }{ 500 } \)

The item values to be included in 3 - 5 are 4,3,3,4.

Formula

A diameter is the largest chord. Diameter\(=2\times \)Radius.

Perimeter of WEF
=DE + EF + FD
=-AB + \(1\over2\)BC + \(1\over2\)CA = \(1\over2\)(AB+BC+CA)
= \(1\over2\) (3 + 4 + 4) = 5.5 cm.

\(\angle A+\angle B+\angle X=180^{ 0 }\)
\(\Rightarrow 50^{ 0 }+\angle B+\angle C=180^{ 0 }\)
\(\Rightarrow \angle B+\angle C=130^{ 0 }\)
\(\angle BOC+\frac { 180^{ 0 }-\angle B }{ 2 } +\frac { 180^{ 0 }-\angle C }{ 2 } =180^{ 0 }\)
\(\Rightarrow \angle BOC=\frac { \angle B+\angle C }{ 2 } =65^{ 0 }\)

x=4 and x=-4 both are parallel to y-axis

Evident

\(x^2-9=(x-3)(x+3) \quad |\quad x-(-3)=x+3\)

(i) P(Number 3)=\(\frac{78}{500}\)
(ii) P(Number >4)=\(\frac{88+97}{500}=\frac{185}{500}\)
(iii) P(Number <4)=\(\frac{89+57+78}{500}\)
\(=\frac{242}{500}\)
(iv) P(Number between 1 and 3)=\(\frac{75}{500}\)

PQ = PR \(\Rightarrow\)  \(\angle\)PQR = \(\angle\)PRQ
(Angles opp. to equal sides) 
ST II QR \(\Rightarrow\) \(\angle\)PST = \(\angle\)PQR
(Corresponding angles)
\(\Rightarrow\) PTS = \(\Rightarrow\)PRQ
(Corresponding angles) 
\(\therefore\) PST = PTS 
\(\Rightarrow\) PS = PT.
(Sides opp. to equal angles)

\(\angle A=57^o\)
and \(\angle ACD=22^o+35^o=57^o\)
\(\angle A=\angle ACD\)
But these are alternate angles
AB||CD
Again \(\angle FEC+\angle ECD=145^o+35^o=180^o\)
EF||CD
Now AB||CD and EF||CD
AB||EF

71,13.66

h = 1 m = 100 cm
2r = 140 cm
\(\Rightarrow\)  \(r=\frac { 140 }{ 2 } cm=70cm\)
 \(\therefore\)  Total surface area of the closed cylindrical tank   
 \(=2\pi r\left( h+r \right) \)
\(\\ =2\times \frac { 22 }{ 7 } \times 70\left( 100+70 \right) \)
\(\\ =74800{ cm }^{ 2 }=\frac { 74800 }{ 100\times 100 } { m }^{ 2 }\)
\(\\ =7.48{ m }^{ 2 }\)
Hence, 7.48 square metres of the sheet are required.

Construction: Draw OL 丄 RS.
Let KR = xm
\(ar\left( \Delta ORS \right) =ar\left( \Delta ORK \right) +ar\left( \Delta SRK \right) \)
\(=\frac { \left( OK \right) \left( KR \right) }{ 2 } +\frac { \left( KS \right) \left( KR \right) }{ 2 }\)
\( \\ =\frac { \left( KR \right) \left( OK+KS \right) }{ 2 } =\frac { \left( KR \right) \left( OS \right) }{ 2 } \)
\(=\frac { \left( x \right) \left( 5 \right) }{ 2 } \ ...........(1)\)

Again, ar \(\left( \Delta ORS \right) \)
\(=\frac { RS\times OL }{ 2 } =\frac { 6\times OL }{ 3 } \)
\(=\frac { 6\times \sqrt { OR^{ 2 }-RL^{ 2 } } }{ 2 } \)                I By Pythagoras Theorem
\(=\frac { 6\times \sqrt { 25-9 } }{ 2 } =\frac { 6\times 4 }{ 2 } =12\quad m\)    .........(2)
From equations (1) and (2),
\(\frac { \left( x \right) \left( 5 \right) }{ 2 } =12\Rightarrow x=\frac { 12\times 2 }{ 5 } =\frac { 24 }{ 5 } =4.8\quad m\)
⇒ KR = 4.8 m
∴ RM = 2KR = 2 x (4.8) = 9.6 m
Hence, the distance between Reshma and Mandip is 9.6 m.

\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​\)
\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​=\left( z+\frac { 3 }{ 2 } \right) \left( z-\frac { 3 }{ 2 } \right) ​​\)| Where \(y^2=z\)
\(={ (z) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) | Using identity III
\(={ z }^{ 2 }-\frac { 9 }{ 4 } ={ ({ y }^{ 2 }) }^{ 2 }-\frac { 9 }{ 4 } \) | Substituting the value of z
\(={ y }^{ 4 }-\frac { 9 }{ 4 } .\)

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.

Proof: QO is bisector of \(\angle\)PQR
\(\angle\)OQR = \(\frac{1}{2}\)\(\angle\)PQR = \(\frac{1}{2}\) =\(\angle\)Q
RO is bisector \(\angle\)ORQ
\(\therefore\) \(\angle\)ORQ =\(\frac{1}{2}\) \(\angle\)PRQ = \(\frac{1}{2}\) \(\angle\)R
In \(\angle\)OQR
\(\angle\)QOR + \(\angle\)OQR + \(\angle\)ORQ = 180°
(Angle sum property)
\(\angle\)QOR + \(\frac{1}{2}\) \(\angle\)Q + \(\frac{1}{2}\) \(\angle\)R = 180°
\(\angle\)QOR = 180°- \(\frac{1}{2}\)(\(\angle\)Q + \(\angle\)R)
But in \(\angle\)PQR
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180°
\(\angle\)Q + \(\angle\)R = 180°- \(\angle\)P
\(\angle\)QOR = 180°- \(\frac{1}{2}\) (180°- \(\angle\)P)
= 180°-90° + \(\frac{1}{2}\)\(\angle\)P
= 90° + \(\angle\)P Hence Proved.
These type of rallies spread awareness among people for not to kill girl child and helping in equalising sex ratio.

(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.