10.5 cm

5400

55,46,52

Probability of getting head \(=\frac{455}{1200}\)
\(=\frac{91}{240}\)
Probability of getting tail\(=\frac{745}{1200}=\frac{149}{240}\)

Definition of data.

Let the base radius of the cylinder be r cm. Then,
\(2\pi r=22\)
\(\\ \Rightarrow \ 2\times \frac { 22 }{ 7 } \times r=22\)
\(\\ \Rightarrow r=\frac { 7 }{ 2 } \)
\( h=3m\)
\(\therefore \)   Total surface area 
 \(=2\pi rh+\pi { r }^{ 2 }\)
\(\\ =2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times 3+\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \)
\(\\ =66+38.5\)
\(\\ =104.5{ m }^{ 2 }\)

Total surface area
\(=\pi RL+\pi { R }^{ 2 }\)
\(\\ =\pi \left( \frac { r }{ 2 } \right) \left( 2l \right) +\pi { \left( \frac { r }{ 2 } \right) }^{ 2 }\)
\(\\ =\pi rl+\frac { \pi { r }^{ 2 } }{ 4 }\)
\( \\ =\frac { \pi r }{ 4 } \left( 4l+r \right) \)

Number of observations (n) = 10, which is even,
\(\therefore\) Median
\(=\frac { { \left( \frac { n }{ 2 } \right) }^{ th }observation+{ \left( \frac { n }{ 2 } +1 \right) }^{ th }observation }{ 2 } \)
\(\Rightarrow 65=\frac { 5^{ th }observation+6^{ th }observation }{ 2 } \)
\(\Rightarrow\)   \(65 = \frac {x+\left(x+2\right)}{2}\)
\(\Rightarrow\)  2x + 2 = 130
\(\Rightarrow\)  2x = 130 - 2 = 128 
\(\Rightarrow\)  \(x = \frac {128}{2}=64\)

ㄥ1=ㄥ2=\(\frac{1}{2}\)ㄥA
ㄥ3=ㄥ4=\(\frac{1}{2}\)ㄥC
In ΔCXB,
ㄥ3+ㄥX+ㄥB=180° ...(i)
(By Angle sum property of a Δ)
In ΔDAY,
ㄥ1+ㄥY+ㄥD=180° ...(ii)
(By Angle sum property of a Δ)
Adding (i) and (ii),
ㄥ3+ㄥX+ㄥB+ㄥ1+ㄥY+ㄥD=180°+180°
i.e., ㄥX+ㄥY+ㄥ3+ㄥ1+ㄥB+ㄥD=180
i.e., ㄥX+ㄥY+\(\frac{1}{2}\)ㄥC+\(\frac{1}{2}\)ㄥA+ㄥB+ㄥD=360...(iii)
But, ㄥA+ㄥB+ㄥC+ㄥD=360...(iv)
(Angle sum property of a quadrilateral)
From (iii) and (iv),
∠X+ㄥY+\(\frac{1}{2}\)ㄥC+\(\frac{1}{2}\)ㄥA+ㄥ​​​​B+ㄥD=ㄥA+ㄥB+ㄥC+ㄥD
i.e., ㄥX+ㄥY=ㄥA-\(\frac{1}{2}\)ㄥA+ㄥC-\(\frac{1}{2}\)ㄥC
=\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥC
ㄥX+ㄥY=\(\frac{1}{2}\)(ㄥA+ㄥC)
Hence Proved.

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA=90°
⇒ ㄥLON=90°
Similarly, ㄥOLM=ㄥLMN=ㄥMNO=90°

ஃ A quadrilateral with all angles, 90° is a rectangle. Also, opposite angles are equal. It is a rectangle.

90

Let E be the event of hitting the boundary.
Then,
\(P(E)=\frac { Number\ of\ times\ the\ batswoman\ hits\ the\ boundary }{ Total\ number\ of\ balls\ she\ plays } \)
 \(=\frac { 6 }{ 30 } =\frac { 1 }{ 5 } =0.2\)
Probability of not hitting the boundary
= 1- Probability of hitting the boundary
= 1- P(E) = 1 - 0.2 = 0.8

In Δs APD and CQB
AD = BC (Opp. sides of a parallelogram)
PD = BQ (Given)
∠ADP=∠QBC
⇒ ΔAPD≡ΔCQB
⇒ AP = CQ (c.p.c.t)
In Δs AQB and CPD
AB = DC, BQ = DP
and ∠AQB=∠PDC
ஃ ΔAQB≡ΔCPB ⇒ AQ=CP
(iii) In quad. APCQ, 
AP = CQ and AQ = CP
⇒ APCQ is a parallelogram.

Writing the given data in ascending order: 2,3,3,4,4,5,6,6,7,7,7,7,7,8,9
Here n = 15, Mean = \(\frac { \sum { x } }{ n } \)
=\(\frac{85}{15}\)= 5.7
Mode = 7, Median = 8th term = 6

(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.

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.

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.