In \(\Delta\)EAB and \(\Delta\)ECD
\(\frac{E A}{E C}=\frac{E B}{E D}\)     [given]
\(\angle A E B=\angle C E D\)   [vertically opposite angles]
\(\therefore\) By SAS rule of similarity
\(\triangle E A B \sim \triangle E C D\)      Hence proved.

10

ABCD is a square
Given: side of the square =14cm
∴ Area of the square = (side)² (14)² = 196 cm²
Radius of the semicircle APD=\({1\over 2}\)(side of square)=\({1\over 2}\times14=7cm\)
Area of the semicircle APD=\({1\over 2}\pi r^2={1\over 2}\times{22\over 7}\times7\times7=11\times7=77cm^2\)
Similarly, area of the semicircle BPC= 77cm² 
Total area of both the semicircles=77 + 77 - 154 cm².
Area of the shaded region = Area of square - area of both semicircles = 196 - 154 -42cm²

Area of portion that horse can graze area of the shaded portion.
Shaded portion is a sector of radius 21 m = length of the rope 
Angle of this sector = angle of the corners of the rectangle = 90°
Area the shaded portion that horse can graze 
= \(\frac { \theta }{ 360° } \times { \pi r }^{ 2 }=\frac { 90° }{ 360° } \times \frac { 22 }{ 7 } \times { (21) }^{ 2 }{ m }^{ 2 }\)
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 21\times 21 \ { m }^{ 2 } \ = \ \frac { 11 }{ 2 } \times 3\times 21 \ { m }^{ 2 }=346.5 \ \ { m }^{ 2 }\)

We know that the corresponding sides of similar triangles are proportional.
∴ PA/PD = AC/DB = PC/PB
⇒ PA/PD = PC/PB
∴ PA.PB = PC.PD

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below

From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than \(\frac{n}{2}\left(\frac{100}{2}=50\right)\) is 78, belonging to interval 35 - 40.
Therefore, median class = 35 - 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
\(\text { Median }=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \)
\(=35+\left(\frac{50-45}{33}\right) \times 5 \)
\(=35+\frac{25}{33}\)
= 35.76
Therefore, median age is 35.76 years.

Area of the segment AYB = Area of sector OAYB - Area of \(\Delta\)OAB                  (1)
Now, area of the sector OAYB = \(=\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2=462 \mathrm{~cm}^2\) (2)
For finding the area of \(\Delta\)OAB, draw OM \(\perp\) AB as shown in Figure.
Note that OA = OB. Therefore, by RHS congruence, \(\Delta\)AMO \(\cong\) BMO.
So, M is the mid-point of AB and \(\angle\)AOM = \(\angle\)BOM = \(\frac{1}{2} \times 120^{\circ}=60^{\circ}\)
Let  OM = x cm
So, from \(\Delta\)OMA, \(\frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ}\)

or, \(\begin{aligned} \frac{x}{21} & =\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right) \\ \end{aligned}\)
or, \(\begin{aligned} x & =\frac{21}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \mathrm{OM} & =\frac{21}{2} \mathrm{~cm} \\ \end{aligned}\)
Also,\(\begin{aligned} \frac{\mathrm{AM}}{\mathrm{OA}} & =\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \mathrm{AM} & =\frac{21 \sqrt{3}}{2} \mathrm{~cm} \end{aligned}\)
Therefore, \(\mathrm{AB}=2 \mathrm{AM}=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{~cm}=21 \sqrt{3} \mathrm{~cm}\)
So, area of \(\begin{aligned} \Delta \mathrm{OAB} & =\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{441}{4} \sqrt{3} \mathrm{~cm}^2 \end{aligned}\)                                                                              (3)
Therefore, are of the segment \(\mathrm{AYB}=\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^2\) [From (1), (2) and (3)]
\(=\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^2\)

(i) 77 cm²
(ii) 44 cm

We have, radius =14 cm
\(\angle\)BAD = 45°
According to the question,
Perimeter = 2 (radius)+ arc BCD
\(\therefore \text { Length of arc }=\frac{\theta}{360^{\circ}} \times 2 \pi r=\frac{45^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 14\)
= 11 cm
\(\therefore\) Perimeter = 2 \(\times\)14 + 11 = 28 + 11 = 39 cm
Hence, perimeter of the sector ABCDA is 39 cm.

In a given figure, the sum of angles of three sectors of a circle = 60° + 80° + 40° = 180°
\(\therefore\) Area of shaded region = Sum of areas of three sectors separately
= Area of sector makes by 180°            ...(i)
Now, length on an arc of sector makes by 180°,
\(\begin{aligned} l & =\frac{\theta}{360^{\circ}} \times 2 \pi r \end{aligned}\)
\(\begin{aligned} =\frac{180^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 7 \quad[\because r=7 \mathrm{~cm} \text { given }] \end{aligned}\)
l = 22 cm
\(\therefore\) From Eq. (i), we get
Area of shaded region = \(\frac{1}{2} l r\)
\(=\frac{1}{2} \times 22 \times 7=77 \mathrm{~cm}^2\)
Hence, area of the shaded region is 77 cm².

Given: D is the mid-point of BC, E is the mid-point of AD. BE produced meets AC at M.
To Prove: BE = 3 EM
Construction: Draw DN ll BM

Proof: In ∆ADN, EM ll DN (Construction)
∆AEM ~ ∆ADN (AA similarity)
\(\frac{EM}{DN}=\frac{AE}{AD}\) (By BPT)
But AD = 2AE
\(\frac{EM}{DN}=\frac{AE}{2AE}\)
DN = 2EM...(i)
In ΔBCM, DN || BM (Construction)
∆CDN ~ ΔCBM
\(\frac{DN}{BM}=\frac{CD}{2CD}\)
BM=2DN...(ii)
From (i) and (ii)
BM = 2(2EM)
BM = 4EM
BE = BM - EM = 4EM - EM
= 3EM.
Hence proved.

Mean x= \(\frac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } =\frac { 6300 }{ 120 } =52.5\)

\(\therefore\) DE || BC
\(\therefore \frac { AD }{ DB } =\frac { AE }{ EC } \)  (By BPT )
\(\Rightarrow \frac { x+2 }{ 3x+16 } =\frac { x }{ 3x+5 } \)
\(\Rightarrow\) (x + 2)(3x + 5) = x(3x + 16)
\(\Rightarrow\)3x² + 5x + 6x + 10 = 3x² + 16x
\(\Rightarrow\) 11x + 10 = 16x 1
\(\Rightarrow\)16x-11x = 10
\(\Rightarrow\) 5x = 10
\(\therefore\)  x = 2

52

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

Let r and R be the radii of each smaller circle and larger circle respectively.
We have, \(d=\frac{1}{4} D\)
\(\Rightarrow r=\frac{1}{4} R \Rightarrow r=\frac{1}{4} \times 16 \Rightarrow r=4 \mathrm{~cm}\)
(i) (c): Area of smaller circle = \(\pi r^{2}\)
\(=\frac{22}{7} \dot{\times} 4 \times 4=50.28 \mathrm{~cm}^{2}\)
(ii) (a): Area oflarger circle = \(\pi R^{2}\)
\(=\frac{22}{7} \times 16 \times 16=\frac{5632}{7}=804.57 \mathrm{~cm}^{2}\)
(iii) (b): Area of the black colour region = Area of larger circle - Area of 4 smaller circles
= 804.57 - 4 x 50.28 = 603.45 cm²
(iv) (d): Area of quadrant of a smaller circle
\(=\frac{1}{4} \times 50.28=12.57 \mathrm{~cm}^{2}\)
(v) (c): Area between two concentric circles
\(=\pi\left(R^{2}-r^{2}\right)=\frac{22}{7}\left(5^{2}-2^{2}\right) \)
\(=\frac{22}{7}(25-4)=\frac{22}{7} \times 21=66 \mathrm{~cm}^{2}\)