Steps of construction:
i) Draw the line segment BC = 8 cm and at point B construct an angle of 45o .i.e XBC = 45^o and AB - AC = 3.5 cm
ii) Cut the line segment BD = 3.5 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.Join AC MBC is the required triangle.

Modified Continues Distribution

(ii) Frequency Polygon.
(iii) No, because the maximum number of leaves have their lengths lying in the original interval 145-153 (or modified interval 144.5-153.5).

(i)

(ii) Political party A won the maximum number of seats.

Given: Line l is the bisector of an angle\(\angle A\) and  \(\angle B\)   is any point on I.BP and BQ are perpendiculars from B to the arms of  \(\angle A\)
To Prove: (i) \(\triangle APB\cong \triangle AQB\)
(ii) BP = BQ or B is equidistant from the arms of \(\angle A\)
Proof: (i) In \(\triangle APB\) and \(\triangle AQB\)
\(\angle BAP=\angle BAQ \)  | l is the bisector of \(\angle A\)
AB = AB | Common
\(\angle BAP=\angle BAQ \)  | Each = 90⁰
BP = BQ or B is equidistant from the arms of \(\angle A\)
\(\triangle APB\cong \triangle AQB\)   | SAS Rule
(ii) \(\triangle APB\cong \triangle AQB\) | Proved in (i) above
BP = BQ  | C.P.C.T

19 25 59 48 35 31 30 32 51
n = 9
n+1/2 = 5
Median = 32

\(\frac{3+5+0+9+x+7x+13}{7}=7\)

Mean \(\frac {2+3+5+7+11}{5}=5.6\)

14 occurs most frequently (4 times).

Maximum number of students (26) have 40 marks.

Required number = 3 + 6 + 8 + 8 = 25.

Highest item value = 3.2
Lowest item value  = 2.1
\(\therefore\) Range = 3.2 -2.1 = 1.1.

s=\(\frac { 20+15+9 }{ 2 } \)=22 cm

ㄥCBA=ㄥCDE=90⁰
ㄥACB=ㄥECD
BC=CD
∴ ΔCBA≅ΔCDE
∴ AB=DE

Theorem

≌ represents congruence.

Steps of Construction:
i) Draw any line OP.
ii) With O as centre and any suitable radius, draw an arc to meet OP at R.
iii) With R as centre and same radius (as in step 2).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)

OA = OB (O is the mid-point of AB)
\(\angle\)AOC = \(\angle\)BOD (Vertically opposite angles)
OC = OD (O is the mid-point of CD)
\(\triangle AOC\cong \triangle BOD\)
\(\Rightarrow\) AC = BD. (By c.p.c.t) Proved.

Join AD.
In \(\triangle\)ABC and \(\triangle\)ACD,
AB = AC   (Given)
BD = CD   (Given)
AD = AD  (Common)
By using SSS Congruency Rule,
\(​​\triangle ABD\cong \triangle ACD\)
\(\therefore \angle ABD=\angle ACD (By c.p.c.t)\)

Since we want to draw a frequency polygon without a histogram, let us find the class-marks of the classes given above, that is of 140 - 150, 150 - 160,....
For 140 - 150, the upper limit = 150, and the lower limit = 140
So, the class-mark = \(\frac{150+140}{2}=\frac{290}{2}=145\)

We can now draw a frequency polygon by plotting the class-marks along the horizontal axis, the frequencies along the vertical-axis, and then plotting and joining the points B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments. We should not forget to plot the point corresponding to the class-mark of the class 130 - 140 (just before the lowest class 140 - 150) with zero frequency, that is, A(135, 0), and the point H (205, 0) occurs immediately after G(195, 2). So, the resultant frequency polygon will be ABCDEFGH.

For finding area of the park, we have
2s = 50 m + 80 m + 120 m = 250 m.
i.e., s = 125 m
Now, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m,
s – c = (125 – 50) m = 75 m.
Therefore, area of the park = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{125 \times 5 \times 45 \times 75} \mathrm{~m}^{2}\)
\(=375 \sqrt{15} \mathrm{~m}^{2}\)
Also, perimeter of the park = AB + BC + CA = 250 m
Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)
= 247 m
And so the cost of fencing = Rs.20 x 247 = RS. 4940

84 cm²

30 cm^2.

For one black colour sheet  a= 4 cm, b = 6 cm
\(\therefore \) Area \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)     | Sheet is an isosceles triangle
 \(=\frac { 4 }{ 4 } \sqrt { 4{ (6) }^{ 2 }-4^{ 2 } } \)  \(=\sqrt { 128 } =8\sqrt { 2 } \) cm²
\(\therefore \) Area of 2 black colour sheets = 2\(\times 8\sqrt { 2 } \) = \( 16\sqrt { 2 } \) cm²
Similarly, area of 2 white colour sheets =  \( 16\sqrt { 2 } \) cm²
\(\therefore \) Total area of black and white colour sheets =  \( 16\sqrt { 2 } \) + \( 16\sqrt { 2 } \) =  \( 32\sqrt { 2 } \) cm²