\((i)\frac { 21 }{ 40 }\) 
\((ii)\frac { 29 }{ 40 }\) 
\((iii)\frac { 29 }{ 40 } \)

Let the given field be in the shape of a trapezium ABCD in which AB=25 m, CD= 10 m, BC = 13 m and AD = 14 m.
From D, draw DE 11 BC meeting AB at E. Also, draw DF \(\bot \) AB.
\(\therefore \) DE = BC =13 m
AE = AB - EB = AB - DC
= 25 - 10 = 15 m

For AED
a = 14 m, b = 13 m, c=  15 m
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 14+13+15 }{ 2 } =\frac { 42 }{ 2 } =21\) m
\(\therefore \)    Area of the \(\Delta \)AED  \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-14)(21-13)(21-15) }\)
\( \\ =\sqrt { 21(7)(8)(6) } =\sqrt { \left( 7\times 3 \right) (7)\left( 4\times 2 \right) \left( 2\times 3 \right) } \)
\(=7\times 3\times 2\times 2=84\) m²
\(\Rightarrow \frac { 1 }{ 2 } \times \)AE\(\times \)DE = 84
\(\Rightarrow \ \frac { 1 }{ 2 } \times \)15\(\times \)DF = 84
\(\Rightarrow \) DF = \(\frac { 84\times 2 }{ 15 } \)
\(\Rightarrow \) DF=  \(\frac { 56 }{ 5 } \) m = 11.2 m
\(\Rightarrow\) Height of the trapezium is 11.2 m.
\(\therefore \) Area of parallelogram EBCD = Base \(\times \) Height
= EB\(\times \) DF = 10 \(\times \)\(\frac { 56 }{ 5 } \) = 112 m²
\(\therefore \) Area of the field = Area of AED + Area of parallelogram EBCD = 84 m² + 112 m² = 196 m².                 

Rs.5376

(x-2)(x-4)(x-3)

0

In \(\triangle ABC\)
\(\angle A+\angle B+\angle C=180^o\) (Angle Sum prop. of \(\triangle)\)
2x-5^o+5x+5^o+3x+50^o=180^o
10x=130^o
x=13^o
\(\angle A=2x-5^o=21^o\)
\(\angle B=5x+5^o=70^o\)
\(\angle C=3x-50^o=89^o\)

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.

\((i)\frac { 21 }{ 100 }\) 
\((ii)\frac { 6 }{ 25 } \) 

\(1500\sqrt { 3 } \) cm²

\((2x-y+z)^2\)

Comparing the given expression with (x + y + z)², we find that
x = 3a, y = 4b and z = 5c.
Therefore, using Identity V, we have
(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

Steps of construction:
i) Draw a line segment XY = 11 cm (As AB+  BC + CA = 11cm)
ii) Construct an angle PXY of  60^o at point X and an angle ㄥQYZ of 45^o at point Y
iii) Bisect ㄥPXY and ㄥQYZ .These bisectors intersect each other at point A
iv) Draw perpendicular bisectors ST of XA and UV of YA.
v) Perpendicular bisector ST intersects XY at B and UV intersects XY at C Join AB, AC MBC is the required triangle.

By triangle inequality property,
In \(\triangle\) ABC, AB + BC>AC   ...(1)
In \(\triangle\)BCD, BC + CD > BD  ...(2)
In \(\triangle\)CDA, CD + DA >AC   ...(3)
In \(\triangle\)DAB, DA +AB > BD   ...(4)
using the fact that sum of two sides of a triangle  is greater than the third side​​​​​​​

Adding (1), (2), (3) and (4), we get
AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD
\(\Rightarrow\) 2(AB + BC + CD + DA) > 2(AC + BD)
Hence,Perimeter > Sum of its diagonals.

\(\angle PBC=x+z\)
\(\Rightarrow 2\angle OBC=x+z\)
\(\angle QCB=x+y\)
\(\Rightarrow 2\angle OCB=x+y\)
\(\therefore \angle BOC+\angle OCB+\angle OBC=180^o\) (Angle sum prop. of \(\triangle)\)
\(\Rightarrow 2\angle BOC+2\angle OCB+2\angle OBC=360^o\) (Multiply by 2 both sides)
\(\Rightarrow 2\angle BOC+x+y+x+z=360^o\)
\(\Rightarrow 2\angle BOC+180^o+x=360^o\)
\(\Rightarrow 2\angle BOC=180^o-x\)
\(\Rightarrow 2\angle BOC=90^o-\frac{1}{2}x\)

Given, \(\angle\)DCA = \(\angle\)ECB
Adding \(\angle\)DCE to both sides, we get
\(\angle\)DCA + \(\angle\)DCE = \(\angle\)ECB + \(\angle\)DCE
\(\Rightarrow\) \(\angle\)ECA = \(\angle\)DCB
In \(\triangle\)ACE and \(\triangle\)BCD,
AC = BC (Given)
\(\angle\)ECA = \(\angle\)DCB (Proved)
\(\angle\)EAC = \(\angle\)DBC (Given)
\(\triangle ACE\cong \triangle BCD\) (By AAS cong.) 
\(\therefore\) BD = AE.

\(LHS=\frac { { \left( 2\sqrt { 5 } +\sqrt { 3 } \right) }^{ 2 }+{ \left( 2\sqrt { 5 } -\sqrt { 3 } \right) }^{ 2 } }{ \left( 2\sqrt { 5 } -\sqrt { 3 } \right) \left( 2\sqrt { 5 } +\sqrt { 3 } \right) } \)
\(=\frac { 4\times 5+3+2\times 2\sqrt { 5 } \times \sqrt { 3 } +4\times 5+3-2\times 2\sqrt { 5 } \times \sqrt { 3 } }{ { \left( 2\sqrt { 5 } \right) }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 } } \)
\(=\frac { 20+3+4\sqrt { 15 } +20+3-4\sqrt { 15 } }{ 20-3 } \)
\(=\frac { 46 }{ 17 } =\frac { 46 }{ 17 } +\sqrt { 15 } \times \left( 0 \right) \)
\(\therefore \frac { 46 }{ 17 } +\sqrt { 15 } \times \left( 0 \right) =a\sqrt { 15 } b=RHS\)
Comparing both sides, we get
a=\(\frac{46}{17}\) , b = 0

\(\therefore\) Quotient = 2x³+x²+x-1
Remainder=0
Hence, (2x+3) is factor of 
p(x)=4x⁴+8x³+5x²+x-3

In right triangle PSQ,  PQ\(\frac { 1 }{ 2 } \)² = PS² + QS²     |By Pythagoras Theorem
= (12)² + (16)²
= 144 + 256 =400
\(\Rightarrow \)  PQ = \(\sqrt { 400 } \) = 20 cm
Now, for  \(\Delta \)PQR
a = 20cm, b = 48cm, c = 52cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+48+52 }{ 2 } =60\) cm
\(\therefore \)  Area of  \(\Delta \)PQR \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-48)(60-52) }\)
\( \\ =\sqrt { (60)(40)(12)(8) } \)
 \(=\sqrt { \left( 6\times 10 \right) \left( 4\times 10 \right) \left( 6\times 2 \right) \left( 8 \right) } \)
\(=6\times 10\times 8=480\) cm²
Area of \(\Delta \) PSQ = \(\frac { 1 }{ 2 } \)\(\times \)Base\(\times \)Altitude
=\(\frac { 1 }{ 2 } \)\(\times \)16\(\times \)12=96 cm²
\(\therefore \) Area of the shaded portion =Area of \(\Delta \)PQR - Area of \(\Delta \)PSQ
= 480 - 96 = 384 cm²

Let AB and CD two lines intersecting at O

This leads to two pairs of vertically  opposite angles,namely
(i)  \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
We are to prove that
(i)   \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
\(\therefore \) Ray OA stands on line CD 
Therefore (i)   \(\angle \)AOC and \(\angle \)AOD=\(180^{ 0 }\)
|Linear Pair Axiom
From (1) and (2) 
\(\angle \)AOC and \(\angle \)AOD= \(\angle \)AOD and \(\angle \)BOD \(\angle \)AOC and \(\angle \)BOD
\(\Rightarrow \) Similarly we can prove that 
\(\angle \)AOD and \(\angle \)BOC