Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

The given numbers are 306 and 657.

The factor form of given numbers are

It is clear that 3 and 3 are common factors. So, the HCF of both numbers is

If a and b are two numbers then

Therefore the LCM of both numbers is 22338.

The square root of 3  is represented using the square root or the radical symbol “√”, and it is written as √3. The value of √3 is approximately equal to 1.732. This value is widely used in mathematics. Since root 3 is an irrational number, which cannot be represented in the form of a fraction. It means that it has an infinite number of decimals. Here, we are going to have a look at the value of root and the long division method to find the value of root 3.

1. 

Let √2 be a rational number 

Therefore, √2= p/q  [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get 
                   p²= 2q²                                                                                    ...(1)
Clearly, 2 is a factor of 2q²
⇒ 2 is a factor of p²                                                                    [since, 2q²=p²]
⇒ 2 is a factor of p

 Let p =2 m for all m ( where  m is a positive integer)

Squaring both sides, we get 
            p²= 4 m²                                                                                          ...(2)
From (1) and (2), we get 
           2q² = 4m²      ⇒      q²= 2m²
Clearly, 2 is a factor of 2m²
⇒       2 is a factor of q²                                                      [since, q² = 2m²]
⇒       2 is a factor of q 

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

     Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

(i) 2x² – 3x + 5 = 0 
Here, a = 2, b = -3, c = 5 
b² – 4ac = (-3)² – 4(2) (5
= 9 – 40 = -31 
b² – 4ac = -31 < 0 
It has no real roots.

(iii) 2x² – 6x + 3 = 0 

Here, a = 2, b = -6, c = 3 

b² – 4ac = (-6)² – 4(2)(3) = 36 – 24 = 12 

∴ b² – 4ac = 12 > 0. 

∴ It has two distinct roots. 2x² – 6x + 3 = 0 Here, a = 2, b = -3, c = 3  

Given: ABCD be a parallelogram circumscribing a circle with centre O.

To prove: ABCD is a rhombus.

We know that the tangents drawn to a circle from an exterior point are equal in length.

Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.

Adding the above equations,

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

2AB = 2BC

(Since, ABCD is a parallelogram so AB = DC and AD = BC)

AB = BC

Therefore, AB = BC = DC = AD.

Hence, ABCD is a rhombus.

Firstly, consider that the given circle will touch the sides AB and AC of the triangle at point E and F respectively. 

Let AF = x

Now, in ABC,

CF = CD = 6cm  (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C)

BE = BD = 8cm  (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B)

AE = AF = x  (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A)

Now, AB = AE + EB 

= x + 8

Also, BC = BD + DC = 8 + 6 = 14 and CA = CF + FA = 6 + x

Now, we get all the sides of a triangle so its area can be find out by using Heron's formula as:

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

⇒ Semi-perimeter = s = (28 + 2x)/2 = 14 + x

Again, area of triangle is also equal to the . Therefore,

Area of ΔOBC = 

Area of ΔOCA =

Area of ΔOAB =

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

On squaring both sides, we get

Either x+14 = 0 or x − 7 =0

Therefore, x = −14and 7

However, x = −14 is not possible as the length of the sides will be negative.

Therefore, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm

BASIC PROPORTIONALITY THEOREM PROOF

If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then  it divides the two sides in the same ratio. 

Given : In a triangle ABC, a straight line l parallel to BC, intersects AB at D and AC at E.  

To prove : AD/DB  =  AE/EC

Construction :

Join BE, CD. 

Draw EF ⊥ AB and DG ⊥ CA

Proof :

Step 1 :

Because EF ⊥ AB, EF is the height of the triangles ADE and DBE. 

Area (ΔADE)  =  1/2 ⋅ base ⋅ height  =  1/2 ⋅ AD ⋅ EF

Area (ΔDBE)  =  1/2 ⋅ base ⋅ height  =  1/2 ⋅ DB ⋅ EF

Therefore, 

Area (ΔADE) / Area (ΔDBE)  :

=  (1/2 ⋅ AD ⋅ EF) / (1/2 ⋅ DB ⋅ EF)

Area (ΔADE) / Area (ΔDBE)  =  AD / DB -----(1)

Step 2 : 

Similarly, we get

Area (ΔADE) / Area (ΔDCE)  :

=  (1/2 ⋅ AE ⋅ DG) / (1/2 ⋅ EC ⋅ DG)

Area (ΔADE) / Area (ΔDCE)  =  AE / EC -----(2)

Step 3 : 

But ΔDBE and ΔDCE are on the same base DE and between the same parallel straight lines BC and DE. 

Therefore, 

Area (ΔDBE)  =  Area (ΔDCE) -----(3)

Step 4 : 

From (1), (2) and (3), we can obtain

AD / DB  =  AE / EC

Hence, the theorem is proved. 

3Median =Mode + 2Mean

Given, Mean = 30

Mode = 40

Now, 3Median = Mode + 2Mean

=> 3Median = 40 + 2(30)

=> 3Median = 40 + 60

=> 3Median = 100

=> Median = 100/3

=> Median = 33.3

Hence,Median=33.3

Let 

first number         =  x

since this no are consecutive so

second number   =  x+1

now according to question

x (x+1)                    =   110

x² + x                     = 110

x² + x - 110             = 0

by using factorization method

x² + 11x - 10x -110  =  0

x(x+11) -10(x+10)    =  0

(x+11) (x-10)            = 0

∵ x = -11 and x       = 10

The two numbers would be 10 and 11.

10 x 11 = 110

10 + 11 = 21

a) the length of the broken part can be figured out by the pythagoras theoram-

so, suppose the length of the broken part is x

15 square +20 square = x square

= 225 + 400 = x  square

thus, x square = 625

hence, x= sqrt of 625

therefore x= 25m

b) the heinght of the full tree is 15+ 25= 40m

c) the length of the hypotenuse is 25m

d) the area of the triangle= (15*20)/4 = 300/4 = 75m sq.

a) the length of the broken part can be figured out by the pythagoras theoram-

so, suppose the length of the broken part is x

15 square +20 square = x square

= 225 + 400 = x  square

thus, x square = 625

hence, x= sqrt of 625

therefore x= 25m

b) the heinght of the full tree is 15+ 25= 40m

c) the length of the hypotenuse is 25m

d) the area of the triangle= (15*20)/4 = 300/4 = 75m sq.

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x

We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x

Now semi perimeter of circle s,
⇒ 2s = AB + BC + CA
         = x + 8 + 14 + 6 + x
         = 28 + 2x
⇒s = 14 + x

Area of ΔABC = √s (s - a)(s - b)(s - c)

                         = √(14 + x) (14 + x - 14)(14 + x - x - 6)(14 + x - x - 8) 

                         = √(14 + x) (x)(8)(6)

                         = √(14 + x) 48 x ... (i)

also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

                                 = 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

                                 = 2×1/2 (4x + 24 + 32) = 56 + 4x ... (ii)

Equating equation (i) and (ii) we get,

√(14 + x) 48 x = 56 + 4x

Squaring both sides,

48x (14 + x) = (56 + 4x)²

⇒ 48x = [4(14 + x)]²/(14 + x)

⇒ 48x = 16 (14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

⇒ x = 7 cm

Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm

Steps of Construction :
(i) Draw a line segment AB = 8 cm.
(ii)    Draw a circle with centre A and radius 4 cm. Draw another circle with centre B and radius 3 cm.
(iii)    Let M be the mid-point of AB. Taking M as centre and AM as radius draw a circle which intersects the circles at P, Q, R and S.


Fig,
(iv) Join AP, AQ, BR and BS. These are required tangents.

Let α and β are the roots of given Quadratic polynomial.
Given:
α+β=3,αβ=2
Quadratic polynomial:
x²−(α+β)x+(αβ)
∴ Polynomial is x²−3x+2