Let 3+√2 is an rational number.. such that
3+√2 = a/b ,where a and b are integers and b is not equal to zero ..
therefore,
3 + √2 = a/b
√2 = a/b -3
√2 = (3b-a) /b
therefore, √2 = (3b - a)/b is rational as a, b and 3 are integers..
It means that √2 is rational....
But this contradicts the fact that √2 is irrational..

So, it concludes that 3+√2 is irrational..
hence proved.

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
<a id="proof" name="Proof"></a>

Basic  Proportionality Theorem Proof

Let us now try to prove the basic proportionality theorem statement

Consider a triangle ΔABC, as shown in the given figure. In this triangle, we draw a line PQ parallel to the side BC of ΔABC and intersecting the sides AB and AC in P and Q, respectively.

According to the basic proportionality theorem as stated above, we need to prove:

AP/PB = AQ/QC

Construction

Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure.

Proof

Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)

Similarly, area of ∆PBQ= 1/2 × PB × QN

area of ∆APQ = 1/2 × AQ × PM

Also,area of ∆QCP = 1/2 × QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

area of ΔAPQarea of ΔPBQ = 12 × AP × QN12 × PB × QN = APPB

Similarly, area of ΔAPQarea of ΔQCP = 12 × AQ × PM12 × QC × PM = AQQC ………..(2)

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

Therefore, we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …………..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

AP/PB = AQ/QC

Also, ∆ABC and ∆APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that ∆ABC ~∆APQ.

The MidPoint theorem is a special case of the basic proportionality theorem.

According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.

Consider an ∆ABC.

Conclusion

We arrive at the following conclusions from the above theorem:

If P and Q are the mid-points of AB and AC, then PQ || BC. We can state this mathematically as follows:

If P and Q are points on AB and AC such that AP = PB = 1/2 (AB) and AQ = QC = 1/2 (AC), then PQ || BC.

Also, the converse of mid-point theorem is also true which states that the line drawn through the mid-point of a side of a triangle which is parallel to another side, bisects the third side of the triangle.

Hence, the basic proportionality theorem is proved.

Given: Triangle ABC and ∆PQR in which AD and PM are medians drawn on sides BC and QR respectively. It is given that

To Prove : ∆ABC ~ ∆PQR
Const : Produce AD to E such that AD = DE and PM to F such that PM = MF.

 

Proof : In ∆ABD and ∆CDE,
AD = DE    [by construction]
∠ADB = ∠CDE
[vertically opposite angles]
and    BD = DC    [AD is a median]
Therefore, by using SAS congruent condition
                
                   [by CPCT]
Similarly, we can prove
                  
                              [by CPCT]
It is given that:

    
          
Therefore, by using SSS congruent condition
                   
                                 ...(i)
Similarly,                         ...(ii)
Adding (i) and (ii), we get
∠1 + ∠3 = ∠2 + ∠4
∠A = ∠P
Now, in ∆ABC and ∆PQR
                     
and           
Therefore, by using SAS similar condition
∆ABC ~ ∆PQR Hence Proved.

16th term in given A.P is 78

Step-by-step explanation:

Given A.P : 3,8,13,18,....,78

First term (a) = 3 ,

common difference (d)= 

= $8-3$

= $5$

Therefore,

16th term in given A.P is 78

Answer:

The coordinates of point P are 

Step-by-step explanation:

The line segment joining the points A(3,4) and B(7,9).

Let point P divides the line segment AB internally in he ratio 2:3.

The section formula is

Using the section formula, we get

26        = 2 x 13
91        =7 x 13
HCF    = 13

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.

Edge of the cube = diameter of sphere = 7 cm

Radius of the sphere =  = 3.5 cm

Volume of the sphere = 

= 179.67 

We know that circumference of a circle = 2π r.

We know that diameter = 2r.

Given circumference of a circle exceeds diameter by 16.8cm.

2πr - 2r = 16.8

2r(π - 1) = 16.8

2r(22/7 - 1) = 16.8

2r(22 - 7/7) = 16.8

2r(15/7) = 16.8

30r = 16.8 * 7

30r = 117.6

r = 117.6/30

r = 3.92 cm.

Therefore the radius of the circle = 3.92cm.

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Let given cubic polynomial be

p(x) = 3x³ - 5x² - 11x - 3

i ) If x = 3 , then

p(3) = 3(3)³ - 5(3)² - 11(3) - 3

= 81 - 45 - 33 - 3

= 81 - 81

= 0

ii ) If x = -1 , then

p(-1) = 3(-1)³ - 5(-1)² - 11(-1) - 3

= -3 - 5 + 11 - 3

= -11 + 11

= 0

iii ) If x = -1/3 , then

p(-1/3) = 3(-1/3)³ - 5(-1/3)² - 11(-1/3) - 3

= -3/27 - 5/9 + 11/3 - 3

= -1/9 - 5/9 + 11/3 - 3

= ( -1 - 5 + 33 - 27 )/9

= ( -33 + 33 )/9

= 0

Therefore ,

p(3) = p(-1) = p(-1/3) = 0 .

So, 3 , -1 , -1/3 are zeroes of given

cubic polynomial 3x³ - 5x² - 11x - 3 .

Compare the coefficients of given

cubic polynomial 3x³ - 5x² - 11x - 3

with ax³ + bx² + cx + d , we get

a = 3 , b = -5 , c = -11 , d = -3

Let the zeroes of the given cubic

polynomial p(x) are p =3 , q=-1, r=-1/3

Now ,

p+q+r = 3 - 1 - 1/3

= 2 - 1/3

= ( 6 - 1 )/3

= 5/3

= -b/a

pq+qr+rp

= 3(-1)+(-1)(-1/3)+(-1/3)(3)

= -3 + 1/3 - 1

= -4 + 1/3

= ( -12 + 1 )/3

= -11/3

= c/a

pqr = 3 × ( -1 ) × ( -1/3 )

= 1

=- d/a

Let r cm be the radius of the hemisphere, then r = 1 cm.
Let R be the radius of the cone and h cm be the height. Then

R = 1 cm and h = 1 cm
[It is given that R = h)
Now, Volume of solid
= Volume of hemisphere + volume of cone

class     freq        total
0-6           10         3 * 10
6-12         p           9 * p
12-18       4          4 * 15
18-24       7          7 * 21
24-30       q          27 * q
30-36        4          4 * 33
36-42        1         1 * 39
=====================
total       40        408 + 9p + 27 q
============================
total of frequencies = 26 + p + q = 40   =>    p+q = 14
sum =   408 + 9 p + 27 q = 14.7 * 40 = 588
So finally,     9 p + 27 q = 180       =>  p + 3 q = 20
  =>   14- q + 3 q = 20     =>  q = 3     and    p = 11 .