Given:

Here  in two triangles, sides of one triangle are proportional to the sides of other triangle.

Also their corresponding angles are equal and hence the two triangles are similar.

Construction:

Let us draw lines at E and F making the angles as marked in the diagram and meeting at G.

Proof:

Now, ΔABC and ΔGEF are equiangular and hence similar and so corresponding sides are in proportion.

         AB : BC = GE : EF

But    AB : BC = DE : EF          (Given)

So              GE = DE                  (1)

 Similarly,  AC : CB = GF : FE

But    AB : BC = DF : FE    (Given)

So              GF = DF                  (2)

 EF is common to both triangles DEF and GEF.          (3)

So from Eq (1), (2) and (3) we have

ΔDEF ≅ ΔGEF.

Therefore triangles ABC and DEF are equiangular and hence similar etc.

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Mean of given observations = sum of given observations/ total number of observations
Mean of 25 observations = 27
∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 – (25 × 7) = 675 – 175 = 500
Then, new mean = 500/25 = 20

Thus, the new mean will be 20. 

  Class Interval          Frequency          Cumulative Frequency

     0 - 6                           4                                  4

     6 - 12                         x                                4 + x

    12 - 18                        5                                9 + x

    18 - 24                        y                                9 + x + y

     24 - 30                       1                               10 + x + y
______________________________________________
                                      20                                                 
______________________________________________

10 + x + y = 20

x + y = 20 - 10

x + y = 10

Median = 14.4 

So, Median class is 12 - 18

Median = l + [(N/2 - cf)*i]/f

l = Lower limit of the Median Class = 12

Class Interval = i = 6

Cumulative Frequency (cf) of the class before the Median Class = 4 + x

N = 20

N/2 = 20/2 = 10

f = frequency of the Median Class = 5

⇒ Median = l + [(N/2 - cf)*i]/2

⇒ 14.4 = 12 + {10 - (4 + x)*6]/5

⇒ 14.4 - 12 = {(10 - 4 - x)*6}/5

⇒ 2.4 = {(6 - x)*6}/5

⇒ 2.4 = (36 - 6x)/5

⇒ 2.4*5 = 36 - 6x

⇒ 12 = 36 - 6x

⇒ - 6x = 12 - 36

 - 6x = - 24

⇒ 6x = 24

x = 24/6

x = 4

Since, x + y = 10

4 + y = 10

y = 10 - 4

y = 6

So, the value of x is 4 and the value of y is 6.

Steps of Construction 
Step 1: Draw a circle with O as center and radius 3.5 cm.
Step 2: Mark a point P outside the circle such that OP = 6.2 cm 
Step 3: Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q. 
Step 4: Draw a circle with Q as center and radius PQ (or OQ), to intersect the given circle at the points T and T’. 
Step 5: Join PT and PT’.

Here, PT and PT’ are the required tangents.Read more on Sarthaks.com -  

Let three consecutive positive integers be, n, n + 1 and n + 2. 

When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.  

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3. 

If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. 

If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. 

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.  

⇒ n (n + 1) (n + 2) is divisible by 3. 

Similarly, when a number is divided 2, the remainder obtained is 0 or 1. 

∴ n = 2q or 2q + 1, where q is some integer. 

If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. 

If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. 

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. 

⇒ n (n + 1) (n + 2) is divisible by 2. 

Hence n (n + 1) (n + 2) is divisible by 2 and 3.

∴ n (n + 1) (n + 2) is divisible by 6. 

Let AD be the lighthouse whose height is h metres. 13 and C are the position of two ships which are on opposite sides of lighthouse. The angles of depression of two ships B and C from the top of the lighthouse are α and β respectively.

i.e.,    ∠ABD = α and ∠ACD = β
Let    BD = x m and CD = y m
In right triangle ABD, we jave


In right triangle ACD, we have


Adding (i) and (ii), we get


Hence, the distance between the ship is  metres.
 

The point of observation is at A.

The height of the deck is 10 meters.

Thus, AB = CD = 10 meters

The top and bottom of a hill is E and C.

Therefore the angles of elevation and the depression are .

Let AD = BC = x meters.

Consider the triangle ADE.

Thus,

;

Now consider the triangle ABC.

Substitute the value of x from equation (2) in equation (1), we have

Therefore 

h = 30 meters  .....(4)

The height of the hill is CE

That is,

Substitute the value of h from equation (4) in equation (5), we hve

Thus, the height of the hill is 40 meters and the distance of the hill from the ship is 

1.For that you have to use formula of volume of cone - 1/3 πr²h and you get the answer 19.64 or you can right it 20 cm²

2. For that you have to take volume of cone and volume of hemisphere [ 1/3 πr²h + 2/3πr³]

Step-by-step explanation:

1. volume of cone - 1/3πr²h

= 1/3 × 22/7 × 2.5² × 9

= (22× 6.25) ÷ 7

= 19.64 or 20cm³

2. volume of cone + volume of hemispherical end

= 1/3πr²h + 2/3πr³

= 1/3 × 22/7 × 2.5² × 9 + 2/3 × 22/7 × 2.5³

= 137.5/7 + 687.5/21

= (412.5 + 687.5) ÷ 21

= 1100 ÷ 21

= 52.3 cm³

(sin30+cos30) - (sin60+cos60)

sin30 = 1/2

sin60= √3/2

cos30= √3/2

cos60= 1/2

=> (1/2+√3/2) - (√3/2+1/2)

=> 1+√3/2 - √3+1/2

=> 1+1+√3-√3 /2

=> 2/2

=>1

AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm.

The tangents at A and B intersect at P.
CONSTRUCTION :  Join OP and OA. Let OP and AB intersect at M.
Let PA = x cm and PM = y cm.
Now, PA=PBPA=PB
and OP is the bisector of ∠APB∠APB   [∵∵  two tangents to a circle from an external point are equally inclined to the line segment joining the centre to that point.
Also, OP⊥ABOP⊥AB and OP bisects AB at M  [∵∵ OP is the right bisector of AB]
∴∴ AM = MB = 9.629.62cm
=4.8cm.=4.8cm.
In right △AMO△AMO, we have
OA=6cmOA=6cm
and AM=4.8cm.AM=4.8cm.
∴∴  OM = OA2−AM2−−−−−−−−−−√OA2−AM2
=62−4.82−−−−−−−√=62−4.82
=12.96−−−−√=12.96
=3.6cm=3.6cm.
In right △PAO△PAO, we have
AP^{2 =} PM² + AM²
⇒x2=y2+(4.8)2⇒x2=y2+(4.8)2
⇒x2=y2+23.04⇒x2=y2+23.04...(i)
In right △PAO△PAO, we have
OP² = PA² + OA² [Note ∠PAO=90∘∠PAO=90∘, since AO is the radius at the point of contact]
⇒(y+3.6)2⇒(y+3.6)2
=x2+62=x2+62
⇒y2+7.2y+12.96⇒y2+7.2y+12.96
=x2+36=x2+36
⇒7.2y=46.08⇒7.2y=46.08 [using (i)]
⇒y=6.4cm⇒y=6.4cm
and
x² = (6.4)² + 23.04
= 40.96 + 23.04 = 64
⇒x=64−−√=8⇒x=64=8
∴∴ PA=8cm.

Example

Each numeral on a die is equally likely to occur when the die is tossed.

Sample space of throwing a die: { 1, 2, 3, 4, 5, 6 }

From this we can make conclusions such as the following:

• Getting a 3 on the toss of a die and getting a 5 on the toss of a die are equally likely events, since the probabilities of each event are equal.
   
• Getting an even number on the toss of a die and getting an odd number on the toss of a die are equally likely events, since the probabilities of each event are equal.
   
• Getting a 1, 2 or 3 on the toss of a die and getting a 4, 5 or 6 on the toss of a die are equally likely events, since the probabilities of each event are equal.

Step-by-step explanation:

We know that 

We are given that 

On comparing

Base = 7

Perpendicular = 8

To find hypotenuse we will use Pythagoras theorem

Now we are supposed to find 

Substitute the values

Hence