Radius of circle = 12 cm
θ = 120°
Area of the segment= {(π/360) × θ - sin θ/2 cos θ/2} r²
Area of the segment={(π) × 120/360 - sin 120/2 cos 120/2 } 12²
= {(π) × ⅓ - sin 60° cos 60° } ×144
= (π/3 - ½ × √3/2) × 144
= (π/3 × 144 - 144 × √3/4
= 48π - 36√3
= 12(4π - 3√3)
= 12( 4 × 3.14 - 3 × 1.73)
[π = 3.14 , √3= 1.73]
= 12 (12.56 - 5.19)
= 12 × 7.37
= 88.44 cm²
Hence, the area of the corresponding segment of the circle is 88.44 cm².

Radius of the circle = 15 cm

ΔAOB is isosceles as two sides are equal.

∴ ∠A = ∠B

Sum of all angles of triangle = 180°

∠A + ∠B + ∠C = 180°

⇒ 2 ∠A = 180° - 60°

⇒ ∠A = 120°/2

⇒ ∠A = 60°

Triangle is equilateral as ∠A = ∠B = ∠C = 60°

∴ OA = OB = AB = 15 cm

Area of equilateral ΔAOB = √3/4 × (OA)² = √3/4 × 15² 

                                          = (225√3)/4 cm² = 97.3 cm²

Angle subtend at the centre by minor segment = 60°

Area of Minor sector making angle 60° = (60°/360°) × π r² cm²

                                                                                     = (1/6) × 15² π  cm² =  225/6 π  cm²

                                                  =  (225/6) × 3.14 cm² = 117.75  cm²

Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB

                                            = 117.75  cm² - 97.3 cm² = 20.4 cm²

Angle made by Major sector = 360° - 60° = 300°

Area of the sector making angle 300° = (300°/360°) × π r² cm²

                                                   = (5/6) × 15² π  cm² =  1125/6 π  cm²

                                                  =  (1125/6) × 3.14 cm² = 588.75  cm²

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB

                                            = 588.75  cm² + 97.3 cm² = 686.05 cm²

Not correct . as the ratio of the squars of the corresponding sides of two similar triangles is equal to the ratio of their areas. so it should be 9:25 . 

The size of each section is to be selected in such a way, It divides both 576 and 488 exactly.

So you have to find HCF of 576 and 448 girls.

576 = 2 x 2 x 2 x 2 x 2 x 2x 3 x 3
448 = 2 x 2 x 2 x 2 x 2 x 2 x 7
Highest Common Factors = 2 x 2 x 2 x 2 x 2 x 2 = 64

(The Highest Common Factor (HCF) is the largest positive integer that divides all the given numbers. Hence the HCF is the biggest size for forming groups.)

Hence, number of sections required;
(576 ÷ 64) + (448 ÷ 64)
= 9 + 7 = 16.

Therefore, Total number of sections thus formed are 16.

Let the speed of the second train be x km hr. Then, the speed of the first train is (x + 5) km/hr.
Let O be the position of the railway station from which the two trains leave.

 

Distance travelled by the first train in 2 hours
= OA = Speed x Time
= 2 (x + 5) km
Distance travelled by the second train in 2 hours
= OB = Speed x Time
= 2x km
By using pythagoras theorem, we have,
AB² = OA² + OB²
⇒ 50² = {2(x + 5)}²+ {2x}²
⇒ 2500 = 4(x + 5)² + 4x²
⇒ 8x² + 40x – 2400 = 0
⇒ x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20) (x – 15) = 0
⇒ x = –20 or x = 15
⇒ x = 15
[∵ cannot be negative]
Hence, the speed of the second train is 15 km/hr and the speed of the first train is 20 km/hr.

Radius (r) = 14cm

θ = 90°

= OA = OB

Area of minor segment (ANB)

= (area of ANB sector)−(area of ΔAOB)

Area of major segment (other than shaded)

= area of circle – area of segment ANB

Clearly, the distance of the point P(-3, 4) from x -axis is 4 units.

In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Fig. 7.8

Let A(3, 4), B(6, 7), C(9, 4) and D(6, 1) be the given points. Then

]

We see that,
AB = BC = CD = DA
and    AC = BD
Therefore, ABCD is a square.
Hence, Champa is correct.

The series can be written as

(5 + 9 + 13 + .... + 81) + [(- 41) + (- 39) + (- 37) + (- 35) ... (- 5) + (- 3)]

For the series (5 + 9 + 13 + ... + 81)

a = 8

d = 9 - 5 = 5

and a(n) = 81

Then, a(n) = 5 + (n - 1)4 = 81

or, (n - 1)4 = 76

⇒ n = 20

S(n) = 20/2(5 + 81)

= 860

For Series (- 41) + (- 39) + (- 37) + (- 35) ... (- 5) + (- 3)

a(n) = - 3

a = - 41

d = 2

Then, a(n) = - 41 + (n - 1)(2)

⇒ n = 20

Now, S(n) = 20/2[- 41 + (- 3)]

= - 440

Sum of the series = 860 - 440

= 420

Hence, the sum of the following series is 420.

We use trigonometry to find the length.

Tan 30° = 3 cm/Tangent
Tangent = 3 √3 cm  as tan 30° = 1/√3

Suresh is having a garden near Delhi. In the garden, there are different types of trees and flower plants. One day due to heavy rain and storm one of the trees got broken as shown in the figure.
The height of the unbroken part is 15 m and the broken part of the tree has fallen at 20 m away from the base of the tree.
Using the Pythagoras answer the following questions:

1. What is the length of the broken part?
   1. 15 m
   2. 20 m
   3. 25 m
   4. 30 m
2. What was the height of the full tree?
   1. 40 m
   2. 50 m
   3. 35 m
   4. 30 m
3. In the formed right-angle triangle what is the length of the hypotenuse?
   1. 15 m
   2. 20 m
   3. 25 m
   4. 30m
4. What is the area of the formed right angle triangle?
   1. 100 m²
   2. 200 m²
   3. 60 m²
   4. 150 m²
5. What is the perimeter of the formed triangle?
   1. 60 m
   2. 50 m
   3. 45 m
   4. 100 m

Answer:

1)C

2)A

3)C

4)D

5)A

Given equation 
x2+px+q
Also, given p,q are the roots of the equation.
p+q=−p  
2p=−q    .....(1)
And   pq=q
q(p−1)=0
⇒q=0 or p=1
So, by (1), q=0⇒p=0  
Hence, the values of p are 0,1.

If d is the hcf of 48 and 72 so the value of d will be the hcf of later
factor of 48 is
2,3,4,6,8,12,24 and 48
factor of 72 is
2,3,4,6,8,9,12,24,36 and 72
so the hcf is 24 and d= 24
or
d = HCF ( 48, 72 )

48 = 2 * 2 * 2 * 2 * 3

72 = 2 * 2 * 2 *3 * 3

so HCF = 2 * 2 * 2 * 3

➡ 24

Answer :- d = HCF ( 48, 72 ) = 24

Let 2+√3 is a rational number.
A rational number can be written in the form of p/q.
2+√3=p/q
√3=p/q-2
√3=(p-2q)/q
p,q are integers then (p-2q)/q is a rational number.
But this contradicts the fact that √3 is an irrational number.
So,our supposition is false.
Therefore,2+√3 is an irrational number.
Hence proved.

(−5) + (−8)+ (−11) + ... + (−230) .

Common difference of the A.P. (d) = a₂ - a₁ 

=-8-(-5)

=-8+5

=-3

So here,

First term (a) = −5

Last term (l) = −230

Common difference (d) = −3

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

a_n =  a + (n-1) d

So, for the last term,

   - 230 = -5 + ( n-1) (-3) 

   - 230 = -5-3n + 3

-23 +2 = -3n

-228-3=n

           n = 76

Now, using the formula for the sum of n terms, we get

Sn=762[2(-5)+(76-1)(-3)]

      = 38 [-10+(75)(-3)]

      =38 (-10-225)

      = 38(-235)

      = -8930

Therefore, the sum of the A.P is  S_n = -8930 

 156=13×3×2²

Given A.P. 2, 7, 12, …..

First term a = 2

Common difference d = 7 – 2 = 5

nth term an = a + (n – 1)d

∴ a10 = 2 + (10 – 1) × 5

= 2 + 9 × 5

= 2 + 45

= 47

Hence, 10th term of given series 47.

The given AP is 1/15, 1/12, 1/10, .........

The given AP is 1/15, 1/12, 1/10, .........  

Answer:

The value of k must be less than or equal to 4.

Step-by-step explanation:

The given quadratic equation is

A quadratic equation  has real roots if

We have,

Divide both sides by 4.

Therefore the value of k must be less than or equal to 4.

the given AP

2,7,12,17.........

the first term(a) =2

common difference = 5

the nth term = a+(n-1)d

=2+(n-1)5

=2+5n-5

=5n-3

therefore 5n - 3 is the nth term of the given AP

Now  T₂₄₀  = 5 x (240) - 3

therefore 240th term is 1197

If the factors of denominator of the given rational number is of form 2ⁿ 5^m ,where n and m are non negative integers, then the decimal expansion of the rational number is terminating otherwise non terminating recurring.
935/10500
935/ (2²× 3× 5³× 7)
Here, the factors of the denominator 10500  are 2²× 3× 5³× 7, which is not in the form 2ⁿ 5^m .
Hence, 935/10500 has non terminating repeating decimal expansion.