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Posted by Theerthapramod Theertha 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago

PS = 12 cm
As PQ = QR =RS
∴ PQ =QR =RS = 1/3 x PS = 1/3 x 12 = 4 cm.
QS = 2 PQ
QS = 2 x 4 = 8 cm
∴ Area of shaded region = Area of semicircle with PS as diameter + Area of semicircle with PQ as diameter – Area of semicircle with QS as diameter.
= ½ [ 3.14 x 6 2 + 3.14 x 2 2 - 3.14 x 4 2 ]
= ½ [ 3.14 x 36 + 3.14 x 4 – 3.14 x 16 ]
= ½ [ 3.14 ( 36 + 4 – 16)]
= ½ ( 3.14 x 24 ) = ½ x 75.36
∴ Area of shaded region = 37.68 cm2
Posted by Arpitha A 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
Let
A ( 5 , -2)
B (6 , 4)
C ( 7 , -2)
Length of each sides
AB = BC = √37
Hence verified that these vertices are of an isosceles triangles
Posted by Raju Sen 6 years, 1 month ago
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Posted by Urmila Panchal 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
To prove : 5-√3 is irrational
Proof:
Let us assume that 5 -√3 is rational.
Let ,
5 - √3 = r , where "r" is rational
5 - r = √3
Here,
LHS is purely rational.But,on the other hand ,RHS is irrational.
This leads to a contradiction.
Hence,5-√3 is irrational
Posted by Kishan Kumar Kumar 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
Let the fraction be x/y
Given that
x+2/y =1/2 -----> equ i
x/y-1 =1/3 -----> equ ii
From i and ii
2(x+2) =1(y) -------> equ i
3(x) =1(y-1) -------> equ ii
=> 2x+4=y
3x=y-1
Subtract i and ii. we get
2x - y = -4
3x - y = -1
(-) (+) = (+)
______________
-x = -3
_____________
=> x=3
subtitute x=3 in equ i. we get
2(3+2)=y
=> 2(5) =y
=> y=10
Therefore the value of x and y are 3 and 10 respectively.
Posted by Shehzad Ansari 6 years, 1 month ago
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Miss. 1234 6 years, 1 month ago
Posted by Jagat Limboo 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
Let the side of one square be x
Perimeter of this square = 4x
Given, Difference of perimeter of 2 squares = 64 m
Thus, Perimeter of the other square = (64 + 4x) m
And, each side of this second square = = (16 +x) m
According to the problem, sum of the areas of two squares is 640
Since side of a square cannot be negative, each side of the square = 8 m.
And, each side of second square = (16+8) m = 24 m
Posted by Ss H 6 years, 1 month ago
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Posted by Dilkash Khan 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
Step-by-step explanation:
Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.
As we know that
The mid-points of two diagonals must be same.
So, we will apply "Mid point rule":
Mid point of AC is given by
Mid point of BD is given by
Now, Mid point of AC = Mid point of BD
Hence, The value of x = 6 and y = 3.
Posted by Riya Shree Gupta 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
You can get the idea on how the questions will come in board exams:
Click on the below link to get the sample paper issued by CBSE :
Basic paper:
<a href="http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsBasic_SQP.pdf">http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsBasic_SQP.pdf</a>
Standard paper:
<a href="http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsStandard_SQP.pdf">http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsStandard_SQP.pdf</a>
Posted by Vishal Kumar 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
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.
Posted by ? ... 6 years, 1 month ago
- 4 answers
Gaurav Seth 6 years, 1 month ago
You can get the idea on how the questions will come in board exams:
Click on the below link to get the sample paper issued by CBSE :
Basic paper:
<a href="http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsBasic_SQP.pdf">http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsBasic_SQP.pdf</a>
Standard paper:
<a href="http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsStandard_SQP.pdf">http://cbseacademic.nic.in/web_material/SQP/ClassX_2019_20/MathematicsStandard_SQP.pdf</a>
Dhiraj Kumar 6 years, 1 month ago
Posted by Ann Mary Johny 6 years, 1 month ago
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Gaurav Seth 6 years, 1 month ago
Given:
The aeroplane when 3000 m high passes vertically above another aeroplane.
Angle of elevation of the aeroplane’s at the same observation = 60°and 45°
To find:
How much high is the one plane to other = ?
Solution:
The height at which the aeroplane (1) is from the ground is of height = 3000m.
The observer is standing at point A from where he observe both the planes one at D and another at C
The height at which the aeroplane (2) is from the ground is of height = h meters.
To find the height difference between the two planes, we first have to find the height of the second plane
The angle for the second plane is 45, which means
Now equating the value of the length between the planes and the observer
The height of the second plane from the ground is 1732m
Therefore, the Distance between the planes or the second plane is (3000 – 1732) = 1268 m lower than the first plane,
The height that the first plane is from second plane is 1268 m.
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Aadya Singh ? 6 years, 1 month ago
3Thank You