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Ratan Raj 5 years, 3 months ago
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Shreejay Kabra 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
taking x instead of tita
tan x = 8/15
tan x = PQ/QR
PQ/QR =8/15
16/QR = 8/15
16 x 15/8=QR
therefore ,QR = 30 m
PQ2 + QR2 =PR2
162 + 302 = PR2
256 + 900= PR2
1156=PR2
PR= root 1156
PR = 34 m
Posted by Tanushree Das 5 years, 3 months ago
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Posted by Palak Goyal 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
Correct option is C
AC = AB
Step-by-step explanation:
Given that AB is chord of bigger circle
Also, it is tangent to a smaller circle touching at C
Then, AB is perpendicular to OC and C is midpoint of AB
There AC = BC
Here, from the options
a) AB = r1
b) AB = r2
c) AC = BC
d) AB = r1 + r2
Correct option is C
Hence, AC = AB
Posted by Renubala Sahu 5 years, 3 months ago
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Meenu Batra 5 years, 3 months ago
Posted by Ipsita Kabi 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.
Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.
The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.
Deleted syllabus of CBSE Class 10 Mathematics
D
Posted by Vanshika Jaiswal 5 years, 3 months ago
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Posted by Atul Kumar 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
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
Posted by Atul Kumar 5 years, 3 months ago
- 1 answers
Yogita Ingle 5 years, 3 months ago
Given: ABCD is a rhombus
To prove: AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Proof:
We know, diagonals of a rhombus bisect at right angles.
Therefore, from triangle AOB,
AB2 = AO2 + OB2
4AB2 = AC2 + BD2
Thus, AB2 + BC2 + CD2 + DA2 = 4AB2 = AC2 + BD2
[As AB = BC = CD = DA]
Posted by Gaurav Pandey 5 years, 3 months ago
- 2 answers
Gaurav Seth 5 years, 3 months ago
Let height of the pedestal BD be h metres, and angle of elevation of C and D at a point A on the ground be 60° and 45° respectively.
It is also given that the height of the statue CD be 1.6 m
i.e., ∠CAB = 60°,
∠DAB = 45° and CD = 1.6m
In right triangle ABD, we have
In right triangle ABC, we have
Comparing (i) and (ii), we get
Hence, the height of pedestal 
Posted by Gaurav Pandey 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
Let AB and BC be the tower and water tank and D be the point of observation.
Given:
BC = 20m
∠BDC = 45° and ∠ADC = 60°
Let AB = h m
In ΔBDC
tan 45° = P/B = BC/DC
1 = 20/DC [ tan 45°=1]
DC= 20 m
Now in ΔADC
tan 60°= P/B = AC /DC = (AB + BC)/DC
√3 = (h+20)/20
20√3 = h+20
20√3 - 20 = h
20(√3-1)= h
h= 20(√3-1) m
h = 20 (1.73 -1)
h = 20 × .73 = 14.6 m
Hence, the height of the tower is 20(√3-1) or 14.6 m.
Posted by Ånanya Priya Gupta 5 years, 3 months ago
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Gaurav Pandey 5 years, 3 months ago
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Gaurav Pandey 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
the given cubic polynomial is:
product of its zeroes =
let the third zero be a.
thus the third zero is 3/2.
Posted by Sangeeta Solanki 5 years, 3 months ago
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Posted by Sonpal Singh Rathore 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
NCERT Solutions for Class 10 Maths Exercise 1.1
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Ans. We have to find the HCF (616, 32) to find the maximum number of columns in which they can march.
To find the HCF, we can use Euclid’s algorithm.
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
<hr />NCERT Solutions for Class 10 Maths Exercise 1.1
4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Ans. Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
Where 
are some positive integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
<hr />NCERT Solutions for Class 10 Maths Exercise 1.1
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Ans. Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
\ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms.
We have three cases.
Case 1: When a = 3q,
Where m is an integer such that m =3q3
Case 2: When a = 3q + 1,
Where m is an integer such that 
Case 3: When a = 3q + 2,
Where m is an integer such that 
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Posted by Harsh Raithatha 5 years, 3 months ago
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Posted by Khushi Gangwani 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.
The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.
CBSE stated in its notification that the deleted portions of the CBSE syllabus should not be considered for internal assessments or for the year-end board exams.
Posted by Khushi Gangwani 5 years, 3 months ago
- 3 answers
Bhargav Navin 5 years, 3 months ago
Bhargav Navin 5 years, 3 months ago
Posted by Riya ? 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
"If alpha and beta are the zeroes of the polynomial"
Taking this into account
1/alpha - 1/beta = 1/0 - 1/0
= infinity - infinity
which is not defined
if incase
"If alpha and beta are the solutions of the polynomial"
x^2 + x - 2 =0
(x+2)(x -1) = 0
=> x= -2 or x =1
therefore (alpha,beta) = (1,-2) or (-2,1)
1/alpha - 1/beta = +(3/2) or -(3/2)
OR
Posted by The_Mr. Ankittt 5 years, 3 months ago
- 3 answers
Harsh Raithatha 5 years, 3 months ago
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Adarsh Awasthi 5 years, 3 months ago
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