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Yogita Ingle 4 years, 10 months ago
Volume of sphere = 4/3 π R*(3)
Here R = 2r
Volume of sphere = 4/3 π (2r)*3
Volume =( 4/3 )8π r*(3)
Volume of sphere = 32/3 π r*3
Aditya Kumar 4 years, 10 months ago
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Vikas Sharma 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
m²-7m-30
= m2 - 10m + 3m - 30
= m(m - 10) + 3 (m - 10)
= (m -10) (m + 3)
Posted by Jyoti Bondu 4 years, 10 months ago
- 2 answers
Gaurav Seth 4 years, 10 months ago
If a, b, c are sides of triangle then
s=(a+b+c)/2
Given:
s-a=4 cm
s-b=8 cm
s-c=12 cm
Adding these we get
s+s+s-a-b-c = 24
3s-(a+b+c) =24
3s-2s = 24
s = 24
Now,
s-a=4
24 - a =4
a = 24 - 4 = 20 cm
s-b=8
24 - b = 8
b = 24 - 8 = 16 cm
s - c = 12
24 - c = 12
c = 24 -12 = 12 cm
Posted by Mohan Ray 4 years, 10 months ago
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Gaurav Seth 4 years, 10 months ago
Given the sum of pair of opposite angles of a quadrilateral is 180° we have to prove that the quadrilateral is cyclic.
Let us assume that the quadrilateral ABCD is not cyclic i.e Let the point D does not lie on the circle which makes the quadrilateral non-cyclic. Now, let us do a construction such that join CD' where D' is the point of intersection of side AD with the circle.
Now, ABCD' is cyclic
⇒ ∠3 + ∠4 = 180°
Now, it is given that the sum of pair opposite angles of a quadrilateral ABCD is 180°
Therefore, ∠2 + ∠4 = 180°
From above two equations we get
∠3 + ∠4 = ∠2 + ∠4
⇒ ∠3 = ∠2
Now, in triangle CDD', by external angle property
∠3 = ∠1 + ∠2
⇒ ∠1 = 0 , hence the side CD' and CD coincides
⇒ Point D lies on circle
Hence, our supposition is wrong quadrilateral ABCD is cyclic.
Posted by Priyanshu Verma 4 years, 10 months ago
- 1 answers
Gaurav Seth 4 years, 10 months ago
Given -
V = 179 2/3 cu.cm = 179.67 cu.cm
Now,
Volume of sphere is given by formula -
V = 4/3 πr^3
179.67 = 4/3 × 3.142 × r^3
r^3 = 179.67 / 4.189
r^3 = 42.89
r = 3.5 cm
Now, surcace area of sphere is given by -
A = 4πr^2
A = 4 × 3.142 × 3.5^2
A = 154 sq.cm
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Arpan Kaur Dhillon 4 years, 10 months ago
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