Math, asked by sandhyababy194, 2 months ago

If the volume of a right circular cone of height 9cm is 48<br />\pi<br />cm3. find the diameter of the base <br /><br />​

Answers

Answered by harshavardhan3901086
0

Answer:

Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.Let Y be a subspace of a metric space (X, d). Then ⊆ is open in Y if and only if =∩ for

some open subset G of X.

Step-by-step explanation:

Answered by 12avanirajput
0

Step-by-step explanation:

Given:−

Radius of Circle = 25 cm

Distance of its Chord (AB) From Centre (P) = 4cm

-----------------------

\large \bf \clubs \: To \: Find :-♣ ToFind:−

Length of Chord

-----------------------

\large \bf \clubs \: Solution :-♣ Solution:−

Let ,

Centre of the circle b = P

AB is chord

N is the centre of the chord .

\bf \large In \: \triangle \: PNB :In△PNB:

Applying Pythagoras Theorem :

\bf (PB)^2 = (PN)^2+(NB)^2(PB)

2

=(PN)

2

+(NB)

2

:\longmapsto {25}^{2} = {4}^{2} + PB^2 = \text{(NB})^2:⟼25

2

=4

2

+PB

2

=(NB)

2

:\longmapsto \text{(NB} {)}^{2} = {25}^{2} - {4}^{2}:⟼(NB)

2

=25

2

−4

2

:\longmapsto \text{(NB)}^2 = 625 - 16:⟼(NB)

2

=625−16

\text{(NB)}^2 = 609(NB)

2

=609

:\longmapsto \text{NB} = \sqrt{609}:⟼NB=

609

\purple{ \large :\longmapsto \underline {\boxed{{\bf NB = 24.67 \: cm} }}}:⟼

NB=24.67cm

As N is the midpoint of the chord AB

So ,

\begin{gathered} \text{Length of Chord AB = 2NB} \\ \\ = 2 \times 24.67\end{gathered}

Length of Chord AB = 2NB

=2×24.67

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