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