A stone is dropped into a pond, the ripples forming concentric circles which expand. At what rate is the area of one of these circles increasing when the radius is m and inceasing at the rate of 0.5ms-1?
Answers
∴Lengthofrope=10m
\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}
Step−by−stepexplanation:
\begin{lgathered}\green{\underline \bold{Given :}} \\ \tt: \implies Height \: of \: pole(AB) = 8 \: m \\ \\ \tt: \implies Distance \: between \: leg \: of \: pole \: and \: end \:point \: of \: rope(BC) = 8 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Length \: of \: rope(AC) = ?\end{lgathered}
Given:
:⟹Heightofpole(AB)=8m
:⟹Distancebetweenlegofpoleandendpointofrope(BC)=8m
ToFind:
:⟹Lengthofrope(AC)=?
• According to given question :
\begin{lgathered}\bold{Using \: phythgoras \: theorem} \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2} \\ \\ \tt: \implies {AC}^{2} = {8}^{2} + {6}^{2} \\ \\ \tt: \implies {AC}^{2} = 64 + 36 \\ \\ \tt: \implies {AC}^{2} =100 \\ \\ \tt: \implies {AC}^{2} = {10}^{2} \\ \\ \tt: \implies AC = \sqrt{ {10}^{2} } \\ \\ \tt: \implies AC = \sqrt{10 \times 10} \\ \\ \tt: \implies AC= \sqrt{2 \times 5 \times 2 \times 5} \\ \\ \tt: \implies AC= \sqrt{2 \times 2 \times 5 \times 5} \\ \\ \tt: \implies AC = 2 \times 5 \\ \\ \green{\tt: \implies AC= 10 \: m} \\ \\ \green{\tt{ \therefore length \: of \: rope \: is \: \: 10 \: m}}\end{lgathered}
Usingphythgorastheorem
:⟹h
2
=p
2
+b
2
:⟹(AC)
2
=(AB)
2
+(BC)
2
:⟹AC
2
=8
2
+6
2
:⟹AC
2
=64+36
:⟹AC
2
=100
:⟹AC
2
=10
2
:⟹AC=
10
2
:⟹AC=
10×10
:⟹AC=
2×5×2×5
:⟹AC=
2×2×5×5
:⟹AC=2×5
:⟹AC=10m
∴lengthofropeis10m