Question

answer me with solution please​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Area of Circle has been used We see that we are given a circular park around which a uniform road is there. Using the radius, we can find the area of park. Then we can add the radius of the park with breadth of road to get the radius of whole circular plot combining the park and road. Then we can subtract the area of circular plot with the area of circular park to get area of road.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Area\;of\;Circle\;=\;\bf{\pi r^{2}}}}}$

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★ Solution :-

Given,

» Radius of circular park = r = 49 m

» Breadth of road = 7 m

» Circular plot = Road + Park

» Radius of circular plot = R = 49 + 7 = 56 m

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~ For the Area of Park ::

We know that,

$\\\;\sf{\rightarrow\;\;Area\;of\;Circle\;=\;\bf{\pi r^{2}}}$

By applying values we get

$\\\;\sf{\Longrightarrow\;\;Area\;of\;Park\;=\;\bf{\dfrac{22}{7}\:\times\:(49)^{2}}}$

$\\\;\sf{\Longrightarrow\;\;Area\;of\;Park\;=\;\bf{\dfrac{22}{7}\:\times\:2401}}$

$\\\;\sf{\Longrightarrow\;\;Area\;of\;Park\;=\;\bf{22\:\times\:343}}$

$\\\;\bf{\Longrightarrow\;\;Area\;of\;Park\;=\;\bf{\green{7546\;\;cm^{2}}}}$

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~ For the Area of Circular Plot ::

We know that,

$\\\;\sf{\rightarrow\;\;Area\;of\;Circle\;=\;\bf{\pi r^{2}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;Area\;of\;Circular\;Plot\;=\;\bf{\dfrac{22}{7}\:\times\:(56)^{2}}}$

$\\\;\sf{\Longrightarrow\;\;Area\;of\;Circular\;Plot\;=\;\bf{\dfrac{22}{7}\:\times\:3136}}$

$\\\;\sf{\Longrightarrow\;\;Area\;of\;Circular\;Plot\;=\;\bf{22\:\times\:448}}$

$\\\;\bf{\Longrightarrow\;\;Area\;of\;Circular\;Plot\;=\;\bf{\blue{9856\;\;cm^{2}}}}$

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~ For the Area of Road ::

We know that,

$\\\;\sf{\mapsto\;\;\red{Area\;of\;Road\;=\;\bf{Ar.\;of\;Circular\;Plot\;-\;Ar.\;of\;Park}}}$

Now by applying values, we get

$\\\;\sf{\mapsto\;\;Area\;of\;Road\;=\;\bf{9856\;-\;7546}}$

$\\\;\sf{\mapsto\;\;\orange{Area\;of\;Road\;=\;\bf{2310\;\;cm^{2}}}}$

$\\\;\underline{\boxed{\tt{Area\;\:of\;\:Road\;=\;\bf{\purple{2310\;\;cm^{2}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Circle\;=\;2\pi r}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Semi\:-\;Circle\;=\;\pi r}$

$\\\;\sf{\leadsto\;\;Area\;of\;Semi\:-\;Circle\;=\;\dfrac{\pi r^{2}}{2}}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Quadrant\;=\;\dfrac{\pi r}{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Quadrant\;=\;\dfrac{\pi r^{2}}{4}}$

Answer 2

$\huge{AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf Given= \begin{cases} &\sf{radius \: of \: circular \: park \: = 49 \: m} \\ &\sf{width \: of \: road \: around \: park = 7 \: m} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{area \: of \: road} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf Formula \: used - : \begin{cases} &\sf{Area \: of \: circle = \pi \: {r}^{2} } \\ \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \: \underline{ ❥︎ Step :- 1}$

❥︎ Radius of park, r = 49 m

$\sf \:  ⟼Area_{(park)} = \pi \: {r}^{2}$

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$\bf \: \underline{ ❥︎ Step :- 2}$

❥︎ Width of road, h = 7 m

❥︎ So, Radius of circular plot, R = r + h = 49 + 7 = 56 m

$\sf \:  ⟼Area_{(plot)} = \pi \: {R}^{2}$

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$\bf \: \underline{ ❥︎ Step :- 3}$

❥︎ To find the area of road.

$\bf \:Area_{(road)} = Area_{(plot)} - Area_{(park)}$

$\sf \:  ⟼Area_{(road)} = \pi \: {R}^{2} - \pi \: {r}^{2}$

$\sf \:  ⟼Area_{(road)} = \pi \: ({R}^{2} - {r}^{2} )$

$\sf \:  ⟼Area_{(road)} = \pi \:(R + r)(R - r)$

$\sf \:  ⟼Area_{(road)} = \dfrac{22}{7} \times (56 + 49)(56 - 49)$

$\sf \:  ⟼Area_{(road)} = \dfrac{22}{ \cancel{7}} \times (105)( \cancel{7})$

$\sf \:  ⟼Area_{(road)} = 2310 \: {m}^{2}$

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