Question

A display board is in the shape of a regular octagon. 8
If the cost of putting a border around its edges at the
rate of 3 per metre is 96, find the length of one side
of the board.
​

Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Perimeter of Octagon has been used. We know all the edges of a regular Octagon are equal. This means that its perimeter will be eight times the length of its each edge. We are given the total cost of bordering the edges. Now length to be bordered will be equal to the Perimeter of Octagon. We can apply values in formula and find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Length\;to\;be\;bordered_{(Perimeter)}\;=\;\bf{\dfrac{Total\;cost}{Rate_{(in\;per\;m)}}}}}$

$\\\;\boxed{\sf{8\;\times\;Length\;of\;Each\;Edge\;=\;\bf{Perimeter\;of\;Octagon}}}$

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★ Solution :-

Given,

» Total cost of bordering = Rs. 96

» Rate of bordering per m = Rs. 3

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~ For the Perimeter of Display Board ::

We know that

Length to be Bordered = Perimeter of Display Board

Then,

$\\\;\;\;\sf{:\Longrightarrow\;\;Length\;to\;be\;bordered_{(Perimeter)}\;=\;\bf{\dfrac{Total\;cost}{Rate_{(in\;per\;m)}}}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;Length\;to\;be\;bordered_{(Perimeter)}\;=\;\bf{\dfrac{96}{3}}}$

$\\\;\;\;\bf{:\Longrightarrow\;\;Length\;to\;be\;bordered_{(Perimeter)}\;=\;\bf{32\;\;m}}$

$\\\;\underline{\boxed{\tt{Perimeter\;\;of\;\;Display\;\;Board\;\;=\;\bf{32\;\;m}}}}$

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~ For the Length of Each Side of the Display Board ::

We know that, Perimeter of Octagon is gives as,

$\\\;\;\;\sf{:\mapsto\;\;8\;\times\;Length\;of\;Each\;Edge\;=\;\bf{Perimeter\;of\;Octagon_{(Display\;Board)}}}$

$\\\;\;\;\sf{:\mapsto\;\;8\;\times\;Length\;of\;Each\;Edge\;=\;\bf{32}}$

$\\\;\;\;\sf{:\mapsto\;\;8\;\times\;Length\;of\;Each\;Edge\;=\;\bf{32}}$

$\\\;\;\;\sf{:\mapsto\;\;Length\;of\;Each\;Edge\;=\;\bf{\dfrac{32}{8}}}$

$\\\;\;\;\bf{:\mapsto\;\;Length\;of\;Each\;Edge\;=\;\bf{4\;\;m}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;length\;of\;one\;side\;of\;display\;board\;is\;\boxed{\bf{4\;\;m}}}}}}$

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★ More to know :-

• Let's verify our answer ::

For verification, we need to simple apply the value we got into equation we formed. Then,

$\\\;\tt{\rightarrow\;\;Perimeter\;\;of\;\;Display\;\;Board\;=\;\bf{8\;\times\;Length\;\;of\;\;Each\;\;Side}}$

$\\\;\tt{\rightarrow\;\;32\;=\;\bf{8\;\times\;4}}$

$\\\;\tt{\rightarrow\;\;32\;=\;32}$

Clearly, LHS = RHS.

So our answer is correct. Hence, Verified.

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• Other Formulas ::

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Square\;=\;4\;\times\;Side}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Rectangle\;=\;2(Length\;+\;Breadth)}$

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

$\\\;\sf{\leadsto\;\;Area\;of\;Square\;=\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Rectangle\;=\;Length\;\times\;Breadth}$

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