Question

4. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance be
their feet is 12 m, find the distance between their tops.​

Answer 1

Answer :-

$\:\\\large{\underbrace{\underline{\sf{What\;the\;Question\;Says\;:-}}}}$

Here the concept of Pythagoras Theorem has been used. According to this, the Hypotenuse square is equal to the sum of the squares of Base and Perpendicular.

We see that we need to find the distance between the tops of the poles. Now the distance between the top of the shorter pole and the point equal to that of top of shorter pole will be equal to the distance between their feet. This is because it forms are rectangle and opposite sides of rectangle are equal.

Now we can apply there Pythagoras Theorem.

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★ Formula used :-

$\:\\\large{\boxed{\sf{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}}$

$\:\\\large{\boxed{\sf{Height\;=\;\bf{Height\;of\;Larger\;Pole\;-\;6}}}}$

$\:\\\large{\boxed{\sf{Base\;=\;\bf{Distance\;between\;their\;feet}}}}$

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★ Questions :-

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance be their feet is 12 m, find the distance between their tops.

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★ Solution :-

Given, (also using the figure)

» Height of the shorter pole = 6 m = ED

» Height of the longer pole = 11 m = AC

» Distance between their feets = 12 m = DC

» Angle ABC = 90°

From figure we see, ∆ ABE forms a Right angled triangle since perpendicular is assumed to be forming 90° with base. So ∆ABE is a right angled triangle and Pythagoras Theorem can be applied there.

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~ For the Height of the ∆ABE :-

Clearly, AB is the height of this triangle.

$\:\\\;\;\;\;\large{\sf{:\rightarrow\;\;\:Height_{(AB)}\;=\;\bf{Height\;of\;Larger\;Pole\;-\;6}}}$

$\:\\\;\;\;\;\large{\sf{:\rightarrow\;\;\:Height_{(AB)}\;=\;\bf{AC\;-\;BC}}}$

$\:\\\;\;\;\;\large{\sf{:\rightarrow\;\;\:Height_{(AB)}\;=\;\bf{11\;-\;6}\;\:=\;\:\underline{\underline{5\;\:cm}}}}$

$\:\\\large{\boxed{\boxed{\tt{Height\;\;of\;\;\Delta\;ABE\;=\;\bf{5\;\;cm\;\;(AB)}}}}}$

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~ For the Base of the ∆ABE :-

Clearly, BE is the base of this Triangle

$\:\\\;\;\;\;\large{\sf{:\rightarrow\;\;\:Base_{(BE)}\;=\;\bf{Distance\;between\;their\;feet}}}$

$\:\\\;\;\;\;\large{\sf{:\rightarrow\;\;\:Base_{(BE)}\;=\;\bf{12\;\;cm}}}$

$\:\\\large{\boxed{\boxed{\tt{Base\;\;of\;\;\Delta\;ABE\;=\;\bf{12\;\;cm\;\;(BE)}}}}}$

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~ For the Hypotenuse of the ∆ABE :-

Clearly, distance between the tops of the poles is equal to the Hypotenuse of this Triangle.

Thus, Hypotenuse of this Triangle is AE.

$\:\\\quad\large{\sf{:\Longrightarrow\;\;\:(Hypotenuse)^{2} _{\tiny{(AE)}}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}$

$\:\\\quad\large{\sf{:\Longrightarrow\;\;\:(Hypotenuse)^{2} _{\tiny{(AE)}}\;=\;\bf{(BE)^{2}\;+\;(AB)^{2}}}}$

$\:\\\quad\large{\sf{:\Longrightarrow\;\;\:(Hypotenuse)^{2} _{\tiny{(AE)}}\;=\;\bf{(12)^{2}\;+\;(5)^{2}}}}$

$\:\\\quad\large{\sf{:\Longrightarrow\;\;\:(Hypotenuse)^{2} _{\tiny{(AE)}}\;=\;\bf{144\;+\;25}}}$

$\:\\\quad\large{\sf{:\Longrightarrow\;\;\:(Hypotenuse)^{2} _{\tiny{(AE)}}\;=\;\bf{169}}}$

$\:\\\quad\large{\sf{:\Longrightarrow\;\;\:(Hypotenuse)_{(AE)}\;=\;\bf{\sqrt{169}\;\:=\;\:\underline{\underline{13\;\;cm}}}}}$

From figure we know, that Hypotenuse is the Distance between the tops of the poles.

$\;\\\large{\underline{\underline{\rm{Thus,\;distance\;between\;the\;tops\;of\;the\;poles\;is\;\;\boxed{\bf{13\;\;cm}}}}}}$

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★ More to know :-

$\;\\\sf{\leadsto\;\;\;Area\;of\;\Delta\;=\;\dfrac{1}{2}\:\times\:Base\:\times\:Height}$

$\;\\\sf{\leadsto\;\;\;Area\;of\;Scalene\;Triangle\;=\;\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}$

$\;\\\sf{\leadsto\;\;\;Semiperimeter\;of\;\Delta,\;s\;=\;\dfrac{Perimeter\:of\:\Delta,\;a\:+\:b\:c}{2}}$

$\;\\\sf{\leadsto\;\;\;Perimeter\;of\;Equilateral\;Triangle\;=\;3\:\times\:Side}$

$\;\\\sf{\leadsto\;\;\;Perimeter\;of\;Isoceles\;Triangle\;=\;(2\:\times\:Equal\:Sides)\;+\;Base}$

$\;\\\sf{\leadsto\;\;\;Perimeter\;of\;Scalene\;Triangle\;=\;Sum\;of\;all\;Sides}$