Question

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

I will mark him brainliest who will solve this​

Answer 1

Answer:

$\huge \bf \colorbox {blue}{solution}$

$\bf \: formula \: applied$

$\huge \rm \: {h}^{2} = {b}^{2} + {p}^{2}$

Here,

H = Hypnoteuse

B = Base

P = Height

Firstly find the height of Pole A and B

Let the A and B be y

y = 11-6

y = 5 m

Now,

DC = EB = 12 CM

Now by using Pythagoras theorem find the Hypnoteuse.

$\sf \implies \: {h}^{2} = {b}^{2} + {p}^{2}$

$\sf \implies \: {h}^{2} = {12}^{2} + {5}^{2}$

$\sf \implies \: {h}^{2} = 144 + 25$

$\sf \implies \: {h}^{2} = 169$

$\sf \implies \: h \: = \sqrt{169}$

$\sf \implies \: h \: = 13m$

$\sf \: therefore \: distance \: = 13 \: m$

Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• length if the ladders is 6m and 11m respectively
• Distance btw. their bases is 12m

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• Distance btw. their tops = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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From the figure we can observe that ;

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AB = 11 m

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CD = 6m

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BD = 12m

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CD = BE = 6m --- ( i )

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AB = BE + EA

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11m = 6m + AE ( from eq ( i ))

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AE = 11m - 6m

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AE = 5m

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BD = EC = 12 m

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Triangle AEC is an right angle triangle

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By applying Pythagoras theoram :-

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$\sf{\red{\boxed{\bold{( hypotenuse ) ^2 = (base)^2 + ( height)^2 }}}}$

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(AC)² = ( AE ) ² + ( EC )²

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(AC)² = 5² + 12²

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(AC)² = 25 + 144

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(AC)² = 169

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AC = √169

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AC = 13

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______________________

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Distance btw the tops of the ladders is 13m