Question

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

Answer 1

Given:-

• Height of first pole (AB) = 6 m

• Height of second pole (CD) = 11 m

• Distance between their feet (AC) = 12 m

To Find:-

• Distance between their tops (BD) =?

Solution:-

Draw BE ∥ AC then,

• CE = AB = 6 m

• BE = AC = 12 m

∴ DE = (CD -CE)

• DE = (11 - 6)

• DE = 5 m

Now,

$\rm\boxed\star\purple{\underline{\underline{{By\:using\:Pythagoras\:theorem:-}}}}$

In right ∆BED,we have

$\rm\:B{D}^{2}=B{E}^{2}+D{E}^{2}$

$\rm\implies\:B{D}^{2}={(12)}^{2}+{(5)}^{2}$

$\rm\implies\:B{D}^{2}=(144+25)$

$\rm\implies\:B{D}^{2}=169$

$\rm\implies\:BD=\sqrt{169}\:$

$\rm\implies\:BD=13\:m$

Hence,the distance between their tops will be 13 m.

Answer 2

Answer:

Given: Two poles AB=11 m and CD=6m

Distance between both poles BD=12 m

Then, create a line segment CE parallel to line BD

So, AE=AB−CD=11−5=6 m

In right angled △AEC,

(AC)2=(AE)2+(AE)2=(12)2+(5)2=144+25=169

⇒AC=13 m

So, the distance between their tops  is 13 m.