Question

Two poles of heights 6m and 11m stand on a plane ground. If the distance between the feet of thepoles is 12m. find the distance b/w their top​

Answer 1

Given:

$\bigstar$Height of first pole (AB) = 6 m

$\bigstar$Height of second pole (CD) = 11 m

$\bigstar$Distance between their feet (AC) = 12 m

To Find:

$\bigstar$Distance between their tops (BD) = ?

Solution:

$\bigstar$Draw BE parallel to AC

$\bigstar$Then,

$\bigstar$CE = AB = 6 m

$\bigstar$BE = AC = 12 m

$\bigstar$DE = (CD -CE)

$\bigstar$DE = (11 - 6)

$\bigstar$DE = 5 m

$\bigstar$Now,

$\bigstar$Using Pythagoras Theorem,

$\bigstar$In triangle BED, we have

$\bigstar$BD² = BE² + DE²

$\bigstar$BD² = (12)² + (5)²

$\bigstar$BD² (144 + 25)

$\bigstar$BD² = 169

$\bigstar$BD = √169

$\bigstar$BD = 13 m

$\bigstar$Hence, their distance between their top is 13 m

$\bigstar$Hence Proved

Answer 2

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 parallel to AC

Then

CE = AB = 6 m

BE = AC = 12 m

DE = (CD -CE)

DE = (11 - 6)

DE = 5 m

Now,

★ Using Pythagoras Theorem,

★ In triangle BED, we have

★ BD² = BE² + DE²

★ BD² = (12)² + (5)²

★ BD² (144 + 25)

★ BD² = 169

★ BD = √169

★ BD = 13 m