Question

to pose 25 metre and 16 metre high stand upright in a playground if their feet are 15 metre apart find the distance between their tops​

Answer 1

Step-by-step explanation:

$\huge \sf \underline{\green{ \underline{given : }}}$

• Height of first pole AB = 25 metre
• Height of second pole CD = 16 metre
• Distance between feet pole AC = 15 metre

$\sf \huge \underline{\underline{ \green{ \: \: To \: find : } }}$

• Distance between the tops of the pole BD

$\sf \huge \underline{ \green{\underline{Solution : }}}$

Let we draw a line BE perpendicular to DC i.e BE ⊥ DC

Since, AC is also perpendicular to DC as pole is vertical to ground

So, BE = AC = 15

Similarly, AC = EC = 7.5

$\red{now}$

$\bf \: DE = DC - EC$

$\bf \: DE = 16 - 7.5$

$\bf \: DE = 8.5m$

$\sf \: Since \: \angle BED = 90° \: as \: BE ⊥ DC$

$\sf \: \triangle BED \: is \: right \: angled \: triangle$

Using Pythagoras theorem is right angle triangle AEB

$\bf \underline {\boxed{ ( Hypothenuse) ² = ( Height) ² + ( Base) ²}}$

$\bf \: (BD) ² = (DE) ² + ( BE) ²$

$\bf \: (BD) ² = {(8.5)}^{2} + {(25)}^{2}$

$\bf \: (BD) ² = 72.25 + 625$

$\bf \: (BD) ² = 697.25$

$\bf \: BD = \sqrt{697.25}$

$\bf \: BD = \sqrt{ {(26.41)}^{2} }$

$\bf \: BD = 26.41$

$\tt \blue{\boxed{ Hence \: the \: distance \: between \: tops \: pole \: = 26.41 \: metre}}$