Question

A tree is 15 feet tall stand in front of 35 feet - high building.If the tree is 15 feet away from the building.Find the distance between their top

Answer 1

Answer:

answer for the given problem is given

Answer 2

i) Height of the Tree (AB) = 15 ft,

ii ) Height of the Building (CE) = 35 ft,

iii ) DE = CE - CD

= 35 ft - 15 ft

= 20 ft

iv ) Distance between base of the tree to base of the building (BC) = AD = 15 ft

Let distance between the tops = AE

$In \triangle ADE , \angle D = 90 \degree$

$AD = 15 \:ft, \: DE = 20 \:ft$

/* By Phythagorean theorem*/

$AE^{2} = AD^{2} + DE^{2}$

$= 15^{2} + 20^{2}$

$= 225 + 400$

$= 625$

$\implies AE = \sqrt{625}$

$= \sqrt{ 25^{2}}$

$= 25 \: ft$

Therefore.,

$\red{ Distance \: between \:the \: tops }$

$\green { = 25 \:ft }$

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