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# The distance between the tops of two trees 20m and 28m high is 17m. The horizontal distance between the two tree is$(a)$ 11m$(b)$ 31m$(c)$ 15m$(d)$ 9m  Verified
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Hint - In this type of problem, they usually make right angled triangles and we can use the Pythagoras theorem to solve.

Given that
AB = 20m
AE = 17m
CE = 28m
$\because$ CE = AB + DE
$\therefore$ DE = 8m

With the given data, diagram can be drawn like this Now we apply the Pythagoras theorem in the ADE triangle.
Then
${\text{A}}{{\text{E}}^2} = {\text{A}}{{\text{D}}^2} + {\text{D}}{{\text{E}}^2}$
${17^2} = {{\text{x}}^2} + {8^2}$
${{\text{x}}^2} = 289 - 64$
${{\text{x}}^2} = 225$
${\text{x = }} \pm {\text{15}}$
$\because$ distance never be negative then
X = 15m
Therefore, the horizontal distance between the two trees is 15m.

Note - In this problem, first let assume the variable for unknown and then apply the Pythagoras theorem and get the value of unknown which is the horizontal distance between the two trees.