Question

two trees are 13m and 25m high.if the distance between their tops in 13m find the distance between their foot​

Answer 1

Answer:

two trees are 13m and 25m high.if the distance between their tops in 13m find the distance between their foot

Answer 2

Given:

• Two trees of measure: 25 m and 18 m (AD & CE)
• The distance between their tops is 13 m (AC)

To find:

• The distance between the foots of Tree A and Tree B (DE)

Solution:

According to the figure (attachment), it is clear that, DE = BC and AB, BC, AC forms a right-angled triangle.

Right-angled Triangle:

• AB = 12 cm (25-18)
• AC = 13 cm
• BC = ?

Using Pythagoras theorem, we can find the measure of BC.

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\: \\ \Large{\boxed{\sf{{a}^{2} + {b}^{2} = {c}^{2}}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

• a = Leg = AB
• b = Base = BC
• c = Hypotenuse = AC

Now substituting the measures:

$\sf = > {(AB)}^{2} + {b}^{2} = {(AC)}^{2}$

$\sf = > {12}^{2} + {b}^{2} = {13}^{2}$

$\sf = > {b}^{2} = \sqrt{ {13}^{2} - {12}^{2}} \\ \:\:\: = \sqrt{169 - 144 } \\ = \sqrt{25} = 5 \: m$

$\rightarrow{\bold{\blue{{b}^{2}=5\:m=BC=DE}}}$

Therefore, the distance between their foots is $\large{\underline{\sf{\red{5\: metres}}}}$.

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