Ok I have a problem with this exercise , please again , provide with clear and correct solution and explanation about it !
I would really appricate it !! :)
thank you :)
Answers
Step-by-step explanation:
Question :-
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Given :-
height of the pole :- 6m
length of the shadow :- 4cm
length of shadow of tower :- 28m
To find :-
height of tower
Solution :-
in triangle ABC and triangle DEF
angle b = angle E = 90°
therefore triangle ABC ~ triangle DEF
triangle ABC ~ triangle DEF
we know that if two triangle are similar ration of their sides are in proportion
so,
\begin{gathered}:\implies\sf \: \frac{ab}{de} = \frac{bc}{ef} \\ \\ \\ :\implies\sf \: \frac{6}{de} = \frac{4}{28} \\ \\ \\ :\implies\sf \: 6 \times 28 = de \times 4 \\ \\ \\ :\implies\sf \: \frac{6 \times 28}{4} = de \\ \\ \\ \end{gathered}
:⟹
de
ab
=
ef
bc
:⟹
de
6
=
28
4
:⟹6×28=de×4
:⟹
4
6×28
=de
\begin{gathered}:\implies\sf \: 6 \times 7 = de \\ \\ \\:\implies\sf \: de = 42 \\ \\ \\ \end{gathered}
:⟹6×7=de
:⟹de=42
∴ the height of the tower is 42
Answer:
Ok I have a problem with this exercise , please again , provide with clear and correct solution and explanation about it !
I would really appricate it !! :)
thank you :)