Question

a vertical pole of length 6m 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.​

Answer 1

Answer:

The height of the tower is 42 m.

Step-by-step explanation:

Height of pole=AB=6 m  

Length of shadow of pole =BC=4 m  

Length of shadow of tower=EF=28 m

 

In △ABC and △DEF  

∠B=∠E=90°  both 90°  as both are vertical to ground  

∠C=∠F  (same elevation in both the cases as both shadows are cast at the same time)

∴ △ABC ≅ △DEF by AA similarity criterion

We know that if two triangles are similar, ratio of their sides are in proportion  

So, ${\mathsf{\dfrac{AB}{DE} = \dfrac{BC}{EF}}}$

$\Longrightarrow {\mathsf{\dfrac{6}{DE} = \dfrac{4}{28}}}$

DE = 6 $\times$ 7 = 42 cm

Hence the height of the tower is 42 m.

Answer 2

Answer:

The length of pole : Length of its casted shadow =

The length of tower : Length of its shadow

=> 6:4 = Length of tower :28

=> Length of tower = 6/4*28 m = 42 m. (ans)