Question

A vertical pole of length 6 m casts a shadow 4 m 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

To Find :-

• The height of the tower.

Solution :-

Given,

• Length of the vertical pole = 6 m

• Shadow of the pole = 4 m

• Length of the shadow of the tower = 28 m

Let the height of the tower be " h " m.

Now,

In ΔABC and ΔDFE,

↪∠C = ∠E (Angle of elevation)

↪∠B = ∠F (Both angle's are 90°)

↪ΔABC ~ ΔDFE (By AA similarity criteria)

We know that,

The corresponding sides of two similar triangles are proportional.

↪AB/DF = BC/EF

[ Put the values ]

↪6/h = 4/28

↪h = (6 ×28)/4

↪h = 6 × 7

↪h = 42 m

Therefore,

The height of the tower is 42 m.

Answer 2

Step-by-step explanation:

• we wil draw 2 triangle
• ∆ ABC and ∆ DEF
• Now ∆ ABC is a representative of pole of 6m perpendicular and 4m base
• And ∆ DEF is a representative of tower of base 28m and we have to find perpendicular
• Now we will proof that these 2 triangles are congruent
• so we will write,
• In ∆ ABC and ∆DEF
• Angle B = Angle E ( each 90°)
• Angle C = Angle F ( Shadows cast at same time )
• Hence, ∆ ABC ~ ∆ DEF

If 2 triangles are similar,

ratio of their sides are in proportion

So AB÷DE = BC÷EF

6÷DE = 4÷28

6×28 = DE × 4

( 6×28 ) ÷ 4 = DE

DE = 42