Question

a vertical stick 20 m long cast a shadow 10 cm long on the ground . at the same time the tower casts a shadow 50 m long on the ground what is the height of the tower is

Answer 1

20m=10cm i.e. 0.1m

20m=0.1m

X m=50m

X+0.1/20=50

X+0.1=50*20

X=1000-0.1

X=999.9

Answer 2

Given:

• Length of vertical stick = 15 cm

• Length of stick's shadow = 12 cm.

• Length of tower's shadow is 50 m.

To Find:

• Height of the tower = ?

SOLUTION:

Let AB be the vertical stick and let AC be its shadow. its shadow.

Then, AB = 15 cm = 0.15 m and AC = 12 cm = 0.12 m. (Given)12 cm = 0.12 m. (Given)

Let DE be the vertical tower and let DF be its shadow. its shadow.

Then, DF = 50 m (given) its shadow.

Then, DF = 50 m (given)

Let DE = x m.

$\rm Now \: in \: \triangle BAC \: and \: \triangle EDF, we \: get$

$\rm \longrightarrow\angle BAC = \angle EDF = 90 \degree$

$\rm \longrightarrow\angle ACB = \angle DFE$

[angular elevationof the sun at the same time]

$\rm \therefore\triangle BAC \sim \triangle EDF$

$\rm \implies \frac{AB}{DE}=\frac{AC}{DF}....(i)</p><p>$

where,

• AB = 0.15 m

• DE = x m

• AC = 0.12 m

• DF = 50 m

put these values in eq.(i)

$\rm \implies \frac{0.15}{x}=\frac{0.12}{50}$$\rm \implies x=\frac{(0.15 \times 50)}{0.12}$

$\rm \implies x=\frac{7.5}{0.12}$

$\rm \implies x=\frac{ \cancel{7.5}}{ \cancel{0.12}} = 62.5m$

$\rm Hence, the \: height \: of \: the \: tower \: is \underline{\boxed{ \rm62.5m}}$