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Q15  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.

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CD = pole

AB = tower

Shadow of pole = DF

Shadow of tower = BE

In \triangle ABE\, \, and\, \triangle CDF,

\angle CDF=\angle ABE            ( Each 90 \degree)

\angle DCF=\angle BAE             (Angle of sun at same place )      

\triangle ABE\, \, \sim \, \triangle CDF,          (AA similarity)

 \frac{AB}{CD}=\frac{BE}{QL}

\Rightarrow \frac{AB}{6}=\frac{28}{4}

\Rightarrow AB=42  cm

Hence, the height of the tower is 42 cm.

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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