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.
Given: The length of a vertical pole = 6m
The shadow cast by the pole on the ground = 4m
The shadow cast by the tower = 28m
To Find: Height of the tower
Solution: Let AB and CD be a tower and a pole respectively.
Let the shadow of BE and DF be the shadow of AB and CD respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
⇒\(\angle\)DCF = \(\angle\)BAE
\(\angle\)DFC = \(\angle\)BEA
\(\angle\)CDF = \(\angle\)ABE (Tower and pole are vertical to the ground)
∴ ∆ABE ∼ ∆CDF (AAA similarity criterion)
⇒\(\frac{AB}{CD}=\frac{BE}{DF}\)
⇒\(\frac{AB}{6m}=\frac{28}{4}\)
⇒AB = 42m
Therefore, the height of the tower will be 42 meters.
Fill in the blanks using the correct word given in the brackets :
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)



| Case No. | Lens | Focal Length | Object Distance |
|---|---|---|---|
| 1 | \(A\) | 50 cm | 25 cm |
| 2 | B | 20 cm | 60 cm |
| 3 | C | 15 cm | 30 cm |