Question

Question 15 A vertical pole of a length 6 m 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.

Class 10 - Math - Triangles Page 141

Answer 1

Length of the vertical pole = 6m (Given)

Shadow of the pole = 4 m (Given)

Let Height of tower = h m

Length of shadow of the tower = 28 m (Given)

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (By AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒ h = 6×28/4

⇒ h = 6 × 7

⇒ h = 42 m

Hence, the height of the tower is 42 m.

Answer 2

Let AB be the Apple of length 6 m.

The length of its shadow BC =4m which sun makes with ground & Ф be the angle.

Also at the same time another Tower casts Shadow NM of length 28 m.

Let PM = h m be the height of the tower and Ф be the angle the sun which sun ray makes with ground.

In ∆ABC & ∆PMN ,

∠ABC = ∠PMN (each 90°)
∠ACB = ∠PNM. (each Ф)
∴∆ABC~∆PMN. (by AA similarity)

Now,
AB/PM = BC/MN              (corresponding sides of two similar Δ's are proportional) 

= 6/h = 4/ 28

h = (28×6)/4
h= 42 m

Hence , the height of the tower is 42 m
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Hope this will help you.. .