# 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

**Solution:**

The ratio of any two corresponding sides in two equiangular triangles is always the same.

AB is the pole = 6 m

BC is the shadow of pole = 4 m

PQ is the tower = ?

QR is the shadow of the tower = 28 m

In ΔABC and ΔPQR

∠ABC = ∠PQR = 90º (The objects and shadows are perpendicular to each other)

∠BAC = ∠QPR (Sunray fall on the pole and tower at the same angle, at the same time)

⇒ ΔABC (AA criterion)

The ratio of any two corresponding sides in two equiangular triangles is always the same.

⇒ AB/BC = PQ/QR

⇒ 6/4 = PQ / 28

⇒ PQ = (6 × 28 )/4

⇒ PQ = 42 m

Hence, the height of the tower is 42 m.

**Video Solution:**

## 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

### NCERT Class 10 Maths - Chapter 6 Exercise 6.3 Question 15:

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

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. Hence the height of the tower is 42 meters