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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.

Solution :

Given:

1. A vertical pole $AB$ of length $6\text{ m}$ casts a shadow $BC$ of length $4\text{ m}$ on the ground.

2. A tower $DE$ of unknown height $h$ casts a shadow $EF$ of length $28\text{ m}$ on the ground at the same time.

To Find:

The height of the tower ($DE = h$).

6m 4m A B C h 28m D E F

Step 1: Establishing Geometric Similarity

Let $\triangle ABC$ be the triangle formed by the pole and its shadow, and $\triangle DEF$ be the triangle formed by the tower and its shadow.

In $\triangle ABC$ and $\triangle DEF$:

1. $\angle B = \angle E = 90^\circ$ (Since both the pole and the tower are vertical to the ground).

2. $\angle C = \angle F$ (Since the sun's rays fall at the same angle at the same time for both objects).

Therefore, by the AA (Angle-Angle) Similarity Criterion, $\triangle ABC \sim \triangle DEF$.

Step 2: Applying the Property of Similar Triangles

Since the triangles are similar, the ratios of their corresponding sides are equal:

$\frac{AB}{DE} = \frac{BC}{EF}$

Step 3: Substituting Known Values

Given $AB = 6\text{ m}$, $BC = 4\text{ m}$, and $EF = 28\text{ m}$. Let $DE = h$.

$\frac{6}{h} = \frac{4}{28}$

Step 4: Solving for $h$

Simplify the fraction on the right side:

$\frac{4}{28} = \frac{1}{7}$

Now, the equation is:

$\frac{6}{h} = \frac{1}{7}$

By cross-multiplication:

$h \times 1 = 6 \times 7$

$h = 42$

Final Answer: The height of the tower is 42 m.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


CBSE Solutions for Class 10 Mathematics Triangles


Chapters in CBSE - Class 10 Mathematics


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