KnowledgeBoat Logo
|

Mathematics

Rohan is trying to find the height of a tower shown below. He is using the properties of similar triangles. He observes a lamp post near it of height 5 m casting a shadow of 4 m on the ground. At the same time, he himself is casting a shadow of 1 m on the ground.

Based on the above information, answer the following questions :

(i) What is the height of the tower ?

(ii) What is Rohan's height ?

(iii) What will be the length of the shadow of the lamp post when the tower casts a shadow of 35 m ?

Rohan is trying to find the height of a tower shown below. He is using the properties of similar triangles. He observes a lamp post near it of height 5 m casting a shadow of 4 m on the ground. At the same time, he himself is casting a shadow of 1 m on the ground. Similarity (With Applications to Maps and Modules), Concise Mathematics Solutions ICSE Class 10.

Similarity

2 Likes

Answer

Rohan is trying to find the height of a tower shown below. He is using the properties of similar triangles. He observes a lamp post near it of height 5 m casting a shadow of 4 m on the ground. At the same time, he himself is casting a shadow of 1 m on the ground. Similarity (With Applications to Maps and Modules), Concise Mathematics Solutions ICSE Class 10.

Since the lamp post, tower, and Rohan are all vertical to the ground and sunlight makes the same angle, the triangles formed are similar.

(i) Thus,

△ABC ∼ △DEF

We know that,

In similar triangles, corresponding sides are proportional (in the same ratio).

ABBC=DEEF54=DE28DE=54×28=5×7=35 m.\Rightarrow \dfrac{AB}{BC} = \dfrac{DE}{EF} \\[1em] \Rightarrow \dfrac{5}{4} = \dfrac{DE}{28} \\[1em] \Rightarrow DE = \dfrac{5}{4} \times 28 = 5 \times 7 = 35 \text{ m}.

Hence, the height of the tower = 35 m.

(ii) Thus,

△ABC ∼ △PQR

In similar triangles, corresponding sides are proportional (in the same ratio).

ABBC=PQQR54=PQ1PQ=54×1=54=1.25 m.\Rightarrow \dfrac{AB}{BC} = \dfrac{PQ}{QR} \\[1em] \Rightarrow \dfrac{5}{4} = \dfrac{PQ}{1} \\[1em] \Rightarrow PQ = \dfrac{5}{4} \times 1 = \dfrac{5}{4} = 1.25 \text{ m}.

Hence, Rohan's height = 1.25 m.

(iii) When tower casts a shadow of 35 m,

Due to similarity,

Height of towerLength of shadow of tower=Height of lampLength of shadow of lamp3535=5Length of shadow of lamp1=5Length of shadow of lampLength of shadow of lamp=5 m.\Rightarrow \dfrac{\text{Height of tower}}{\text{Length of shadow of tower}} = \dfrac{\text{Height of lamp}}{\text{Length of shadow of lamp}} \\[1em] \Rightarrow \dfrac{35}{35} = \dfrac{5}{\text{Length of shadow of lamp}} \\[1em] \Rightarrow 1 = \dfrac{5}{\text{Length of shadow of lamp}} \\[1em] \Rightarrow \text{Length of shadow of lamp} = 5 \text{ m}.

Hence, length of shadow of lamp = 5 m.

Answered By

3 Likes


Related Questions