Question

❥ Heya ❥

A vertical stick which is 15 cm long casts a 12 cm long shadow on the ground. At the same time , a vertical tower casts a 50 m long shadow on the ground. Find the height of the tower.

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Answer 1

Height = 62.5 m

Step-by-step explanation:

Given:

• Length of vertical tower is 15 cm.
• Length of shadow is 12 cm.
• Length of another shadow is 50 m.

To Find:

• What is the height of the tower ?

Solution: Let AB be the vertical stick and let AC be its shadow.

First change metre into cm.

• 100 cm = 1 metre
• 15 cm = 15/100 = 0.15 m
• 12 cm = 12/100 = 0.12 m

So AB = 0.15 m {tower}

AC = 0.12 m {tower}

Let DE be the vertical tower of x metre and DF be its shadow.

Now in ∆BAC and ∆EDF we have,

$\implies{\rm }$ ∠BAC = ∠EDF = 90°

$\implies{\rm }$ ∠ACB = ∠DFE { angular elevation of the sun at the same time}

∴ ∆BAC ~ ∆EDF

Since these two triangles are similar therefore their corresponding sides will be in same ratio.

$\implies{\rm }$ AB/DE = AC/DF

$\implies{\rm }$ 0.15/x = 0.12/50

$\implies{\rm }$ 0.15 × 50 = 0.12 × x

$\implies{\rm }$ 7.5 = 0.12x

$\implies{\rm }$ 7.5/0.12 = x

$\implies{\rm }$ 62.5 = x

Hence, the height of tower is 62.5 m.

Answer 2

✔ Given :-

• Length Of Vertical Stick = 15cm = 0.15m.

• Length Of Shadow = 12cm = 0.12m.

• Length Of Another Shadow = 50cm = 0.50m.

✔ To Find :-

• Height Of Tower.

✔ Solution :-

Let,

AB be the vertical stick.

AC be its shadow.

DE be the vertical tower.

DF be its shadow.

Now,

In ∆BAC and ∆EDF,

∠BAC = ∠EDF = 90°

∠ACB = ∠DFE

∴ ∆BAC is similar to ∆EDF.

We Can Say that,

➨ AB/DE = AC/DF

So, Put the values.

➨ 0.15/x = 0.12/50

➨ 0.15 × 50 = 0.12 × x

➨ 0.12x = 7.5

➨ x = 7.5/0.12

➨ x = 62.5

Hence, Height of the tower is 62.5 m.