Question

A vertical pole of a length 6cm cast a shadow of 4m long on ground and at the same time a tower cast a shadow 28m long. Find the hight of a tower

Answer 1

Answer:

Step-by-step explanation:

Given :-

Length of the vertical pole = 6 m

Shadow of the pole = 4 m

Length of shadow of the tower = 28 m

To Find :-

Height of the tower.

Solution :-

Let Height of tower = h m

According to the given Question,

In ΔABC and ΔDEF,

⇒  ∠C = ∠E (each Ф)

⇒  ∠B = ∠F = (each Ф)

∴ ΔABC ~ ΔDEF (By AA similarity criterion)

AB/DF = BC/EF (corresponding sides of two similar Δ's are proportional)

⇒ AB/DF = BC/EF

⇒ 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

$\huge{\underline{\boxed{\red{\mathsf{Answer:-}}}}}$

Length of pole = 6 m

Shadow of pole = 4 m

Length of shadow of tower = 28 m

_____________________

Let the length of tower be "x" m .

_________________________

So,

In Δ ABC and Δ XYZ

we have

∠C = ∠Y

∠B = ∠Z

$\mathsf{\implies}{{\triangle} ABC {\sim} {\triangle} XYZ}$

So,

Property of similar Triangles

We know that

$\large{\mathsf{{\frac{AB}{XZ}} \: = \: {\frac{BC}{YZ}}}}$

_____________[Put Values]

⇒ 6/h = 4/28

⇒ h = 6 × 28 / 4

⇒ h = 168/4

⇒ h = 42 m

$\huge{\mathsf{h \: = \: 42 \: m}}$

Height of Tower is 42 m