Question

an electric pole cast a shadow of length 20 m at a time when a tree 6 m high casts a shadow of length 8 m.Find the height of the pole.

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Answer 1

Given:-

An electric pole casts a shadow of 20 m.

Height of tree = 6 m.

Tree casts a shadow of length = 8 m.

Find:-

Height of the electric pole.

Solution:-

Let height of electric pole be "h" m.

$\underbrace{\sf{Shadow\:=\:20\:m\:and\:height\:=\:h\:m}}_{\sf{for\:electric \:pole}}$

$\underbrace{\sf{Shadow \:=\:8\:m\:and\:height\:=\:6\:m}}_{\sf{for\:tree}}$

Now, for tree and electric pole..

=> Height : Shadow

=> $\sf{6\: :\: 8}$

=> $\sf{h\: :\: 20}$

Cross-multiply them

=> $\sf{6 \:\times\: 20 \:=\: h \:\times\: 8}$

=> $\sf{120\: = \:8h}$

=> $\sf{8h\: =\: 120}$

=> $\sf{h \:=\:\frac{ 120}{8}}$

=> $\sf{h \:= \:15}$

•°• Height of electric pole is 15 m.

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Height of the pole = 15 m

$\bold{\underline{\underline{Step\:-\:by\:-step\:explanation:}}}$

Given :

• An electric pole cast a shadow of length 20 m (AC)
• A tree 6 m high (ED)
• Shadow casted by tree is 8 m (EC)

To find :

• Height of the pole (AB)

Solution :

In Δ ABC and Δ EDC,

$\bold{\frac{AB}{ED}}$ = $\bold{\frac{\:\:\:\:BC}{\:\:\:\:DC}}$ = $\bold{\frac{\:\:\:\:AC}{\:\:\:\:EC}}$$\bold{\underbrace{\sf{Corrseponding\:sides\:of\:triangle}}}$

In the above pairs of corresponding sides, we have the values of $\bold{\frac{AB}{ED}}$ and $\bold{\frac{AC}{EC}}$, which implies that we are in no need for the second pair, $\bold{\frac{BC}{DC}}$

•°• Substituting the values of AB, ED, AC and EC ,

$\bold{\frac{AB}{ED}}$ = $\bold{\frac{AC}{EC}}$

$\bold{\frac{x}{6}}$ = $\bold{\frac{20}{8}}$

Cross multiplying,

$\bold{x\times\:8\:=\:20\times\:6}$

$\bold{8x\:=\:120}$

$\bold{x=\frac{120}{8}}$

$\bold{x\:=\frac{30}{2}}$

$\bold{x=15}$

•°• Height of the pole = 15 m