Question

A vertical pole BC of length 6m casts a shadow CA 4m long, on the same time a tower DE casts a shadow EA 28m long.
Find arABCE/ar(ADE)​

Answer 1

Solution :-

Height of pole = BC = 6m

Length of shadow of pole = AC = 4m

Length of shadow of tower = EA = 28m

In △ABC and △ADE

∠C = ∠E = 90°  as both are vertical to ground

∠A = ∠A (Same elevation in both the cases as both shadows are cast at the

same time)

∴ △ABC ∼ △ADE by AA similarity criterion

We know that if two triangles are similar, ratio of their sides are in

proportion.

$So,\ \frac{BC}{DE}=\frac{AC}{AE}$

$=>\frac{6}{DE}=\frac{4}{28}$

$=>DE=42$

$So, \ area(ABC)=\frac{1}{2}.(AC).(BC)=\frac{1}{2}.4.6=12$

$And,\ area(ADE)=\frac{1}{2}.(AE).(DE)=\frac{1}{2}.28.42=588$

So, required ratio = 12 : 588 = 49

For diagram, refer given picture in the attachment.

Remarks :-

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Answer 2

Answer:

hey here is your solution in above pics

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