Question

In figure \[DE\parallel BC,AD = 1cm\] and \[BD = 2cm\]. What is the ratio of \[ar(\vartriangle ABC):ar(\vartriangle ADE)\]?

Answer 1

Hint:Here we use the concept of ratio to find the length of the required side and then since lines are parallel and are cut by two transversals, we use the property that states corresponding angles are equal when parallel lines are cut by transversal. Making both the triangles similar and we can find the ratio of area of two triangles.
* If two similar triangles have a ratio of their sides as \[a:b\]then the ratio of areas of two similar triangles will be \[{a^2}:{b^2}\].

Complete step-by-step answer:
We are given \[AD = 1cm,BD = 2cm\]
Therefore, the length \[AB = AD + DB = (1 + 2)cm = 3cm\]
Now ratio of length of sides \[AD:AB = 1:3\]
Looking at the parallel lines \[DE\parallel BC\] cut by a transversal \[AB\], then corresponding angles made by transversal are equal .
Therefore, \[\angle ADE = \angle ABC\]
Also, at the parallel lines \[DE\parallel BC\] cut by a transversal \[AC\], then corresponding angles made by transversal are equal .
Therefore, \[\angle AED = \angle ACB\]
Now we can say \[\vartriangle ABC\] is similar to \[\vartriangle ADE\]( since both set or corresponding angles are equal and \[\angle A\] is common
Which means both the triangles have the same shape but different sizes.
Now we find ratio of \[ar(\vartriangle ABC):ar(\vartriangle ADE)\]
Since, we know if two similar triangles have a ratio of their sides as \[a:b\]then the ratio of areas of two similar triangles will be \[{a^2}:{b^2}\].
Therefore, substitute \[a = 1,b = 3\]
Ratio \[ar(\vartriangle ABC):ar(\vartriangle ADE) = {1^2}:{3^2} = 1:9\]

Note: Students are likely to make mistake of writing that ratio of area of similar triangles is same as ratio of sides of similar triangles, but we know area is in square units and length is in simple units so the ratio of sides of similar triangles cannot be equal to ratio of area of similar triangles.