Answer
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Hint: Here the given is the relation between the triangles \[\Delta ABC\] and $\Delta BDE$. We have to find the ratio of the areas of given triangles. By using some triangle properties to find it.
SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
Complete step-by-step answer:
It is given that, \[ABC\] and \[BDE\] are two equilateral triangles such that \[D\] is the midpoint of \[BC\]
We need to find the ratio of \[\dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}}\].
Since we have to find the ratio of the areas of \[\Delta ABC\] and \[\Delta BDE\], we first need to prove these triangles are similar.
We know that for any equilateral triangle the sides are equal.
Since, \[ABC\] and \[BDE\] are two equilateral triangles such that \[D\] is the mid-point of \[BC\].
In $\Delta ABC$,
\[ \Rightarrow AB = BC = CA\]
In $\Delta BDE$,
\[ \Rightarrow BE = DE = BD = \dfrac{1}{2}BC\]
Thus we have their sides would be in the same ratio.
\[ \Rightarrow \dfrac{{AB}}{{BE}} = \dfrac{{AC}}{{DE}} = \dfrac{{BC}}{{BD}}\]
Hence by SSS similarity,
\[ \Rightarrow \Delta ABC \sim \Delta BDE\]
We know that if two triangle are similar,
Ratio of areas is equal to square of ratio of its corresponding sides
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = {\left( {\dfrac{{BC}}{{BD}}} \right)^2}\]
Since \[D\] is the midpoint of \[BC\]
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = {\left( {\dfrac{{\dfrac{1}{1}BC}}{{\dfrac{1}{2}BC}}} \right)^2}\]
Cancelling the common term \[BC\],
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = {\left( {1 \times \dfrac{2}{1}} \right)^2}\]
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = \dfrac{4}{1}\]
Hence, the areas of triangles \[\Delta ABC\] and \[\Delta BDE\] is: \[4:1\] .
(C) is the correct option.
Note: If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar. The corresponding sides of similar triangles are in proportion.
We have used the following theorem.
Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
Complete step-by-step answer:
It is given that, \[ABC\] and \[BDE\] are two equilateral triangles such that \[D\] is the midpoint of \[BC\]
We need to find the ratio of \[\dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}}\].
Since we have to find the ratio of the areas of \[\Delta ABC\] and \[\Delta BDE\], we first need to prove these triangles are similar.
We know that for any equilateral triangle the sides are equal.
Since, \[ABC\] and \[BDE\] are two equilateral triangles such that \[D\] is the mid-point of \[BC\].
In $\Delta ABC$,
\[ \Rightarrow AB = BC = CA\]
In $\Delta BDE$,
\[ \Rightarrow BE = DE = BD = \dfrac{1}{2}BC\]
Thus we have their sides would be in the same ratio.
\[ \Rightarrow \dfrac{{AB}}{{BE}} = \dfrac{{AC}}{{DE}} = \dfrac{{BC}}{{BD}}\]
Hence by SSS similarity,
\[ \Rightarrow \Delta ABC \sim \Delta BDE\]
We know that if two triangle are similar,
Ratio of areas is equal to square of ratio of its corresponding sides
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = {\left( {\dfrac{{BC}}{{BD}}} \right)^2}\]
Since \[D\] is the midpoint of \[BC\]
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = {\left( {\dfrac{{\dfrac{1}{1}BC}}{{\dfrac{1}{2}BC}}} \right)^2}\]
Cancelling the common term \[BC\],
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = {\left( {1 \times \dfrac{2}{1}} \right)^2}\]
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta BDE)}} = \dfrac{4}{1}\]
Hence, the areas of triangles \[\Delta ABC\] and \[\Delta BDE\] is: \[4:1\] .
(C) is the correct option.
Note: If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar. The corresponding sides of similar triangles are in proportion.
We have used the following theorem.
Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
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