
In the given figure, DE is parallel to BC
\[{\text{(i)}}\] If \[{\text{DE = 4 cm, BC = 6 cm}}\] and \[{\text{Area(}}\vartriangle {\text{ADE) = 16 c}}{{\text{m}}^{\text{2}}}\], find the area of \[\vartriangle {\text{ABC}}\].
\[{\text{(ii)}}\] If \[{\text{DE = 4 cm, BC = 8 cm}}\] and \[{\text{Area(}}\vartriangle {\text{ADE) = 25 c}}{{\text{m}}^{\text{2}}}\], find the area of \[\vartriangle {\text{ABC}}\].
\[{\text{(iii)}}\] If \[{\text{DE:BC = 3:5}}\]. Calculate the ratio of the areas of \[\vartriangle {\text{ADE}}\] and the area of BCED.
Answer
509.6k+ views
Hint- Here, we will be using the properties of congruent triangles.
Given, DE is parallel to BC.
From the figure, in \[\vartriangle {\text{ADE}}\] and \[\vartriangle {\text{ABC}}\]
\[\angle {\text{A = }}\angle {\text{A}}\] (Common angle)
Since, DE is parallel to BC therefore the below angles will be equal because they are corresponding angles.
\[ \Rightarrow \angle {\text{D = }}\angle {\text{B}}\] (Corresponding angles) and (Corresponding angles)
Therefore, by AAA congruence criteria, the triangles ADE and ABC are congruent
i.e., \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}}\]
\[{\text{(i)}}\] Given, \[{\text{DE = 4 cm, BC = 6 cm}}\] and \[{\text{Area(}}\vartriangle {\text{ADE) = 16 c}}{{\text{m}}^{\text{2}}}\]
Since, for two congruent triangles, the ratio of their areas will be equal to the square of the ratio of their corresponding sides.
As, \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{{\text{DE}}}}{{{\text{BC}}}}} \right)^2} \Rightarrow \dfrac{{16}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{\text{4}}}{{\text{6}}}} \right)^2} = \dfrac{{16}}{{36}}\]
\[ \Rightarrow {\text{Area(}}\vartriangle {\text{ABC)}} = \dfrac{{36 \times 16}}{{16}} = 36{\text{ c}}{{\text{m}}^2}\]
Therefore, the area of triangle ABC is \[36{\text{ c}}{{\text{m}}^2}\].
\[{\text{(ii)}}\] Given, \[{\text{DE = 4 cm, BC = 8 cm}}\] and \[{\text{Area(}}\vartriangle {\text{ADE) = 25 c}}{{\text{m}}^{\text{2}}}\]
Since, for two congruent triangles, the ratio of their areas will be equal to the square of the ratio of their corresponding sides.
As, \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{{\text{DE}}}}{{{\text{BC}}}}} \right)^2} \Rightarrow \dfrac{{25}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{\text{4}}}{8}} \right)^2} = \dfrac{{16}}{{64}}\]
\[ \Rightarrow {\text{Area(}}\vartriangle {\text{ABC)}} = \dfrac{{25 \times 64}}{{16}} = 100{\text{ c}}{{\text{m}}^2}\]
Therefore, the area of triangle ABC is \[{\text{100 c}}{{\text{m}}^2}\].
\[(iii)\] Given, \[{\text{DE:BC = 3:5}} \Rightarrow \dfrac{{{\text{DE}}}}{{{\text{BC}}}} = \dfrac{3}{5}\]
Since, for two congruent triangles, the ratio of their areas will be equal to the square of the ratio of their corresponding sides.
As, \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{{\text{DE}}}}{{{\text{BC}}}}} \right)^2} = {\left( {\dfrac{3}{5}} \right)^2} = \dfrac{9}{{25}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ADE) + Area(BCED)}}}} = \dfrac{9}{{25}}\]
Reciprocating the above equation, we get
\[
\Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE) + Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} + \dfrac{{{\text{Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} \\
\Rightarrow {\text{1}} + \dfrac{{{\text{Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} \Rightarrow \dfrac{{{\text{Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} - 1 = \dfrac{{16}}{9} \\
\]
Now again reciprocating the above equation, we get
\[ \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(BCED)}}}} = \dfrac{9}{{16}}\]
Therefore, the ratio of triangle ADE to the quadrilateral BCED is \[{\text{9:16}}\].
Note- These type of problems are mostly solved with the help of geometry. Hence, congruency rule is used to obtain the relation between known and unknown data. By these relations, the required value of the unknown can be computed easily.
Given, DE is parallel to BC.
From the figure, in \[\vartriangle {\text{ADE}}\] and \[\vartriangle {\text{ABC}}\]
\[\angle {\text{A = }}\angle {\text{A}}\] (Common angle)
Since, DE is parallel to BC therefore the below angles will be equal because they are corresponding angles.
\[ \Rightarrow \angle {\text{D = }}\angle {\text{B}}\] (Corresponding angles) and (Corresponding angles)
Therefore, by AAA congruence criteria, the triangles ADE and ABC are congruent
i.e., \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}}\]
\[{\text{(i)}}\] Given, \[{\text{DE = 4 cm, BC = 6 cm}}\] and \[{\text{Area(}}\vartriangle {\text{ADE) = 16 c}}{{\text{m}}^{\text{2}}}\]
Since, for two congruent triangles, the ratio of their areas will be equal to the square of the ratio of their corresponding sides.
As, \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{{\text{DE}}}}{{{\text{BC}}}}} \right)^2} \Rightarrow \dfrac{{16}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{\text{4}}}{{\text{6}}}} \right)^2} = \dfrac{{16}}{{36}}\]
\[ \Rightarrow {\text{Area(}}\vartriangle {\text{ABC)}} = \dfrac{{36 \times 16}}{{16}} = 36{\text{ c}}{{\text{m}}^2}\]
Therefore, the area of triangle ABC is \[36{\text{ c}}{{\text{m}}^2}\].
\[{\text{(ii)}}\] Given, \[{\text{DE = 4 cm, BC = 8 cm}}\] and \[{\text{Area(}}\vartriangle {\text{ADE) = 25 c}}{{\text{m}}^{\text{2}}}\]
Since, for two congruent triangles, the ratio of their areas will be equal to the square of the ratio of their corresponding sides.
As, \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{{\text{DE}}}}{{{\text{BC}}}}} \right)^2} \Rightarrow \dfrac{{25}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{\text{4}}}{8}} \right)^2} = \dfrac{{16}}{{64}}\]
\[ \Rightarrow {\text{Area(}}\vartriangle {\text{ABC)}} = \dfrac{{25 \times 64}}{{16}} = 100{\text{ c}}{{\text{m}}^2}\]
Therefore, the area of triangle ABC is \[{\text{100 c}}{{\text{m}}^2}\].
\[(iii)\] Given, \[{\text{DE:BC = 3:5}} \Rightarrow \dfrac{{{\text{DE}}}}{{{\text{BC}}}} = \dfrac{3}{5}\]
Since, for two congruent triangles, the ratio of their areas will be equal to the square of the ratio of their corresponding sides.
As, \[\vartriangle {\text{ADE}} \cong \vartriangle {\text{ABC}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ABC)}}}} = {\left( {\dfrac{{{\text{DE}}}}{{{\text{BC}}}}} \right)^2} = {\left( {\dfrac{3}{5}} \right)^2} = \dfrac{9}{{25}} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ADE) + Area(BCED)}}}} = \dfrac{9}{{25}}\]
Reciprocating the above equation, we get
\[
\Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE) + Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} + \dfrac{{{\text{Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} \\
\Rightarrow {\text{1}} + \dfrac{{{\text{Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} \Rightarrow \dfrac{{{\text{Area(BCED)}}}}{{{\text{Area(}}\vartriangle {\text{ADE)}}}} = \dfrac{{25}}{9} - 1 = \dfrac{{16}}{9} \\
\]
Now again reciprocating the above equation, we get
\[ \Rightarrow \dfrac{{{\text{Area(}}\vartriangle {\text{ADE)}}}}{{{\text{Area(BCED)}}}} = \dfrac{9}{{16}}\]
Therefore, the ratio of triangle ADE to the quadrilateral BCED is \[{\text{9:16}}\].
Note- These type of problems are mostly solved with the help of geometry. Hence, congruency rule is used to obtain the relation between known and unknown data. By these relations, the required value of the unknown can be computed easily.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

