
In the figure D and E are points on the sides AB and AC respectively of a $\vartriangle ABC$ such that ${\text{DE||BC}}$ and divides $\vartriangle ABC$ into two parts, equal in area, find $\dfrac{{BD}}{{AB}}$.

Answer
513.9k+ views
Hint: In this problem we have been with a triangle ABC such that the line segment DE inside this triangle is parallel to BC. So BCED forms a trapezium now and it is given that this DE divides this triangle into two equal areas. Thus use the respective formula for area of triangle and area of trapezium along with triangle congruence postulates to reach the answer.
Complete step-by-step answer:
It is given that the area of triangle ADE is equal to the area of trapezium BCED.
$ \Rightarrow {\text{Area}}\left( {\Delta ADE} \right) = {\text{Area}}\left( {{\text{trapezium BCED}}} \right)$
Now add area of triangle ADE both sides we have
$ \Rightarrow {\text{Area}}\left( {\Delta ADE} \right) + {\text{Area}}\left( {\Delta ADE} \right) = {\text{Area}}\left( {\Delta ADE} \right) + {\text{Area}}\left( {{\text{trapezium BCED}}} \right)$
In above equation R.H.S part is the total area of triangle ABC
\[ \Rightarrow 2{\text{Area}}\left( {\Delta ADE} \right) = {\text{Area}}\left( {\Delta ABC} \right)\]
\[ \Rightarrow \dfrac{{{\text{Area}}\left( {\Delta ADE} \right)}}{{{\text{Area}}\left( {\Delta ABC} \right)}} = \dfrac{1}{2}\] …………………. (1)
Now in triangle ADE and in triangle ABC we have
$\angle ADE = \angle B$ [$\because DE||BC{\text{ & }}\angle ADE = \angle B$ (Corresponding angles)]
And, $\angle A = \angle A$ [common angle]
Therefore \[{\text{Area}}\left( {\Delta ADE} \right) \approx {\text{Area}}\left( {\Delta ABC} \right)\] (similar triangles)
Therefore according the property of similar triangles
$\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}} = \dfrac{{DE}}{{BC}} = \sqrt {\dfrac{{{\text{Area}}\left( {\Delta ADE} \right)}}{{{\text{Area}}\left( {\Delta ABC} \right)}}} $
Now from equation (1) we have
$\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}} = \dfrac{{DE}}{{BC}} = \sqrt {\dfrac{1}{2}} = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow AD = \dfrac{{AB}}{{\sqrt 2 }}$……………………………. (2)
Now from figure we can say that
$BD = AB - AD$
Now from equation (2)
$BD = AB - AD = AB - \dfrac{{AB}}{{\sqrt 2 }}$
$ \Rightarrow \dfrac{{BD}}{{AB}} = 1 - \dfrac{1}{{\sqrt 2 }}$.
So, this is the required ratio of BD to AB.
Thus, this is the required answer.
Note: Whenever we face such types of geometry questions the key point is to understand the diagrammatic representation of the data provided in the question. Having a good understanding of various triangle congruence postulates like ASA, AAA, SSS along with formulas of various sections like triangle and trapezium always helps in getting on the right track to reach the answer.
Complete step-by-step answer:
It is given that the area of triangle ADE is equal to the area of trapezium BCED.
$ \Rightarrow {\text{Area}}\left( {\Delta ADE} \right) = {\text{Area}}\left( {{\text{trapezium BCED}}} \right)$
Now add area of triangle ADE both sides we have
$ \Rightarrow {\text{Area}}\left( {\Delta ADE} \right) + {\text{Area}}\left( {\Delta ADE} \right) = {\text{Area}}\left( {\Delta ADE} \right) + {\text{Area}}\left( {{\text{trapezium BCED}}} \right)$
In above equation R.H.S part is the total area of triangle ABC
\[ \Rightarrow 2{\text{Area}}\left( {\Delta ADE} \right) = {\text{Area}}\left( {\Delta ABC} \right)\]
\[ \Rightarrow \dfrac{{{\text{Area}}\left( {\Delta ADE} \right)}}{{{\text{Area}}\left( {\Delta ABC} \right)}} = \dfrac{1}{2}\] …………………. (1)
Now in triangle ADE and in triangle ABC we have
$\angle ADE = \angle B$ [$\because DE||BC{\text{ & }}\angle ADE = \angle B$ (Corresponding angles)]
And, $\angle A = \angle A$ [common angle]
Therefore \[{\text{Area}}\left( {\Delta ADE} \right) \approx {\text{Area}}\left( {\Delta ABC} \right)\] (similar triangles)
Therefore according the property of similar triangles
$\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}} = \dfrac{{DE}}{{BC}} = \sqrt {\dfrac{{{\text{Area}}\left( {\Delta ADE} \right)}}{{{\text{Area}}\left( {\Delta ABC} \right)}}} $
Now from equation (1) we have
$\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}} = \dfrac{{DE}}{{BC}} = \sqrt {\dfrac{1}{2}} = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow AD = \dfrac{{AB}}{{\sqrt 2 }}$……………………………. (2)
Now from figure we can say that
$BD = AB - AD$
Now from equation (2)
$BD = AB - AD = AB - \dfrac{{AB}}{{\sqrt 2 }}$
$ \Rightarrow \dfrac{{BD}}{{AB}} = 1 - \dfrac{1}{{\sqrt 2 }}$.
So, this is the required ratio of BD to AB.
Thus, this is the required answer.
Note: Whenever we face such types of geometry questions the key point is to understand the diagrammatic representation of the data provided in the question. Having a good understanding of various triangle congruence postulates like ASA, AAA, SSS along with formulas of various sections like triangle and trapezium always helps in getting on the right track to reach the answer.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
