Question

In $ \Delta ABC $ point D is on side AB and point E on side AC such that quadrilateral. BCED is a trapezium. If $ DE:BC = 3:5 $ then find the ratio of area of $ \Delta ADE\,\,and\,\,Quadrilateral.BCED. $

Answer 1

Hint: In this we first prove two triangle formed by joining point D and E similar and then using area theorem to form an equation or ratio of area of two triangles and then writing big triangle as a sum of small triangle and trapezium and then solving to find ratio of required areas to get solution of given problem.

Complete step-by-step answer:
Given, D and E are points on sides AB and AC of triangle ABC such that the quadrilateral. BCED so formed is a trapezium.

Since, the quadrilateral BCED is a trapezium.
Therefore, line DE is parallel to BC.
Now, in triangle ADE and triangle ABC. We have
 $
  \angle ADE = \angle ABC\,\,\,\,\,\,\left( {corresponding\,\,angles} \right) \\
  \angle AED = ACB\,\,\,\,\,\,\,\,\,\,\,\left( {corresponding\,\,angles} \right) \\
  $
Therefore, $ \Delta ADE \sim \Delta ABC $
Now, by area theorem. We have a ratio of area of two similar triangles equal to the ratio of squares or their corresponding sides.
Therefore, we have,
 $ \dfrac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta ABC} \right)}} = {\left( {\dfrac{{DE}}{{BC}}} \right)^2} $
But it is given that ratio of sides DE to BC is $ 3:5 $ or we can write it as:
 $ \dfrac{{DE}}{{BC}} = \dfrac{3}{5} $
Using this value of $ \dfrac{{DE}}{{BC}} $ in the above formed equation. We have
\[
  \dfrac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta ABC} \right)}} = {\left( {\dfrac{3}{5}} \right)^2} \\
   \Rightarrow \dfrac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta ABC} \right)}} = \dfrac{9}{{25}} \;
 \]
Taking reciprocal of above formed equation. We have
 $ \dfrac{{ar\left( {\Delta ABC} \right)}}{{ar\left( {\Delta ADE} \right)}} = \dfrac{{25}}{9} $
Also, from the figure we see that the area of $ \Delta ABC $ can be written as the sum of triangle and quadrilateral.
Therefore, we have,
 $ \Delta ABC = \Delta ADE + Quadrilateral.(BCDE) $
Using this in the above formed equation. We have,
 $
  \dfrac{{\Delta ADE + Quadrilateral.(BCDE)}}{{ar\left( {\Delta ADE} \right)}} = \dfrac{{25}}{9} \\
   \Rightarrow \dfrac{{\Delta ADE}}{{\Delta ADE}} + \dfrac{{Quadrilateral.(BCDE)}}{{\Delta ADE}} = \dfrac{{25}}{9} \\
   \Rightarrow 1 + \dfrac{{Quadrilateral.(BCDE)}}{{\Delta ADE}} = \dfrac{{25}}{9} \\
   \Rightarrow \dfrac{{Quadrilateral.(BCDE)}}{{\Delta ADE}} = \dfrac{{25}}{9} - 1 \\
   \Rightarrow \dfrac{{Quadrilateral.(BCDE)}}{{\Delta ADE}} = \dfrac{{25 - 9}}{9} \\
   \Rightarrow \dfrac{{Quadrilateral.(BCDE)}}{{\Delta ADE}} = \dfrac{{16}}{9} \\
  or\,\,we\,\,can\,\,write\,\,above\,\,\,equation\,\,as: \\
  \dfrac{{\Delta ADE}}{{Quadrilateral.(BCDE)}} = \dfrac{9}{{16}} \;
  $
Therefore, from above we see that the ratio of area of triangle ADE to quadrilateral or trapezium is $ \dfrac{9}{{16}} $ .
So, the correct answer is “ $ \dfrac{9}{{16}} $ ”.

Note: To find required ratio of areas we first prove small triangle formed by joining points D and E and big triangle ABC similar and then using area theorem and using value of ratio given to form an equation and then writing triangle ABC in term of sum of triangle and quadrilateral and then simplifying to get ratio of area of triangle and trapezium and hence required solution of given problem