Question

if D, E, F are the mid-points of sides BC, CA and AB respectively of ΔABC, Then the ratio of the areas of ΔDEF and ΔABC is​

Answer 1

Given Question :-

• If D, E, F are the mid-points of sides BC, CA and AB respectively of ΔABC, Then the ratio of the areas of ΔDEF and ΔABC is _____

Answer

Given :-

• D, E, F are the mid-points of sides BC, CA and AB respectively of ΔABC

To Find :-

• ar(ΔDEF) : ar(ΔABC)

Concept Used :-

Midpoint Theorem

• The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

SSS Similarity

• SSS Similarity Theorem: If all three pairs of corresponding sides of two triangles are proportional, then the two triangles are similar.

Area Ratio 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.

$\large\underline{\bold{Solution :- }}$

Given that

• D is the midpoint of BC

• E is the midpoint of AC

• By midpoint theorem,

$\rm :\implies\:DE = \dfrac{1}{2}AB$

$\rm :\implies\:\dfrac{DE}{AB} = \dfrac{1}{2} - - (1)$

Again,

• E is the midpoint of AC

• F is the midpoint of AB

• By midpoint theorem,

$\rm :\implies\:EF = \dfrac{1}{2}BC$

$\rm :\implies\:\dfrac{EF}{BC} = \dfrac{1}{2} - - (2)$

Again,

F is the midpoint of AB

D is the midpoint of BC

By midpoint theorem,

$\rm :\implies\:DF = \dfrac{1}{2}AC$

$\rm :\implies\:\dfrac{DF}{AC} = \dfrac{1}{2} - - (3)$

So,

• From equation (1), (2) and (3), we concluded that

$\rm :\longmapsto\:\dfrac{DE}{AB} = \dfrac{EF}{BC} = \dfrac{DF}{AC} = \dfrac{1}{2}$

So,

• By SSS Similarity Theorem, we have

$\rm :\longmapsto \: \triangle \: ABC \: \sim \: \triangle \: DEF$

Now,

• By using Area Ratio Theorem, we have

$\rm :\longmapsto\:\dfrac{ar(\triangle \: DEF )}{ar( \triangle \: ABC)} = \dfrac{ {DE}^{2} }{ {AB}^{2} }$

$\rm :\longmapsto\:\dfrac{ar( \triangle \: DEF)}{ar( \triangle \: ABC)} = { \bigg(\dfrac{DE}{AB} \bigg)}^{2}$

$\rm :\longmapsto\:\dfrac{ar(\triangle \: DEF)}{ar( \triangle \: ABC)} = { \bigg(\dfrac{1}{2} \bigg)}^{2} = \dfrac{1}{4}$

$\bf :\longmapsto\: \boxed{ \bf \: \dfrac{ar( \triangle \: ABC)}{ar(\triangle \: DEF)} =\dfrac{4}{1}}$