Question

Points D, E and F are the midpoints of the sides of Δ ABC. Show that area of Δ DEF is one-quarter of the area of Δ ABC.

Answer 1

Appropriate Question :-

Points D, E and F are the midpoints of the sides AB, BC, CA respectively of Δ ABC. Show that area of Δ EFD is one-quarter of the area of Δ ABC.

$\large\underline{\sf{Solution-}}$

Given that,

In Δ ABC,

• D is the midpoint of AB

• E is the midpoint of BC

• F is the midpoint of CA

We know,

Midpoint Theorem :- The midpoint theorem states that the line segment joining the midpoints of the two sides of a triangle is parallel to third side and equals to half of it.

Now,

• D is the midpoint of AB.

and

• E is the midpoint of BC

⇛ By using Midpoint Theorem,

$\rm :\longmapsto\:DE = \dfrac{1}{2}AC$

$\red{\bf\implies \:\boxed{ \rm{ \: \dfrac{DE}{AC} = \dfrac{1}{2} \: \: - - - - (1) }}}$

Again,

• D is the midpoint of AB.

and

• F is the midpoint of AC

⇛ By using Midpoint Theorem,

$\rm :\longmapsto\:DF = \dfrac{1}{2}BC$

$\red{\bf\implies \:\boxed{ \rm{ \: \dfrac{DF}{BC} = \dfrac{1}{2} \: \: - - - - (3) }}}$

Again,

• E is the midpoint of BC.

and

• E is the midpoint of AC

⇛ By using Midpoint Theorem

$\rm :\longmapsto\:EF = \dfrac{1}{2}AB$

$\red{\bf\implies \:\boxed{ \rm{ \: \dfrac{EF}{AB} = \dfrac{1}{2} \: \: - - - - (4) }}}$

So, from equation (2), (3) and (4), we concluded that

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

$\bf\implies \: \triangle \: ABC \sim \triangle \:EFD \: \: \: \: \: \: \: \{SSS \}$

So, By Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

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

$\rm :\longmapsto\:\dfrac{ar(\triangle \:EFD)}{ar(\triangle \:ABC)} = {\bigg[\dfrac{1}{2} \bigg]}^{2}$

$\rm :\longmapsto\:\dfrac{ar(\triangle \:EFD)}{ar(\triangle \:ABC)} = \dfrac{1}{4}$

$\bf\implies \:ar(\triangle \:EFD) = \dfrac{1}{4}ar(\triangle \:ABC)$

Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem :-

If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

Answer:

Step-by-step explanation:

1. for triangle DEF, the height is 1/2 h of triangle ABC, the base is also 1/2 b of triangle ABC.

-> area = bh/(2)(2)(2) = bh/8

2. triangle ABC = bh/2

triangle DEF = bh/8

THEREFORE triangle DEF is 1/4 of the area of triangle ABC