# Ex.6.4 Q5 Triangles Solution - NCERT Maths Class 10

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## Question

$$D, E$$ and $$F$$ are respectively the mid-points of sides $$AB, BC$$ and $$CA$$ of $$\Delta \,ABC.$$ Find the ratio of the areas of $$\Delta \,DEF$$ and $$\Delta \,ABC.$$

Diagram

## Text Solution

Reasoning:

As we know that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it – (mid-point theorem).

Steps:

In $$\Delta ABC$$, $$D\, {\rm{}and}\, F$$ are the midpoints of $$AB, AC$$ respectively.
\begin{align} \Rightarrow DE\parallel BC \end{align} and \begin{align} DE = \frac{1}{2}BC \end{align}        (by mid-point theorem)

Again $$,E$$ is mid-point of $$BC$$
\begin{align} \Rightarrow DF \parallel BE \,\text{and}\, DF = BE\end{align}

In quadrilateral $$DFEB,$$

$$DF \parallel BE$$ and $$DF= BE$$

$$\therefore$$ $$DFEB$$ is a parallelogram

$$\Rightarrow \angle B = \angle F$$       (1)    (opposite angles of a parallelogram are equal)

Similarly, we can prove that

$$DFCE$$ is a parallelogram

$$\Rightarrow \angle C$$ $$=$$ $$\angle D$$         (2)    (opposite angles of a parallelogram are equal)

Now ,In $$\Delta DEF$$ and $$\Delta ABC$$

$$\angle DEF = \angle ABC$$ (from 1)

$$\angle EDF = \angle ACB$$ (from 2)

$$\Rightarrow \,\,\Delta DEF \sim \Delta CAB$$       [AA Criterion]

The ratio of the areas of two similar triangles is equal to the square of corresponding sides.

\begin{align}\frac{{Area\,of\,\Delta DEF}}{{Area\,of\,\Delta ABC}} = \frac{{{{(DE)}^2}}}{{{{(AC)}^2}}} = \frac{{{{(EF)}^2}}}{{{{(AB)}^2}}} = \frac{{{{(DF)}^2}}}{{{{(BC)}^2}}}\end{align}

\begin{align} \frac{\text {Area of } \Delta D E F}{\text {Area of } \Delta A B C} &=\frac{(D F)^{2}}{(BC)^{2}} \\ &=\frac{(\frac12 BC)^{2}}{(BC)^{2}} \quad\\ &=\frac{B C^{2}}{4 B C^{2}}\\ &=\frac{{1}}{4} \end{align}

The ratio of the areas of $$\,\Delta DEF$$ and $$\Delta ABC$$ is $$1:4$$

Alternate  method:

Reasoning:

Mid-Point Theorem : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Steps:

In $$\Delta ABC$$ ,$$D$$ and $$E$$ are midpoints of sides $$AB$$ and $$AC$$

$$\Rightarrow DE\parallel BC$$ and $$\,DE = \frac{1}{2}BC$$ …….. (1)

Now in quadrilateral $$DBFE$$

$$\Rightarrow DE\parallel BC$$ and $$\,DE = BF$$     (from 1)

$$\Rightarrow$$ $$DBFE$$ is a parallelogram

$$\Rightarrow$$ Area of $$\,\Delta DBF$$ $$=$$ Area of $$\,\Delta DEF$$ …... (2)

($$\because$$ diagonal DF divides the parallelogram into two triangle of equal area)

Similarly, we can prove

Area of $$\Delta DBF$$ $$=$$ Area of $$\Delta EFC$$ …. (3)

And area of $$\Delta DEF$$ $$=$$ Area of $$\Delta ADE$$ … (4)

From (2 ) ( 3) and (4)

Area of $$\,\Delta DBF$$ $$=$$ Area of $$\,\Delta DEF$$ $$=$$ Area of $$\,\Delta EFC$$ $$=$$ Area of $$\,\Delta ADE$$ …. (5)

(Things which are equal to the same thing are equal to one another $$–$$ Euclid’s 1staxiom.)

Area of $$\Delta ABC$$ $$=$$ Area of $$\,\Delta ADE$$ $$+$$ Area of $$\Delta DBF+$$ Area of $$\,\Delta EFD$$ $$+$$ Area of$$\,\Delta DEF$$

From (5)

Area of $$\Delta ABC = 4 \times$$ Area of $$\,\Delta DEF$$

\begin{align}\frac{{Area\,\,of\,\,\Delta DEF}}{{Area\,\,of\,\,\Delta ABC}} = \frac{1}{4}\end{align}

Area of $$\Delta DEF$$ $$:$$ Area of $$\Delta ABC$$ $$= 1:4$$

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