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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