Ex.6.4 Q5 Triangles Solution - NCERT Maths Class 10

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\(D, E\) 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 \(\Delta \,ABC.\)


 Video Solution
Ex 6.4 | Question 5

Text Solution


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).


In \(\Delta ABC\), \(D\, {\rm{}and}\, F\) are the midpoints of \(AB, AC\) respectively.

\[\begin{align} \Rightarrow \;\;&DE\parallel BC \;\text{ and }\;  DE = \frac{1}{2}BC \\ &\text{(by mid-point theorem)} \end{align}\]

Again, \(E\) is mid-point of \(BC \)

\[\begin{align} \Rightarrow \;\;DF \parallel BE \;\text{ and }\; DF = BE\end{align}\]

In quadrilateral \(DFEB,\)

\[\begin{align} DF \parallel BE \;\text{ and }\; DF = BE\end{align}\]

\(\therefore \) \(DFEB\) is a parallelogram

\[\begin{align}\Rightarrow \quad &\angle B = \angle F\;\;\dots(1) \\ &\begin{bmatrix} \text{Opposite angles of a } \\ \text{parallelogram are equal}\end{bmatrix} \end{align}\]

Similarly, we can prove that

\(DFCE\) is a parallelogram

\[\begin{align}\Rightarrow \quad& \angle C=\angle D\;\;\dots(2)\\ &\begin{bmatrix} \text{Opposite angles of a } \\ \text{parallelogram are equal}\end{bmatrix} \end{align}\]


In \(\Delta DEF\) and \(\Delta ABC\)

\[\angle DEF = \angle ABC \;\;\dots \text{from (1)}\]

\[\angle EDF = \angle ACB \;\;\dots \text{from (2)}\]

\[\begin{align}\Rightarrow \quad & \Delta DEF \sim \Delta CAB \\ & \text{ [AA Criterion]} \end{align}\]

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

\[\begin{align}\frac{{ \text{ar}\,\Delta DEF}}{{\text{ar}\,\Delta ABC}} & = \frac{{{{(DE)}^2}}}{{{{(AC)}^2}}} \\ & = \frac{{{{(EF)}^2}}}{{{{(AB)}^2}}} \\ & = \frac{{{{(DF)}^2}}}{{{{(BC)}^2}}}\end{align}\]

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

The ratio of the areas of \(\Delta DEF\) and \(\Delta ABC\) is \( 1:4\)

Alternate  method:


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


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

\[ \Rightarrow DE\parallel BC\]


\[\begin{align} DE \!=\! \frac{1}{2}BC \end{align} \dots (1)\]


Now in quadrilateral \(DBFE\)

\[\Rightarrow DE\parallel BC  \text{ and }DE \!=\! BF \dots\text{from (1)}\]

\( \Rightarrow \) \(DBFE\) is a parallelogram

\[ \begin{align} &\Rightarrow \quad  ar\Delta DBF \!=\! ar \Delta DEF \;\;\dots(2) \\ &\begin{bmatrix}\because \text{Diagonal DF divides the parallelogram } \\ \text{into two triangles of equal area} \end{bmatrix}\end{align} \]


Similarly, we can prove

\[ar\Delta DBF \! = \! ar\Delta EFC\;\; \dots (3)\]

\[ar\Delta DEF \! = \! ar\Delta ADE\;\; \dots (4)\]


From \((2 )\) \(( 3)\) and \((4)\)

\[\begin{bmatrix}ar\Delta DBF \! = \! ar\Delta DEF \! =\\ \! ar\Delta EFC \! = \! ar\Delta ADE\end{bmatrix}\;\;\dots(5)\]

Things which are equal to the same thing are equal to one another \(–\) Euclid’s \(1\)st axiom.

\[\begin{align} \begin{bmatrix} ar\Delta ABC = ar\Delta ADE = ar\Delta DBF = \\ar\Delta EFD = ar\Delta DEF \end{bmatrix} \end{align}\]


From \((5)\)

\[\begin{align} ar\Delta ABC & \! = \! 4  \! \times \!  ar\Delta DEF \\\frac{ar\Delta DEF}{ar\Delta ABC} & \! = \!  \frac{1}{4}\\ar\Delta DEF  :  ar\Delta ABC & \! = \!  1 \! : \! 4 \end{align}\]

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