# Ex.9.3 Q5 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

\(D\), \(E\) and \(F\) are respectively the mid-points of the sides \(BC\), \(CA\) and \(AB\) of a \(\Delta ABC\). Show that:

(i) \(BDEF\) is a parallelogram

(ii) \(\begin{align}{ar } (DEF) = \frac { 1 } { 4 } { ar } (ABC)\end{align}\)

(iii) \(\begin{align}{ar } (BDEF) = \frac { 1 } { 2 } { ar } (ABC)\end{align}\)

## Text Solution

**What is known?**

\(D\), \(E\) and \(F\) are respectively the mid-points of the sides \(BC\), \(CA\) and \(AB\) of a \(\Delta \text{ABC}\)

**What is unknown?**

How we can show that

(i) \(BDEF\) is a parallelogram.

(ii)\({ar }\left( {DEF} \right){ = }\frac{{1}}{{4}}{ar }\left( {ABC} \right)\)

(iii) \({ar }\left( {BDEF} \right){ = }\frac{{1}}{{2}}{ar }\left( {ABC} \right)\)

**Reasoning:**

Line joining the mid-points of two sides of a triangle is parallel to the third and half of its length. If one pair of opposite side in quadrilateral is parallel and equal to each other then it is a parallelogram. Diagonals divides the parallelogram in two triangles of equal areas.

**Steps:**

(i) \(F\) is the mid-point of \(AB\) and \(E\) is the mid-point of \(AC\).

\(\begin{align} \therefore \; FE \| BC \text { and } FE = \frac { 1 } { 2 } BC \end{align}\)

Line joining the mid-points of two sides of a triangle is parallel to the third and half of its length.

\[\begin{align} \therefore FE \| BD \,[BD \text { is the part of } BC]\end{align}\]

\(\begin{align} \text { Also, } D \text { is the mid-point of BC. }\end{align}\)

\[\therefore {{BD = }}\frac{{{1}}}{2}{{ BC}}\]

\[\begin{align} FE = BD\end{align}\]

Now \(FE\|BD\) and \(FE = BD\)

Since one pair of opposite side in quadrilateral \(BDEF\) is parallel and equal to each other

Therefore, \(BDEF\) is a parallelogram.

(ii) \(BDEF\) is a parallelogram.

\[\begin{align} \therefore { ar } (\Delta BDF) = {ar } (\Delta DEF) \ldots (i) \end{align}\]

[The diagonal \(DF\) of the parallelogram \(BDEF\) divides it in two triangles \(BDF\) and \(DEF\) of equal area]

Similarly,\(DCEF\) is also parallelogram.

\[\begin{align} \therefore { ar } (\Delta DEF)\!=\! {ar } (\Delta DEC)\; \ldots (ii)\end{align}\]

Also, \(AEDF\) is a parallelogram.

\[\begin{align} \therefore\!{ ar } (\Delta AFE)\!=\!{ar } (\Delta DEF) \;\ldots (iii)\end{align}\]

From equations (i), (ii) and (iii),

\[\begin{align} (\Delta DEF)& = { ar } (\Delta BDF)\\&= { ar } (\Delta DEC) \\&= { ar } (\Delta AFE)\;\ldots (iv) \end{align}\]

Now,

\[\begin{align}&{ar}(\Delta ABC)\\&\!=\!\left[ \begin{array} {ar}(\Delta DEF)\!+\!{ar}(\Delta BDF)+ \\ {ar}(\Delta DEC)\!+\!{ar}(\Delta {AFE})\ \\\end{array} \right]\!\ldots\! ({v})\end{align}\]

\[\begin{align}&{{ar}}(\Delta ABC)\\&\!=\!\left[ \begin{array}{l}{{ ar }}(\Delta DEF)\!+\!{{ ar }}(\Delta DEF)\!+\!\\{{ ar }}(\Delta DEF)\!+\!{{ ar }}(\Delta DEF)\end{array} \right]\end{align}\]

\({\left[ {{{Using }}\left( {{{iv}}} \right){{ \& }}\left( {{v}} \right)} \right]}\)

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

(iii)

\[\begin{align} {ar } (\| gm\; BDEF) &= {ar } (\Delta BDF) + {ar } (\Delta DEF)\\&= {ar } ( \Delta DEF) + {ar } (\Delta DEF) \\\text{ Using } (iv) \end{align}\]

\[\begin{align} {ar } (\| { gm } BDEF) &\!=\!2 { ar }\!\!(\Delta DEF) \\ { ar } (\| { gm } BDEF) &\!=\! 2\!\times\!\frac { 1 } { 4 }\!\! { ar }\!\!(\Delta ABC) \\\! { ar }\!( \| { gm } BDEF) &\!=\! \frac { 1 } { 2 } { ar }\!(\Delta ABC) \end{align}\]