# Ex.7.4 Q6 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

The vertices of a $$\Delta ABC$$ are $$A \;(4, 6)$$, $$B \;(1, 5)$$ and $$C\; (7, 2)$$. $$A$$ line is drawn to intersect sides $$AB$$ and $$AC$$ at $$D$$ and $$E$$ respectively, such that \begin{align}\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{1}{4}\,\,.\end{align}

Calculate the area of the $$\Delta ADE$$ and compare it with the area of $$\Delta ABC$$.

(Recall Converse of basic proportionality theorem and Theorem 6.6 related to Ratio of areas of two similar triangles)

## Text Solution

Reasoning:

Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Theorem 6.6 : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

What is known?

The $$x$$ and $$y$$ coordinates of the vertices of the triangle.

\begin{align}\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{1}{4}\,\,.\end{align}

where $$AB$$ and $$AC$$ are sides of the triangle intersected at $$D$$ and $$E$$.

What is unknown?

The area of the $$\Delta ADE$$ is to be calculated and compared with the area of $$\Delta ABC$$.

Steps:

From the figure,

Given

\begin{align}\frac{{{\text{AD}}}}{{{\text{AB}}}} = \frac{{{\text{AE}}}}{{{\text{AC}}}} = \frac{1}{4}\end{align}

Therefore, $$D$$ and $$E$$ are two points on side $$AB$$ and $$AC$$ respectively such that they divide side $$AB$$ and $$AC$$ in a ratio of $$1:3$$.

By Section formula

\begin{align}P(x, y)=\left[\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right] \quad \dots \dots(1)\end{align}

Coordintaes of Point D \begin{align} &= \left( {\frac{{1 \times 1 + 3 \times 4}}{{1 + 3}},\frac{{1 \times 5 + 3 \times 6}}{{1 + 3}}} \right)\end{align}

\begin{align} &= \left( {\frac{{13}}{4},\frac{{23}}{4}} \right){\text{ }}\\ \end{align}

Coordintaes of Point E \begin{align} &= \left( {\frac{{1 \times 7 + 3 \times 4}}{{1 + 3}},\frac{{1 \times 2 + 3 \times 6}}{{1 + 3}}} \right) \end{align}

\begin{align} &= \left( {\frac{{19}}{4},\frac{{20}}{4}} \right)\end{align}

Area of triangle\begin{align} = \frac{1}{2}\left[ {{{\text{x}}_1}\left( {{{\text{y}}_2} - {{\text{y}}_3}} \right) + {{\text{x}}_2}\left( {{{\text{y}}_3} - {{\text{y}}_1}} \right) + {{\text{x}}_3}\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)} \right]\quad \ldots \left( 2 \right)\end{align}

By substituting the vertices $$A$$, $$D$$, $$E$$ in (2),

Area of $$\Delta ADE$$ \begin{align} &= \frac{1}{2}\end{align}\begin{align}\left[ {4\left( {\frac{{23}}{4} - \frac{{20}}{4}} \right) + \frac{{13}}{4}\left( {\frac{{20}}{4} - 6} \right) + \frac{{19}}{4}\left( {6 - \frac{{23}}{4}} \right)} \right]\end{align}

\begin{align} &= \frac{1}{2}\left[ {3 - \frac{{13}}{4} + \frac{{19}}{{16}}} \right] \end{align}

\begin{align} &= \frac{1}{2}\left[ {\frac{{48 - 52 + 19}}{{16}}} \right] \end{align}

\begin{align} &= \frac{{15}}{{32}}\;\rm {} Square\; units\end{align}

By substituting the vertices $$A$$, $$B$$, $$C$$ in (2)

Area of $$\Delta ADE$$ \begin{align} &= \frac{1}{2}\end{align}\begin{align}\left[ {4\left( {\frac{{23}}{4} - \frac{{20}}{4}} \right) + \frac{{13}}{4}\left( {\frac{{20}}{4} - 6} \right) + \frac{{19}}{4}\left( {6 - \frac{{23}}{4}} \right)} \right] \end{align}

\begin{align} &= \frac{1}{2}\left[ {3 - \frac{{13}}{4} + \frac{{19}}{{16}}} \right] \end{align}

\begin{align} &= \frac{1}{2}\left[ {\frac{{48 - 52 + 19}}{{16}}} \right] \end{align}

\begin{align} &= \frac{{15}}{{32}}{\text{ Square units}}\end{align}

Clearly, the ratio between the areas of $$\Delta ADE$$ and $$\Delta ABC$$ is $$1:16$$.

Alternatively, we know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle[Converse of Basic Proportionality Theorem]. These two triangles so formed (here $$\Delta ADE$$ and $$\Delta ABC$$) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.[Theorem 6.6]

Therefore, ratio between the areas of $$\Delta ADE$$ and

\begin{align}\Delta {\text{ABC}}\,{\text{ = }}\,{\left( {\frac{1}{2}} \right)^2} = \frac{1}{{16}}\end{align}

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